We are sorry to see you leave - Beta is different and we value the time you took to try it out. Before you decide to go, please take a look at some value-adds for Beta and learn more about it. Thank you for reading Slashdot, and for making the site better!

New submitter arsheive (609065) writes with a link to this interesting RPS brainteaser: "How do you play against an opponent who _must_ throw Rock 50% of the time, and how much would you be willing to pay to play against them?"

The odds are even odds. If he must throw rock 50% of the time, and he knows that you know that data, the other 50% of the time he will throw scissors to beat your paper....

Re:Simple.... Odds are even (0)

Anonymous Coward | about 10 months ago | (#46670539)

The odds are even odds. If he must throw rock 50% of the time, and he knows that you know that data, the other 50% of the time he will throw scissors to beat your paper....

Which means that he won't throw anything that gives points against rock so you should play rock all the time. Until he figures that out, then he will have to play paper every now and then.

Actually now that I think about it more, it is much more difficult that it at first seams... The fact that he plays rock 50% of the time really has no bearing on what he plays the other 50% of the time. If you do the 100% paper, you will eventually only win 50% of the time.

After that you need to run a probability set on all the possible combinations, with the unknowns for his paper and scissors, only knowing that they total to 50%.

Anonymous Coward | about 10 months ago | (#46671409)

As interesting as you making a total fool of yourself vs. apk http://tech.slashdot.org/comme... [slashdot.org] ? He utterly destroyed you, "3 digit registered 'luser'" that you clearly are (technically inept noob who libels when shot down is more like it).

Re:Simple.... Odds are even (1)

Anonymous Coward | about 10 months ago | (#46671165)

"more difficult that it at first seams"

Is it harder than spelling seems?

Re:Simple.... Odds are even (0)

Anonymous Coward | about 10 months ago | (#46671559)

"Is it harder than spelling seems?"

Is it harder than putting referenced words in quotation marks?

Re:Simple.... Odds are even (2, Funny)

Anonymous Coward | about 10 months ago | (#46672095)

actually, as each choice beats one and either ties or is beaten by the other two, the odds of winning any random round of RPS is 33%.

But the whole point is that it is not random, since your opponent's choices are constrained. He is forced to play rock 50% of the time. If he plays randomly the other 50%, then you can always play paper and win 2/3rds of the time, lose 1/6 of the time, and tie 1/6th. But then he can adapt, and play rock 50% and scissors the other 50%, resulting in a tie (1/3 win, 1/3 loss, 1/3 tie). But you can adapt to that by playing rock 50% and paper 50%. You will win 50%, lose 25%, and tie 25%.

The Nash Equilibrium [wikipedia.org] is for you to play paper 2/3rds of the time, and rock 1/3rd. His best counter strategy is to play rock 50% (he cannot go lower) and scissors 50%. He cannot do better. If you deviate from 2/3 paper and 1/3 rock, he can adjust his strategy to do better. With the optimal strategy, you will win 1/2, lose 1/3, and tie 1/6.

Anonymous Coward | about 10 months ago | (#46671837)

Actually, 50% Rock, 50% Scissors is not the opponent's equilibrium strategy, nor is it the "best response" to 2/3 Paper 1/3 Rock. All responses involving exactly 50% Rock are equally good against 2/3 Paper 1/3 Rock, but if your opponent plays 50% Rock 50% Scissors you can exploit that with 100% Rock, which does better against that strategy than 2/3 Paper 1/3 Rock. So your opponent would be better off with a strategy that cannot be exploited.

The only strategy for the opponent that cannot be exploited is 1/2 Rock 1/6 Paper 1/3 Scissors. All your strategies are equally good against this unless you stupidly play Scissors with nonzero probability.

His nash strategy (minimax strategy) is to play rock 1/2, scissors 1/3 and paper 1/6 (it is true that rock 1/2 and scissors 1/2 is a best response to the nash strategy, but that does not mean it is optimal against abitrary strategies). I have a long earlier post showing the strategies and so on:)

I'm getting better test results from having around 38.4% rock, rest paper. At this point his scissor and paper strategies become equal and I'm winning (assuming ties are repeated) 61.7% of cases instead of 60% of cases with 1/3 rock rest paper.

I think that trick is finding a place where any mixture of scissors or paper for him is equally bad. At this moment, his choices doesn't matter anymore (of course, unless he choses rock even at free choice). At 1/3 rock 2/3 paper, scissors are clearly better choice for him, so it is not a perfect place.

You seriously think that each player in RPS has a 33% chance of winning each round? Think a little bit about that. Oh, I forgot, this is/.

What do you think the odds are for each player? Keep in mind that there are three possible outcomes for each round: win, lose, or draw.

Re:Simple.... Odds are even (0)

Anonymous Coward | about 10 months ago | (#46671297)

If I choose 2, there are three outcomes:
- you choose 1 - I lose
- you choose 2 - tie
- you choose 3 - I win.

This applies for any mapping between {1,2,3} and {rock,paper,scissors}.

That looks awfully like 33% to me.

Re:Simple.... Odds are even (0)

Anonymous Coward | about 10 months ago | (#46672355)

We each pick a number between 1 and 10^10^10, and then throw a die whomever guessed it wins... I see three outcome I win, you win, we both win, or we both lose. so there is a 25% chance that we both guess the number every single time right?

Always do rock. (0)

Anonymous Coward | about 10 months ago | (#46670449)

You tie 50% of the time, win 25% of the time, and lose 25% of the time. Pay? Nothing.

You tie 50% of the time, win 25% of the time, and lose 25% of the time. Pay? Nothing.

Come again? It is specified that the player forced to play 50% rock (as mandated by a fair coin flip you don't get to see) plays intelligently and will adapt to your play. When he figures out you're always playing rock, he'll always play paper when he doesn't have to play rock. You tie 50% of the time and lose 50% of the time. That's lousy strategy.

You're guaranteed to break even by always playing paper. When the opponent adapts, he'll always play scissors when he doesn't have to play rock, and you win 50% of the time and lose 50% of the time. The question is, can you do better than that?

Of course. You use the rock to smash him in the head while he tries to stab you with the scissors. Your friend then uses the paper to write a letter to your parents about how you died in a stupid fight about statistics.

100% paper strategy will win 50% of the time. Of the remaining 50% of games played, (assuming even distribution of the remaining picks) 25% will be losses and 25% will be tied. Thus, you'd be assured a win-loss-tie ratio of 2-1-1, which is quite good.
If their remaining options are not distributed evenly, this changes things. You'd want to look at their play to see whether there are any discernable patterns, such that you know that Rock will be played for certain every other move, for example. Then you just sync Paper moves to their Rock moves, and play Scissors or Rock randomly for the other half.

The opponent is forced to pick 50% rock, but he has no limitations other than that. If you went 100% paper, you'd not beat me playing 50% rock by more than a tiny sliver of a percent it takes me to realize you are going 50% paper.

And the problem has nothing to do with the someone not just picking rock, but doing so in a very predictable manner. Otherwise, we'd not be talking about someone picking rock 50% of the time, but playing against someone that plays randomly, but tells you what he is picking half the time, never lying. That's a very, very boring brainteaser.

The first naive approach that doesn't assume idiocy on the other side's part is to assume he'll try to guess our play. If we play in a non deterministic fashion, his best options is probably random, but instead of.33/.33/.33, we get.50/.25/.25. Against an opponent doing that, you could go.25/.50/.25, and win quite a bit.

If he is forced to get at least 50% rock after a certain number of plays, then we can change our probabilities depending on how far from the standard distribution our opponent is: For instance, if we were playing to 10 throws, and he never picked rock in the first 5, we'd get him in the last 5, because his moves are forced. To avoid this, someone that has to reach that 50% will probably also change their probabilities after every throw, just to avoid such a 'free' run. One might even consider starting with an over 50% rock probability, because under that set of rules, having overplayed rock lets us play more optimally later, while underplaying rock leads to major losses.

Either way, a more detailed problem specification is required.

From TFA: "At the start of each round an independent judge flips a fair coin and tells your opponent the result but does not tell you. If the coin came up heads your opponent must play rock." The opponent isn't forced to get at least 50% rock after any number of plays.

100% paper strategy will win 50% of the time. Of the remaining 50% of games played, (assuming even distribution of the remaining picks) 25% will be losses and 25% will be tied

No, you are describing another game where the opponent is forced to play rock 50% of the time, paper 25% of the time and scissors 25% of the time.

This game is different.

(I'm repeating my self from another post but many people are making the same mistake)

With no such restriction, random choices on both sides lead to 33% win, 33% draw, 33% loss, right? With the opponent throwing Rock 50% of the time, assuming the other 50% is evenly divided between Paper and Scissors, if I always throw paper I'll win 50% of the time, lose 25% of the time, and draw 25% of the time.

So depending how the betting works, I'd be pretty willing.

With no such restriction, random choices on both sides lead to 33% win, 33% draw, 33% loss, right? With the opponent throwing Rock 50% of the time, assuming the other 50% is evenly divided between Paper and Scissors, if I always throw paper I'll win 50% of the time, lose 25% of the time, and draw 25% of the time.

So depending how the betting works, I'd be pretty willing.

No, you are describing another game where the opponent is forced to play rock 50% of the time, paper 25% of the time and scissors 25% of the time.

This game is different.

Re:For money, you mean? (0)

Anonymous Coward | about 10 months ago | (#46670837)

Your opponent will play an optimal strategy.

You always throw paper. You will win only 50% of the time. The 50% when they are forced to throw rock. Your opponent will throw scissors 50% of the time and beat your paper.

You failed to understand the proposition.

Play paper 50% of the time. (1, Interesting)

Anonymous Coward | about 10 months ago | (#46670499)

Expect them to play scissors a lot to beat your paper. Play rock as often as they play scissors.

Scissors cut paper Paper covers rock Rock crushes lizard Lizard poisons Spock Spock smashes scissors Scissors decapitate lizard Lizard eats paper Paper disproves Spock Spock vaporizes rock Rock crushes scissors

How would you measure 50%? When does the game end?
If my opponent strictly goes by 50 25 25 then Yes, I'd be happy yo play for an infinite amount of money, however given the play I am going to be playing, my opponent may opt/adapt for 50 0 50 instead and we are stalemate again.. So the game is rigged, conditions flawed, how would you measure those 50%?

Re:Impossible (0)

Anonymous Coward | about 10 months ago | (#46670769)

Exactly right. Unless we know the distribution, it's an unanswerable question in general. You could play a million games and never see rock for the first half of those, and yet over the entire game domain, your opponent could still throw rock 50% of the time.

it would do good to RTFA as it says the condition is determined by a judge with a fair coin at the start of each round. So yes there is a chance he will never use rock in N first rounds because s/he is not forced to AND his strategy tells him to use paper or scissors. In other words it is not only his/her strategy that decides.

Assuming we randomly use 50% paper and 50% rock, we get: - on rounds he is forced to play rock, we get half victory and half tie, so 25% win, 25% tie so far - on other rounds he can:
- always use scissors, which will turn into our 25% win, 25% lose - we win overall, 50% win, 25% tie, 25% lose
- always use rock, which will turn into our 25% tie, 25%win - we win overall, 50% win, 50% tie
- always use paper, which will turn into 25% win, 25% lose - we win overall, 50% win, 25% tie, 25% lose
- any kind of random mix of above, which can only move between results above, as our picks are completely random as well

So, he best strategy is to always play scissors or rock, for 50:25:25 result. If ties are repeated, we win 2 out of 3 times.

Any better strategy?

Re:50% paper and 50% rock ? (0)

Anonymous Coward | about 10 months ago | (#46670935)

he best strategy is to always play scissors or rock, for 50:25:25 result. If ties are repeated, we win 2 out of 3 times.

you mean paper or rock. I wonder if any strategy that never plays scissors is a winner. It's hard to answer that because we don't know the opponent's strategy/distribution on the rounds where they're not forced to play rock. 50/50 rock/paper is a winning strategy, as you've shown. But is it optimal? Idk

Re:50% paper and 50% rock ? (0)

Anonymous Coward | about 10 months ago | (#46670951)

Err, what?

always use paper, which will turn into 25% win, 25% lose - we win overall, 50% win, 25% tie, 25% lose

using your terms, random rock/paper against opponent playing paper is 25% tie, 25% lose so we get 25:50:25 result. Net zero.

I think you have an error in there. His best strategy in your example would be to always use paper, in which case he's doing the exact same thing as you are, and so the odds are even.

I think your best strategy would have to involve reacting to his. Start with 100% paper until you see what he's throwing when he's not throwing rock. This gives you at least even odds regardless of what he does.

He'll probably start throwing scissors, at which point you switch to you switch to 50% rock 50% paper. This will have you win 2/4, lose 1/4, draw 1/4.

He should respond by starting to throw paper. so you just go back to 100% paper and win half the time. If he switches back to scissors, you start throwing rock again.

The goal is to use long stretches of paper to for him into 50R:50S (which only breaks even at best), and then switch to 50R:50P to pick up some wins until he switches back 50R:50P (again, he'll only be able to break even). So when he's properly countering you, he can only break even, and every time you switch strategies you should pick up some extra wins. If he tries to go for a midpoint (50:25:25) he'll lose to either of your two strategies.

Yes, you are right - paper will be 25% tie, 25% lose, ending up with 50% tie, 25% win, 25% lose, so still purely random.

I don't think that strategy we are looking for involves "reactions" - this can be always defeated by opponent which overguess you by one step. I would hope to find a strategy which would lead you to win > 50% even if opponent knows it (strategy itself, not the result of random choice at given stage).

Ok, my current guess would be around 1/3 rock and 2/3 paper. Opponent cannot replicate this strategy, because of requirement of having at least 50% rock. If he goes 50% scissors, he will have 1/6 tie, 2/6 lose, 1/6 lose and 2/6 win, so 3:2 in my favor. With 50% paper, he gets 1/6 tie, 2/6 lose, 1/6 win, 2/6 tie, so 2:1 in my favor. 100% rock is obvious lose. I think that mixing scissors and paper will be clearly worse than pure scissors on his part (because with scissors, he has at least 2/3 chances of winning after he got a choice).

From random testing, it seems that optimal ratio of rock is around 38% (not 1/3). Around that point, pure paper and pure scissor strategies for opponent seems to get equal and I'm winning around 62% of cases.

There is probably some interesting math behind that...

I've been having students in my introductory programming courses work on this class of problem for a few years.. They all seem to really enjoy it. I code up bots to play RPS with certain biases just like the OP and they have to program a single player that identifies the bias in an opponent and adjusts its play to give it an advantage. They all routinely can generate solutions that perform far better than random against predictable, dumb bots, but things get very interesting when I throw the students' bots against each other in a throwdown tournament.:)

recalculate every round (1)

Anonymous Coward | about 10 months ago | (#46670827)

If i know the number of rounds in advance (atleast your oponent has to right if he needs to calculate the 50%). I would recalculate the odds after every round taking into account the number of rock move remaining for him...

Like drawing to an inside flush, an "optimized" strategy is not necessarily what the opponent plays. There is no reason inherent in the description to make assumptions about the opponent's other play. They may also be constrained to play paper the other fifty percent of the time, and to play paper , then rock, then paper, then rock. In the real world, don't assume that the minimal description of the problem gives all the important data.

I play a lot of poker, but I've never heard of an "inside flush" draw. I know the inside straight draw, and I suppose you could have an inside straight flush draw, but a flush is just a draw for another like suited card. All flush draws have nine possible outs unless you have seen your opponents cards and know there are fewer. An inside straight has only four outs, which is why it is a bad draw most of the time. An open ended straight has eight outs, which makes it almost as good as the flush draw. Both the flush draw and the open ended straight can be worth playing, depending on the situation with pot odds, and your expectations about your opponents.

there is a factual problem with the summary.TFA says it all so better read it. If not read this (I hope not to mess up it too much).

It is not required of the opponent to play rock 50% of the time. The referee is using a fair coin to determine if your opponent is to play rock or not. If s/he is not forced to play rock s/he is free to chose allowing him also to chose rock 100% of time too if s/he so wishes.

It is not required of the opponent to play rock 50% of the time. The referee is using a fair coin to determine if your opponent is to play rock or not. If s/he is not forced to play rock s/he is free to chose allowing him also to chose rock 100% of time too if s/he so wishes.

It totally is required to play rock 50% of the time. It's not required to play rock exactly 50% of the time.

It is a bad summary, but only because the wording is ambiguous, not that it's factually incorrect. The statement you're objecting to is perfectly correct in one interpretation, and dead wrong in another. Your own counter-statement, "it is not required of the opponent to play rock 50% of the time," is equally ambiguous. In fact, 50% of the time (assuming a fair coin), the opponent is required to play rock, so it's true that "it is required of the opponent to play rock, 50% of the time". Leaving out the comma yields a true sentence (assuming the correct interpretation is chosen of the now even more ambiguous sentence) that contradicts the quoted sentence of yours, assuming you parse your sentence as "it is not required: that the opponent play rock 50% of the time", but does not contradict it at all if your sentence is parsed "it is not required that the opponent play rock, 50% of the time", since 50% of the time, the opponent can choose freely, and thus is 50% of the time, is not required to make any particular choice. So, both the summary and your explanation of what's wrong with it contain statements that not factually incorrect, just ambiguously worded such that a reader might interpret it to mean something that is incorrect rather than something that is correct.

This is a textbook example of why programming computers in plain English would be a monstrously bad idea.

"Never go in against a Sicilian when death is on the line!"

bonus problem: paper,paper (0)

Anonymous Coward | about 10 months ago | (#46671023)

If the opponent plays P or S on the first move, his second move is determined, so there's a greater than random chance he will want to play R first. So the safest thing is to play paper twice. Or: paper first, and if the opponent played rock, then play anything at random.

Lets call the guy with the restriction player 1 and the other player 2. If you think about it player 1 got 3 "pure" strategies (as in: each other strategy he can play can be seen as a mixture of these 3): (1) rock 100%, (2) paper 50%/rock 50% and (3) scissors 50%/rock 50%. Against (1) rock gives 0, paper -1 and scissors 1. Against (2) rock gives 1/2, paper -1/2 and scissors 0. Against (3) rock gives -1/2, paper 0 and scissors 1/2. In each case, the number is the probability of player 1 winning minus player 2 winning.

We see that player 2 should not play scissors, because he will never gain anything from it and clearly the optimal strategy should be able to gain something (we will see that it is 1/6). Then, knowing that, paper 50%/rock 50% is better than rock 100% for player 1: If player 2 plays rock, player 1 gets 1/2 (instead of 0) and if player 2 plays scissors player 1 gains -1/2 (instead of -1).

Hence, we are down to: (2) paper 50%/rock 50% and (3) scissors 50%/ rock 50% vs. rock and paper. If player 1 plays (2) with pr. 1/3 and (3) with probability 2/3, he loses 1/6 against both rock and paper. If player 2 plays rock with probability 1/3 and paper with probability 2/3 he gets 1/6 against (1) and 1/6 against (2). This is optimal since each player have a way to guarantee that player 1 loses 1/6 and player 2 wins 1/6. If either had a better strategy it would break the other players guarantee (note that the given strategy for player 1 wins 1/6 against scissors, again showing that it is a bad strategy for player 2 and player 2's strategy wins 1/3 against rock 100% showing that it is a bad strategy).

Sorry:/ There are some mistakes in the last part. The strategy for player 1 wins 1/2-1/6=1/3 and not 1/6 as claimed. Also, the strategy for player 2 wins 2/3 against pure rock and not 1/3 as claimed. Still, it just makes it even clearer that you should not play rock with probability more than 50% as player 1 and not play scissors at all.

Also, to be more precise, the strategy for player 1 is to play rock with probability 1/2, paper with probability 1/6 and scissors with probability 1/3.

Randomize 2/3 paper, 1/6 rock, 1/6 scissors (1, Insightful)

Anonymous Coward | about 10 months ago | (#46671093)

Over the long term, the strategy must converge to stable, therefore true random can be the only optimized strategy.

50% of time opponent must play R. The remaining 50% of the time they can equally choose R,P,S.

R -> P = 1/2 + 1/3 * 1/2 S -> R = 1/3 * 1/2 P -> S = 1/3 * 1/2

For a guaranteed win, roll a die: 1-4 => P, 5 => R, 6 => S.

By how much? Consider you are random as above and opponent is fixed wlog at 100% R. You win 2/3 of the time and lose 1/6 of the time.

The expected payoff to play is 2/3, the expected cost is 1/6. Payoff - cost = 1/2, so the most you should be willing to pay to play a $100 game is $150.

You should play paper 4/6 of the time, rock 1/6, and scissors 1/6 of the time.

The key (if you RFTA) is that whether or not your opponent plays rock is determined by a coin toss. So really you are playing a compound game. You are playing a coin toss and rock paper scissors (RPS). Since the coin toss determines your opponents move, you can think of it as playing 50% coin toss and 50% RPS. The RPS is a subgame of the coin toss.

Since the coin toss is the dominate game, you play with win that first. But instead of heads/tails, it is paper/other. The answer to the coin toss is a 50/50 guess of heads/tails, so the answer to the paper/other is 50% paper, 50% other.

The "other" is the RPS game. And since the answer to the RPS game is 1/3 rock, 1/3 paper, 1/3 scissors, we know what the solution to the other 50% of the game is.

So the equations are:choice = (Coin Toss) + (RPS) so: paper = 1/2 + 1/3, rock = 0 + 1/3, scissors = 0 + 1/3. Or paper = 4/6, rock = 1/6, scissors = 1/6.

The opponent could respond to this by playing scissors on all non-forced-rock turns. If the opponent plays rock, you win 4/6 of the time and lose 1/6 of the time, but if the opponent plays scissors you lose 4/6 of the time and win 1/6 of the time, so overall you'd be even.

The opponent doesn't have the option to play anything greater than 1/2 scissors because the other 1/2 must be rock. If he uses the "all scissors" response, he can only actually do a 1/2 scissors response. So is we play it out:

1/2 scissors x 4/6 paper = 2/6 = 1/3 victory for the opponent. 1/2 scissors x 1/6 scissors is 1/12 tie. And 1/2 scissors x 1/6 rock is 1/12 lose. So the "all scissors" strategy only nets him 1/3 victory not 4/6.

I listed the chances in the context of the opponent move ("if the opponent plays rock"). The chance of playing rock or playing scissors is 1/2 each (the coin toss), so if you list it as overall chances you get 1/3 win and 1/12 loss (same as you wrote) due to the opponent playing scissors and also 1/3 loss and 1/12 win due to the opponent playing rock; the expected result result is still 0.

There is a flaw in your reasoning. You do not know that your oppoent flipped so you can not condition on it like you do here (you can not play paper all the time if he "flips" rock because you do not know his coin flip). If you think about it you should NEVER play scissors. In the best case for you he plays rock 50% and paper 50% and you get 0 in expectation and clearly you got an advantage so 0 is not good.

The optimal strategy is to play 1/3 rock, 2/3 paper. It gives at least 1/6 against anything he could play. He can similarly ensure that you can not get more than 1/6 a game by playing rock with probability 1/2, paper with probability 1/6 and scissors with probability 1/3. Your strategy would get less than 1/6 against that (more precisely, you get 1/6 if you play either rock or paper and you lose 1/3 if you play scissors. Therefore you get 1/6*5/6-1/3*1/6=1/12 which is less than the 1/6 you get for playing 1/3 rock and 2/3 paper).

We know that the opponent must play rock 1/2 of the time.

If I play paper 4/6 of the time, than I should expect 1/2 of my paper to align with his rock. So 4/6 * 1/2 = 2/6 = 1/3. So I should expect to win 1/3 of the time, plus my winnings on the other combinations. That means 1/3 is the lower bound.

If you play 1/3 rock and 2/3 paper, his response will be 1/2 paper and 1/2 rock. So you are going to get 2/3 * 1/2 = 1/3 for your paper. But your 1/3 rock will never win because he will never play scissors either. But his 1/2 paper will meet your 1/3 rock, giving him 1/2 * 1/3 = 1/6 win. Putting you head by only 1/6.

This is where the two games key comes in. You and I both recognize that 2/3 paper is the right move because 1/2 of his moves will be rock. But by playing the other half as regular RPS with a win/tie/loss of 1/1/1 you can expect the win/loss to cancel out, leaving you with your 1/3 lower bound advantage.

Ok lets see: you play 2/3 paper (I shorten the fraction I hope that is ok:) ) and 1/6 scissors and 1/6 rock. You play against the strategy 1/2 rock, 1/3 scissors, 1/6 paper. Fast version: lets look at a random round in which you play rock: In those you win 1/3 against his scissors and lose 1/6 against paper, thus you get 1/6 on avg. Next, random round in which you play paper: In those you get 1/2 against his rock, and lose 1/3 against his scissors, i.e. again you gain 1/6. Next, random round in which you play scissors: In those you get 1/6 against his paper and lose 1/2 against his rock, i.e. you LOSE 1/3. On avg you play rock 2/3 of the time and get 1/6 in those rounds, scissors in 1/6 of the time and LOSE 1/3 and paper 1/6 of the time and get 1/6. Thus, on avg. 2/3*1/6+1/6*(-1/3)+1/6*1/6=1/12. This is below your lower bound so there is something wrong with it. (the reason is that you lose whenever a bit on avg. whenever you do not play paper).

My strategies, played against each other gives 1/6. Thus, you can not say that yours is better always. I can argue that against ANY strategy mine gets 1/6. You can not get better than 1/12 (because you get that against mine strategy for the other player). Thus, yours can not be optimal sorry:(

Note, I say I can argue that my strategy gives 1/6 on avg. in the sense that I did argue that in a looong post above:/

2/3 Paper 1/3 Rock (0)

Anonymous Coward | about 10 months ago | (#46671125)

The payoff is 16 2/3. The defender should do 1/3 paper 2/3 scissors. (When not forced to play rock.) This is just a standard mini max problem. The payoff matrix looks like: 0 -50 50 100 50 0 -100 0 -50 (With row and columns being in RPS order.)

Re:The bonus question2/3 Paper 1/3 Rock (0)

Anonymous Coward | about 10 months ago | (#46671335)

For the bonus question you can set up a similar payoff matrix. For the case where the restricted player picks rock on the first round the value of the second round will be zero. When rock is not picked in the first round by the resticted player the value of the second game will be 50 to the unrestricted player. So the payoff matrix for the first round that includes the value of the second round looks like: 0 0 100 50 50 0 -50 100 50 (Again rows and columns are in RPS order.) The payoff to the unrestricted player is 33 1/3 for both rounds combined. The unrestricted player should use 2/3 paper 1/3 rock in the first around. In the second round either just paper or 1/3 RPS depending on whether or not the restricted player must play rock. The resticted player should player should play 2/3 rock 1/3 scissors in the first round. There is a program to calculate these results at: http://wolff.to/bruno/strategy... [wolff.to]

Huh? (0)

Anonymous Coward | about 10 months ago | (#46671155)

Nobody seems to be asking what the definition of "50% of the time" is ?

Over what interval/time frame?

If the opponent throws scissors or paper on his first move is he then required to throw rock on his second move?

First, make sure you read TFA, since it explains what the summary doesn't: how the 50% is determined and how the opponent can play in the non-forced turns.

If you play using a deterministic algorithm, for example always play paper, the opponent can figure it out and beat you on all the non-forced turns. At best you'll get an even result.

If you play using a random algorithm, the opponent can figure out the frequencies you're using and compensate for that. For example, if you decide to play paper 50% of the time and rock and scissors 25% of the time, you'd win against an opponent playing rock 50% of the time and paper and scissors 25% of the time. However, if the opponent decides to play rock 50% of the time and scissors the other 50%, the result is even again. If the opponent would be forced to play rock more than 50% of the time, there is no room to compensate and you would win consistently with 100% paper. I think that with 50% rock, there is enough room to respond to any frequency distribution you can come up with, although I have no proof for that.

You could change your algorithms during play, but if there isn't any algorithm that results in an advantage when playing it consistently, gaining an advantage from changing your algorithm would depend on how well your opponent responds to your changes. In other words, you're playing mind games. I don't think the 50% rock restriction is going to be of any help here.

You can get an advantage. The important point is to notice that you should not play scissors ever. You can only get 0 in expectation IF he plays paper 50% and rock 50% and he gets an advantage otherwise and 0 is not good for you:/ See my above post for further details (spoiler: The optimal choice for you is 1/3 rock, 2/3 paper).

Easy win, isn't it? (0)

Anonymous Coward | about 10 months ago | (#46672001)

You play all paper, all the way. You beat him in all the forced turns, giving you an even score. However it will cost him a few lost points on the non-forced turns to figure out your all paper strategy and switch to an all scissor strategy. At that point, haven't you already won? Opponent will only win 50% of the remainder of the games but is already behind, so will never win. For extra points, you can switch to rock at random intervals for 1 turn only, after opponent picks an all-scissor strategy on the non-forced turns. Those will give you a 50/50 chance of a draw (on the forced turns) or a win against scissors (on the non forced turns).

51 to 49 is a win as good as 99 to 1 in RPS.

Re:Easy win, isn't it? (0)

Anonymous Coward | about 10 months ago | (#46672411)

Not true: You win and lose $100 for each game of RPS in this model. So the goal is to maximize your winning.

First, there is some uncertainty about what ratio of your opponent's throws will be paper. Let's call that ratio x. This means that we can define the likelyhood of his throws this way:

HeDoesRock =.5, HeDoesPaper = x, HeDoesScissors =.5 - x

Now you need to figure out the optimal winning ratio of your throwing either rock or paper.

Now take the first derivative of winning margin and set it to zero to find the maximum:
d/dx.5FreqRock - 2FreqRock(x) +FreqPaper(x) = -2FreqRock +FreqPaper = 0

So 2(FreqRock) = FreqPaper, so you should throw paper twice as often as you throw rock. And surprisingly, you can get this result without ever needing to know x, the optimal ratio of your opponent throwing paper. Is that right?

This is right yes:) Well, assuming that you know that he is playing 50% rock (and not more - that is btw. right - he would worse off if he played rock with even higher probability). Also, to be truly formal about it you should argue that it is a maximum and not a minimum you found:)

Anonymous Coward | about 10 months ago | (#46671191)

If I play rock all the time, I tie 50% of the time, lose 25% and win 25%. Perfect tie overall.

If I play scissors all the time, I lose 50% of the time, win 25% and tie 25%. Lose two out of three.

If I play paper all the time, I win 50% of the time, tie 25% and lose 25%. Win twice as much as lose.

Go for paper.

Profit if the opponent is human and visible (0)

Anonymous Coward | about 10 months ago | (#46671299)

If the opponent is human and both players can see each other, the opponent won't be able to completely hide his foreknowledge of when he must throw rock. Learn his body language and tells, then throw paper whenever he inevitably gives away his move. If you're good enough to successfully predict, better than chance, when he must throw rock, then over a large enough number of games you can throw randomly the rest of the time and still make a profit.

If your opponent must throw rock 50% of the time, then you throw paper 100% of the time.

You will win AT LEAST 51% of the time, because you get the 50% gifted to you, and the other non-0% of the times that your opponent throws paper will cause a rematch.

If there were no betting caps, I would progressive bet on paper. Bet 1 dollar paper. If lose, bet 2 dollar. If lose, bet 4 dollar. If lose, bet 8 dollar, et al. Once you win you reset back to zero.
This naturally has drawbacks and potential for huge failure, but I've had good luck with it in blackjack and roulette if betting caps are high and I have about $10000 to play around with.

There are a fair number good ideas here that are getting close to the correct solution here, but there is also a lot of stuff that is pretty far off so I'll be posting a video solution to my youtube channel for this Monday.
If your interested in the solution you can subscribe to my youtube channel and you'll see it when I post it on Monday:)
https://www.youtube.com/channe... [youtube.com]

Re:I'm the author -- video solution coming soon (1)

I'll give a few other hints.
Your optimal strategy never plays scissors.
Your optimal strategy plays rock 50% of the time.
Your optimal strategy does profit on average.

Re:I'm the author -- video solution coming soon (1)

I'll give a few other hints.
Your optimal strategy never plays scissors.
Your optimal strategy plays rock 50% of the time.
Your optimal strategy does profit on average.

ugh, I meant your optimal strategy plays rock 50% of the time. sorry.

Re:I'm the author -- video solution coming soon (1)

"you" are not the guy playing rock 50% of the time. "you" are the guy beating on the poor guy playing rock 50% of the time. The optimal choice is to play rock 1/3 and paper 2/3 (his is to play rock 1/2 and paper 1/6 and scissors 1/3).

Depends on the stakes (0)

Anonymous Coward | about 10 months ago | (#46671581)

Personally, I always throw 'gun' if there is sufficient cash involved.

Since I already explained the optimal solution to the basic question mentioned in the summery lets solve the bonus question too (my solution also matches the solution given in comments on the article side so it should be good (and said to be correct by the author) - note currently no answer with a high score is correct - mine has 1).

The bonus question is that you play two rounds, and your oppoent must play atleast rock once. So, if he plays something not rock in the first round he must play rock in the second and loss (you just play paper). If he plays rock in the first he can play 1/3 all in the second (which leads to a draw like normal). Thus, if he plays rock first it is like normal RPS (because he get 0 in the next). Otherwise you get one free win (for the second round).

Thus, we can model the first game of the bonus question as (where the numbers is the number of rounds he wins on avg given the choice in round 1):
R P S R 0 -1 1 P 0 -1 -2 S -2 0 -1

Where you pick columns and him rows. We see that rock dominates paper for the row player. We get

R P S R 0 -1 1 S -2 0 -1

For the column player, the choice of rock now dominates scissors. We get

R P R 0 -1 S -2 0

Playing rock 1/3 and paper 2/3 for the collumn player gives -2/3 wins on avg. Similarly, the row player can get -2/3 wins on avg by playing rock 2/3 and scissors 1/3.

To figure out what you should do, first assume your opponent is rational, and will make good choices whenever he is able. Since he knows that you will play a paper-heavy strategy to counter his rock-heavy strategy, it would not be rational to voluntarily choose more rocks. That could only make things worse.

But if he tried to exploit your paper-heavy strategy by throwing scissors on turns when he gets a choice, you'd have a perfect strategy against this: All rock. On forced rock, you get a redo, and on non-forced rock, he does scissors and you win. So on first pass, I think that your opponent should favor paper when he can choose. It's not like you'll ever do scissors. That's auto-lose half the time - basically a complete surrender of your advantage. So paper is a safe move for your opponent.

The problem is that if he did 50% rock and 50% paper, then all-paper will be your perfect counter. He won't let you do that, so he'll have to throw in some scissors. Just how many? It looks now like you will both simultaneously have to determine optimal strategies in order to answer that question, and this will require derivatives of two sets of optimal-choice equations - so that you can solve for the two maxima. Sounds like a fun problem!

2/3 paper 1/3 rock (0)

Anonymous Coward | about 10 months ago | (#46671857)

The best strategy is to randomly choose paper 2/3 of the time and randomly choose rock 1/3 of the time. You will be expected to win at least 2/3*1/2 = 1/3 of the time and tie 1/3*1/2 = 1/6 of the time when the opponent randomly chooses rock. When free to choose, the best option for the opponent is then to always choose scissors which results in you winning 1/3*1/2 = 1/6 of the time and losing 2/3*1/2 = 1/3 of the time. So, overall you will be expected to win 1/2 the time, tie 1/6 of the time, and lose 1/3 of the time.

This is a typical game theory type problem. One simply needs to optimize a max-min problem. If the probability of the player choosing rock, paper, scissors is x,y,z, respectively, and the probability of the opponent choosing is a,b,c, then one optimizes (Wins - Losses):

max_{x,y,z}min_{a,b,c} (b-c)*x + (c-a)*y + (a-b)*z subject to x + y + z = 1, x,y,z >= 0, a + b + c = 1, a >= 1/2, b,c >= 0.

There are no solutions in the interior since those points boil down to the original rock paper scissors game. So, one only needs to check the boundary, when b = 1/2 and, thus, c = 1/2 - a. If you fix x,y,z then the extreme points occur at a = 0 and a = 1/2 giving a Win-Loss of x/2 - y/2 when y >= (x+z)/2 and y/2 - z/2 otherwise . Again the interesting behavior only happens at the boundary and, so, setting y = (x+z)/2 gives a Win-Loss of x/4 - z/4 which is maximized when z = 0, and, therefore 1 = x + y = x + x/2 =(3/2)*x, yielding x = 2/3, y = 1/3, and a Win-Loss of 1/6.

Re:2/3 paper 1/3 rock (0)

Anonymous Coward | about 10 months ago | (#46672021)

(Corrected for dyslexia)

The best strategy is to randomly choose paper 2/3 of the time and randomly choose rock 1/3 of the time. You will be expected to win at least 2/3*1/2 = 1/3 of the time and tie 1/3*1/2 = 1/6 of the time when the opponent randomly chooses rock. When free to choose, the best option for the opponent is then to always choose scissors which results in you winning 1/3*1/2 = 1/6 of the time and losing 2/3*1/2 = 1/3 of the time. So, overall you will be expected to win 1/2 the time, tie 1/6 of the time, and lose 1/3 of the time.

This is a typical game theory type problem. One simply needs to optimize a max-min problem. If the probability of the player choosing rock, paper, scissors is x,y,z, respectively, and the probability of the opponent choosing is a,b,c, then one optimizes (Wins - Losses):

max_{x,y,z}min_{a,b,c} (c-b)*x + (a-c)*y + (b-a)*z subject to x + y + z = 1, x,y,z >= 0, a + b + c = 1, a >= 1/2, b,c >= 0.

There are no solutions in the interior since those points boil down to the original rock paper scissors game. So, one only needs to check the boundary, when a = 1/2 and, thus, c = 1/2 - b. If you fix x,y,z then the extreme points occur at b = 1/2 and b = 0 giving a Win-Loss of y/2 - x/2 when x >= (y+z)/2 and x/2 - z/2 otherwise . Again the interesting behavior only happens at the boundary and, so, setting x = (y+z)/2 gives a Win-Loss of y/4 - z/4 which is maximized when z = 0, and, therefore 1 = x + y = y + y/2 =(3/2)*y, yielding y = 2/3, x = 1/3, and a Win-Loss of 1/6.

... is coming monday on the author's youtube channel https://www.youtube.com/channe... [youtube.com] -- he commented with this but is evidently bad at slashdot so I'm hopping this comment gets read even if his doesn't...

It says that rock *must* be thrown 50% of the time. Not that there is a 50/50 chance of the choice being rock for each game. So the number of games *has* to be set out before gaming commences. As you draw closer to the last game, the odds will change to how likely rock will be thrown. This is very similar to counting cards in a blackjack.

Not much of a brain teaser. (0)

Anonymous Coward | about 10 months ago | (#46672131)

Always play paper. Worst case scenario is a perfect stalemate. Over the long term it's not possible for you to lose, only tie.

Well, you could... (0)

Anonymous Coward | about 10 months ago | (#46672137)

doesn't include a consideration of my lifespan. If I only live 100 years my opponent can play scissors every single game and play rock continuously after I am dead and the restriction would still be satisfied. In other words, "having to play rock 50 percent of the time" doesn't give you any relevant information about your opponent's behaviour.

## Simple.... Odds are even (0)

## Fallen Kell (165468) | about 10 months ago | (#46670445)

## Re:Simple.... Odds are even (0)

## Anonymous Coward | about 10 months ago | (#46670539)

The odds are even odds. If he must throw rock 50% of the time, and he knows that you know that data, the other 50% of the time he will throw scissors to beat your paper....

Which means that he won't throw anything that gives points against rock so you should play rock all the time.

Until he figures that out, then he will have to play paper every now and then.

## Re:Simple.... Odds are even (1)

## Fallen Kell (165468) | about 10 months ago | (#46670563)

After that you need to run a probability set on all the possible combinations, with the unknowns for his paper and scissors, only knowing that they total to 50%.

## Re:Simple.... Odds are even (5, Funny)

## narcc (412956) | about 10 months ago | (#46671051)

Actually now that I think about it more

... you realize that you're doing the submitter's homework?

## Re:Simple.... Odds are even (3)

## Tom (822) | about 10 months ago | (#46671193)

...but...but... but it's :-)

interestinghomework!## Re:Simple.... Odds are even (-1)

## Anonymous Coward | about 10 months ago | (#46671409)

As interesting as you making a total fool of yourself vs. apk http://tech.slashdot.org/comme... [slashdot.org] ? He utterly destroyed you, "3 digit registered 'luser'" that you clearly are (technically inept noob who libels when shot down is more like it).

## Re:Simple.... Odds are even (1)

## Anonymous Coward | about 10 months ago | (#46671165)

Is it harder than spelling seems?

## Re:Simple.... Odds are even (0)

## Anonymous Coward | about 10 months ago | (#46671559)

"Is it harder than spelling seems?"

Is it harder than putting referenced words in quotation marks?

## Re:Simple.... Odds are even (2, Funny)

## Anonymous Coward | about 10 months ago | (#46672095)

A hem! I'm a frayed knot.

## Re:Simple.... Odds are even (0)

## ihtoit (3393327) | about 10 months ago | (#46670585)

actually, as each choice beats one and either ties or is beaten by the other two, the odds of winning any random round of RPS is 33%.

## Re:Simple.... Odds are even (2)

## ShanghaiBill (739463) | about 10 months ago | (#46670743)

actually, as each choice beats one and either ties or is beaten by the other two, the odds of winning any random round of RPS is 33%.

But the whole point is that it is

not random,since your opponent's choices are constrained. He is forced to play rock 50% of the time. If he plays randomly the other 50%, then you can always play paper and win 2/3rds of the time, lose 1/6 of the time, and tie 1/6th. But then he can adapt, and play rock 50% and scissors the other 50%, resulting in a tie (1/3 win, 1/3 loss, 1/3 tie). But you can adapt to that by playing rock 50% and paper 50%. You will win 50%, lose 25%, and tie 25%.## Re:Simple.... Odds are even (1)

## Chris Mattern (191822) | about 10 months ago | (#46670891)

But then he adapts by always playing paper when he doesn't have to play rock. Now you win 25%, tie 50%, lose 25%, and you're back to even up.

## Re:Simple.... Odds are even (5, Insightful)

## ShanghaiBill (739463) | about 10 months ago | (#46671263)

The Nash Equilibrium [wikipedia.org] is for you to play paper 2/3rds of the time, and rock 1/3rd. His best counter strategy is to play rock 50% (he cannot go lower) and scissors 50%. He cannot do better. If you deviate from 2/3 paper and 1/3 rock, he can adjust his strategy to do better. With the optimal strategy, you will win 1/2, lose 1/3, and tie 1/6.

Here is my search for the Nash Equilibrium:

#include

struct rps {

double rock;

double paper;

double scissors;

};

static double

eval(struct rps *a, struct rps *b)

{

return

(a->rock * (b->scissors - b->paper)) +

(a->paper * (b->rock - b->scissors)) +

(a->scissors * (b->paper - b->rock));

}

int

main(void)

{

struct rps you;

struct rps him;

him.rock = 0.5;

double worst_best_eval_for_him = 1.0;

double best_rock_for_you = 0;

double best_paper_for_you = 0;

double worst_best_paper_for_him = 0;

double dx = 0.001;

for (you.rock = 0; you.rock best_eval_for_him) {

best_eval_for_him = p;

best_paper_for_him = him.paper;

}

}

if (worst_best_eval_for_him > best_eval_for_him) {

worst_best_eval_for_him = best_eval_for_him;

best_rock_for_you = you.rock;

best_paper_for_you = you.paper;

worst_best_paper_for_him = best_paper_for_him;

}

}

}

printf("worst_best_eval_for_him = %f\n", worst_best_eval_for_him);

printf("best_rock_for_you = %f\n", best_rock_for_you);

printf("best_paper_for_you = %f\n", best_paper_for_you);

printf("worst_best_paper_for_him = %f\n", worst_best_paper_for_him);

return 0;

}

## Re:Simple.... Odds are even (1)

## ShanghaiBill (739463) | about 10 months ago | (#46671291)

Sorry, but Slashdot mangled that code badly because of the angle brackets.

## Re:Simple.... Odds are even (5, Interesting)

## ShanghaiBill (739463) | about 10 months ago | (#46671323)

Sorry, but Slashdot mangled that code badly because of the angle brackets.

Let me try again:

#include <stdio.h>

struct rps {

double rock;

double paper;

double scissors;

};

static double

eval(struct rps *a, struct rps *b)

{

return

(a->rock * (b->scissors - b->paper)) +

(a->paper * (b->rock - b->scissors)) +

(a->scissors * (b->paper - b->rock));

}

int

main(void)

{

struct rps you;

struct rps him;

him.rock = 0.5;

double worst_best_eval_for_him = 1.0;

double best_rock_for_you = 0;

double best_paper_for_you = 0;

double worst_best_paper_for_him = 0;

double dx = 0.001;

for (you.rock = 0; you.rock < 1.0; you.rock += dx) {

for (you.paper= 0; (you.paper + you.rock) < 1.0; you.paper+= dx) {

you.scissors = 1.0 - you.rock - you.paper;

double best_paper_for_him = 0.0;

double best_eval_for_him = -1.0;

for (him.paper = 0; him.paper < 0.5; him.paper += dx) {

him.scissors = 1.0 - him.rock - him.paper;

double p = eval(&him, &you);

if (p > best_eval_for_him) {

best_eval_for_him = p;

best_paper_for_him = him.paper;

}

}

if (worst_best_eval_for_him > best_eval_for_him) {

worst_best_eval_for_him = best_eval_for_him;

best_rock_for_you = you.rock;

best_paper_for_you = you.paper;

worst_best_paper_for_him = best_paper_for_him;

}

}

}

printf("worst_best_eval_for_him = %f\n", worst_best_eval_for_him);

printf("best_rock_for_you = %f\n", best_rock_for_you);

printf("best_paper_for_you = %f\n", best_paper_for_you);

printf("worst_best_paper_for_him = %f\n", worst_best_paper_for_him);

return 0;

}

## Re:Simple.... Odds are even (0)

## Anonymous Coward | about 10 months ago | (#46671837)

Actually, 50% Rock, 50% Scissors is not the opponent's equilibrium strategy, nor is it the "best response" to 2/3 Paper 1/3 Rock. All responses involving exactly 50% Rock are equally good against 2/3 Paper 1/3 Rock, but if your opponent plays 50% Rock 50% Scissors you can exploit that with 100% Rock, which does better against that strategy than 2/3 Paper 1/3 Rock. So your opponent would be better off with a strategy that cannot be exploited.

The only strategy for the opponent that cannot be exploited is 1/2 Rock 1/6 Paper 1/3 Scissors. All your strategies are equally good against this unless you stupidly play Scissors with nonzero probability.

## Re:Simple.... Odds are even (1)

## inhuman_4 (1294516) | about 10 months ago | (#46671971)

Wins for you:

Paper vs rock: 2/3 * 1/2 = 1/3 win

Rock vs scissors: 1/3 * 0 = 0 win

Scissors vs paper: 0 * 1/2 = 0 win

For him:

Paper vs rock: 1/2 * 1/3 = 1/6

Rock vs scissors: 1/2 * 0 = 0

Scissors vs paper: 0 * 2/3 = 0

Your optimal strategy (2/3 paper, 1/3 rock) vs his optimal strategy (1/2 paper, 1/2 rock), results 1/3 win not a 1/2 win.

## Re:Simple.... Odds are even (1)

## Reaper9889 (602058) | about 10 months ago | (#46672003)

His nash strategy (minimax strategy) is to play rock 1/2, scissors 1/3 and paper 1/6 (it is true that rock 1/2 and scissors 1/2 is a best response to the nash strategy, but that does not mean it is optimal against abitrary strategies). I have a long earlier post showing the strategies and so on :)

## Re:Simple.... Odds are even (1)

## abies (607076) | about 10 months ago | (#46672379)

I'm getting better test results from having around 38.4% rock, rest paper. At this point his scissor and paper strategies become equal and I'm winning (assuming ties are repeated) 61.7% of cases instead of 60% of cases with 1/3 rock rest paper.

I think that trick is finding a place where any mixture of scissors or paper for him is equally bad. At this moment, his choices doesn't matter anymore (of course, unless he choses rock even at free choice). At 1/3 rock 2/3 paper, scissors are clearly better choice for him, so it is not a perfect place.

## Re:Simple.... Odds are even (1)

## jklovanc (1603149) | about 10 months ago | (#46670781)

Though of the rounds where there is a winner the odds are 50/50.

## Re:Simple.... Odds are even (-1, Troll)

## wealthychef (584778) | about 10 months ago | (#46670989)

## Re:Simple.... Odds are even (1)

## osu-neko (2604) | about 10 months ago | (#46671253)

You seriously think that each player in RPS has a 33% chance of winning each round? Think a little bit about that. Oh, I forgot, this is /.

What do you think the odds are for each player? Keep in mind that there are three possible outcomes for each round: win, lose, or draw.

## Re:Simple.... Odds are even (0)

## Anonymous Coward | about 10 months ago | (#46671297)

If I choose 2, there are three outcomes:

- you choose 1 - I lose

- you choose 2 - tie

- you choose 3 - I win.

This applies for any mapping between {1,2,3} and {rock,paper,scissors}.

That looks awfully like 33% to me.

## Re:Simple.... Odds are even (0)

## Anonymous Coward | about 10 months ago | (#46672355)

We each pick a number between 1 and 10^10^10, and then throw a die whomever guessed it wins...

I see three outcome I win, you win, we both win, or we both lose.

so there is a 25% chance that we both guess the number every single time right?

## Always do rock. (0)

## Anonymous Coward | about 10 months ago | (#46670449)

You tie 50% of the time, win 25% of the time, and lose 25% of the time. Pay? Nothing.

## Re:Always do rock. (1)

## Chris Mattern (191822) | about 10 months ago | (#46670493)

Come again? It is specified that the player forced to play 50% rock (as mandated by a fair coin flip you don't get to see) plays intelligently and will adapt to your play. When he figures out you're always playing rock, he'll always play paper when he doesn't have to play rock. You tie 50% of the time and lose 50% of the time. That's lousy strategy.

You're guaranteed to break even by always playing paper. When the opponent adapts, he'll always play scissors when he doesn't have to play rock, and you win 50% of the time and lose 50% of the time. The question is, can you do better than that?

## Re:Always do rock. (5, Funny)

## ArcadeMan (2766669) | about 10 months ago | (#46671797)

Of course. You use the rock to smash him in the head while he tries to stab you with the scissors. Your friend then uses the paper to write a letter to your parents about how you died in a stupid fight about statistics.

## 100% paper (0)

## Junior J. Junior III (192702) | about 10 months ago | (#46670475)

## Re:100% paper (1)

## hibiki_r (649814) | about 10 months ago | (#46670639)

The opponent is forced to pick 50% rock, but he has no limitations other than that. If you went 100% paper, you'd not beat me playing 50% rock by more than a tiny sliver of a percent it takes me to realize you are going 50% paper.

And the problem has nothing to do with the someone not just picking rock, but doing so in a very predictable manner. Otherwise, we'd not be talking about someone picking rock 50% of the time, but playing against someone that plays randomly, but tells you what he is picking half the time, never lying. That's a very, very boring brainteaser.

The first naive approach that doesn't assume idiocy on the other side's part is to assume he'll try to guess our play. If we play in a non deterministic fashion, his best options is probably random, but instead of .33/.33/.33, we get .50/.25/.25. Against an opponent doing that, you could go .25/.50/.25, and win quite a bit.

If he is forced to get at least 50% rock after a certain number of plays, then we can change our probabilities depending on how far from the standard distribution our opponent is: For instance, if we were playing to 10 throws, and he never picked rock in the first 5, we'd get him in the last 5, because his moves are forced. To avoid this, someone that has to reach that 50% will probably also change their probabilities after every throw, just to avoid such a 'free' run. One might even consider starting with an over 50% rock probability, because under that set of rules, having overplayed rock lets us play more optimally later, while underplaying rock leads to major losses.

Either way, a more detailed problem specification is required.

## Re:100% paper (4, Informative)

## BlackPignouf (1017012) | about 10 months ago | (#46670773)

From TFA: "At the start of each round an independent judge flips a fair coin and tells your opponent the result but does not tell you. If the coin came up heads your opponent must play rock."

The opponent isn't forced to get at least 50% rock after any number of plays.

## Re:100% paper (1)

## seyyah (986027) | about 10 months ago | (#46670813)

100% paper strategy will win 50% of the time. Of the remaining 50% of games played, (assuming even distribution of the remaining picks) 25% will be losses and 25% will be tied

No, you are describing another game where the opponent is forced to play rock 50% of the time, paper 25% of the time and scissors 25% of the time.

This game is different.

(I'm repeating my self from another post but many people are making the same mistake)

## For money, you mean? (0)

## therealkevinkretz (1585825) | about 10 months ago | (#46670481)

With no such restriction, random choices on both sides lead to 33% win, 33% draw, 33% loss, right? With the opponent throwing Rock 50% of the time, assuming the other 50% is evenly divided between Paper and Scissors, if I always throw paper I'll win 50% of the time, lose 25% of the time, and draw 25% of the time.

So depending how the betting works, I'd be pretty willing.

## Re:For money, you mean? (2)

## seyyah (986027) | about 10 months ago | (#46670801)

With no such restriction, random choices on both sides lead to 33% win, 33% draw, 33% loss, right? With the opponent throwing Rock 50% of the time, assuming the other 50% is evenly divided between Paper and Scissors, if I always throw paper I'll win 50% of the time, lose 25% of the time, and draw 25% of the time.

So depending how the betting works, I'd be pretty willing.

No, you are describing another game where the opponent is forced to play rock 50% of the time, paper 25% of the time and scissors 25% of the time.

This game is different.

## Re:For money, you mean? (0)

## Anonymous Coward | about 10 months ago | (#46670837)

Your opponent will play an optimal strategy.

You always throw paper. You will win only 50% of the time. The 50% when they are forced to throw rock.

Your opponent will throw scissors 50% of the time and beat your paper.

You failed to understand the proposition.

## Play paper 50% of the time. (1, Interesting)

## Anonymous Coward | about 10 months ago | (#46670499)

Expect them to play scissors a lot to beat your paper. Play rock as often as they play scissors.

## put a spin on it (3, Interesting)

## ihtoit (3393327) | about 10 months ago | (#46670575)

Rock, paper, scissors, lizard, Spock!

Scissors cut paper

Paper covers rock

Rock crushes lizard

Lizard poisons Spock

Spock smashes scissors

Scissors decapitate lizard

Lizard eats paper

Paper disproves Spock

Spock vaporizes rock

Rock crushes scissors

## 50% paper? (1)

## istartedi (132515) | about 10 months ago | (#46670635)

I think the winning strategy is to randomly throw 50% paper to cover his rock. I'm just guessing though. No idea how much to pay.

## Gambling Apocalypse Kaiji (0)

## Anonymous Coward | about 10 months ago | (#46670663)

If you actually enjoy this question, watch the Kaiji Anime or read the manga, you won't be disappointed.

If you don't, run far away from it.

## Impossible (1)

## cytg.net (912690) | about 10 months ago | (#46670677)

## Re:Impossible (0)

## Anonymous Coward | about 10 months ago | (#46670769)

Exactly right. Unless we know the distribution, it's an unanswerable question in general. You could play a million games and never see rock for the first half of those, and yet over the entire game domain, your opponent could still throw rock 50% of the time.

## Re:Impossible (1)

## umghhh (965931) | about 10 months ago | (#46670927)

## Edward Scissorhands would hate this opponent (2)

## eman1961 (642519) | about 10 months ago | (#46670709)

## 50% paper and 50% rock ? (0)

## abies (607076) | about 10 months ago | (#46670757)

Assuming we randomly use 50% paper and 50% rock, we get:

- on rounds he is forced to play rock, we get half victory and half tie, so 25% win, 25% tie so far

- on other rounds he can:

- always use scissors, which will turn into our 25% win, 25% lose - we win overall, 50% win, 25% tie, 25% lose

- always use rock, which will turn into our 25% tie, 25%win - we win overall, 50% win, 50% tie

- always use paper, which will turn into 25% win, 25% lose - we win overall, 50% win, 25% tie, 25% lose

- any kind of random mix of above, which can only move between results above, as our picks are completely random as well

So, he best strategy is to always play scissors or rock, for 50:25:25 result. If ties are repeated, we win 2 out of 3 times.

Any better strategy?

## Re:50% paper and 50% rock ? (0)

## Anonymous Coward | about 10 months ago | (#46670935)

he best strategy is to always play scissors or rock, for 50:25:25 result. If ties are repeated, we win 2 out of 3 times.

you mean paper or rock. I wonder if any strategy that never plays scissors is a winner. It's hard to answer that because we don't know the opponent's strategy/distribution on the rounds where they're not forced to play rock. 50/50 rock/paper is a winning strategy, as you've shown. But is it optimal? Idk

## Re:50% paper and 50% rock ? (0)

## Anonymous Coward | about 10 months ago | (#46670951)

Err, what?

always use paper, which will turn into 25% win, 25% lose - we win overall, 50% win, 25% tie, 25% lose

using your terms, random rock/paper against opponent playing paper is 25% tie, 25% lose

so we get 25:50:25 result. Net zero.

## Re:50% paper and 50% rock ? (1)

## artor3 (1344997) | about 10 months ago | (#46671657)

I think you have an error in there. His best strategy in your example would be to always use paper, in which case he's doing the exact same thing as you are, and so the odds are even.

I think your best strategy would have to involve reacting to his. Start with 100% paper until you see what he's throwing when he's not throwing rock. This gives you at least even odds regardless of what he does.

He'll probably start throwing scissors, at which point you switch to you switch to 50% rock 50% paper. This will have you win 2/4, lose 1/4, draw 1/4.

He should respond by starting to throw paper. so you just go back to 100% paper and win half the time. If he switches back to scissors, you start throwing rock again.

The goal is to use long stretches of paper to for him into 50R:50S (which only breaks even at best), and then switch to 50R:50P to pick up some wins until he switches back 50R:50P (again, he'll only be able to break even). So when he's properly countering you, he can only break even, and every time you switch strategies you should pick up some extra wins. If he tries to go for a midpoint (50:25:25) he'll lose to either of your two strategies.

## Re:50% paper and 50% rock ? (1)

## abies (607076) | about 10 months ago | (#46672075)

Yes, you are right - paper will be 25% tie, 25% lose, ending up with 50% tie, 25% win, 25% lose, so still purely random.

I don't think that strategy we are looking for involves "reactions" - this can be always defeated by opponent which overguess you by one step. I would hope to find a strategy which would lead you to win > 50% even if opponent knows it (strategy itself, not the result of random choice at given stage).

## 38% rock and 62% paper ? (1)

## abies (607076) | about 10 months ago | (#46672297)

Ok, my current guess would be around 1/3 rock and 2/3 paper. Opponent cannot replicate this strategy, because of requirement of having at least 50% rock. If he goes 50% scissors, he will have 1/6 tie, 2/6 lose, 1/6 lose and 2/6 win, so 3:2 in my favor. With 50% paper, he gets 1/6 tie, 2/6 lose, 1/6 win, 2/6 tie, so 2:1 in my favor. 100% rock is obvious lose. I think that mixing scissors and paper will be clearly worse than pure scissors on his part (because with scissors, he has at least 2/3 chances of winning after he got a choice).

From random testing, it seems that optimal ratio of rock is around 38% (not 1/3). Around that point, pure paper and pure scissor strategies for opponent seems to get equal and I'm winning around 62% of cases.

There is probably some interesting math behind that...

## Programming Course Topic (4, Interesting)

## NeuroBoy (82993) | about 10 months ago | (#46670765)

I've been having students in my introductory programming courses work on this class of problem for a few years.. They all seem to really enjoy it. I code up bots to play RPS with certain biases just like the OP and they have to program a single player that identifies the bias in an opponent and adjusts its play to give it an advantage. They all routinely can generate solutions that perform far better than random against predictable, dumb bots, but things get very interesting when I throw the students' bots against each other in a throwdown tournament. :)

## recalculate every round (1)

## Anonymous Coward | about 10 months ago | (#46670827)

If i know the number of rounds in advance (atleast your oponent has to right if he needs to calculate the 50%). I would recalculate the odds after every round taking into account the number of rock move remaining for him...

## Like most decisions, it's partly mental (1)

## Antique Geekmeister (740220) | about 10 months ago | (#46670849)

Like drawing to an inside flush, an "optimized" strategy is not necessarily what the opponent plays. There is no reason inherent in the description to make assumptions about the opponent's other play. They may also be constrained to play paper the other fifty percent of the time, and to play paper , then rock, then paper, then rock. In the real world, don't assume that the minimal description of the problem gives all the important data.

## Re:Like most decisions, it's partly mental (1)

## Todd Palin (1402501) | about 10 months ago | (#46671617)

## BAD SUMMARY (1)

## umghhh (965931) | about 10 months ago | (#46670977)

there is a factual problem with the summary.TFA says it all so better read it. If not read this (I hope not to mess up it too much).

It is not required of the opponent to play rock 50% of the time. The referee is using a fair coin to determine if your opponent is to play rock or not. If s/he is not forced to play rock s/he is free to chose allowing him also to chose rock 100% of time too if s/he so wishes.

## Re:BAD SUMMARY (1)

## Hognoxious (631665) | about 10 months ago | (#46671339)

It totally is required to play rock 50% of the time. It's not required to play rock

exactly50% of the time.## Re:BAD SUMMARY (2)

## osu-neko (2604) | about 10 months ago | (#46671535)

there is a factual problem with the summary...

It

isa bad summary, but only because the wording is ambiguous, not that it's factually incorrect. The statement you're objecting to is perfectly correct in one interpretation, and dead wrong in another. Your own counter-statement, "it is not required of the opponent to play rock 50% of the time," is equally ambiguous. In fact, 50% of the time (assuming a fair coin), the opponent is required to play rock, so it's true that "itisrequired of the opponent to play rock, 50% of the time". Leaving out the comma yields a true sentence (assuming the correct interpretation is chosen of the now even more ambiguous sentence) that contradicts the quoted sentence of yours, assuming you parse your sentence as "it is not required: that the opponent play rock 50% of the time", but does not contradict it at all if your sentence is parsed "it is not required that the opponent play rock, 50% of the time", since 50% of the time, the opponent can choose freely, and thus is 50% of the time, is not required to make any particular choice. So, both the summary and your explanation of what's wrong with it contain statements that not factually incorrect, just ambiguously worded such that a reader might interpret it to mean something that is incorrect rather than something that is correct.This is a textbook example of why programming computers in plain English would be a monstrously bad idea.

## Strategy... (1)

## bhoar (1226184) | about 10 months ago | (#46671011)

## bonus problem: paper,paper (0)

## Anonymous Coward | about 10 months ago | (#46671023)

If the opponent plays P or S on the first move, his second move is determined, so there's a greater than random chance he will want to play R first. So the safest thing is to play paper twice. Or: paper first, and if the opponent played rock, then play anything at random.

## Solution (1)

## Reaper9889 (602058) | about 10 months ago | (#46671061)

Lets call the guy with the restriction player 1 and the other player 2.

If you think about it player 1 got 3 "pure" strategies (as in: each other strategy he can play can be seen as a mixture of these 3):

(1) rock 100%,

(2) paper 50%/rock 50% and

(3) scissors 50%/rock 50%.

Against (1) rock gives 0, paper -1 and scissors 1.

Against (2) rock gives 1/2, paper -1/2 and scissors 0.

Against (3) rock gives -1/2, paper 0 and scissors 1/2.

In each case, the number is the probability of player 1 winning minus player 2 winning.

We see that player 2 should not play scissors, because he will never gain anything from it and clearly the optimal strategy should be able to gain something (we will see that it is 1/6). Then, knowing that, paper 50%/rock 50% is better than rock 100% for player 1: If player 2 plays rock, player 1 gets 1/2 (instead of 0) and if player 2 plays scissors player 1 gains -1/2 (instead of -1).

Hence, we are down to:

(2) paper 50%/rock 50% and (3) scissors 50%/ rock 50% vs. rock and paper. If player 1 plays (2) with pr. 1/3 and (3) with probability 2/3, he loses 1/6 against both rock and paper. If player 2 plays rock with probability 1/3 and paper with probability 2/3 he gets 1/6 against (1) and 1/6 against (2). This is optimal since each player have a way to guarantee that player 1 loses 1/6 and player 2 wins 1/6. If either had a better strategy it would break the other players guarantee (note that the given strategy for player 1 wins 1/6 against scissors, again showing that it is a bad strategy for player 2 and player 2's strategy wins 1/3 against rock 100% showing that it is a bad strategy).

## Re:Solution (1)

## Reaper9889 (602058) | about 10 months ago | (#46671289)

Sorry :/ There are some mistakes in the last part. The strategy for player 1 wins 1/2-1/6=1/3 and not 1/6 as claimed. Also, the strategy for player 2 wins 2/3 against pure rock and not 1/3 as claimed. Still, it just makes it even clearer that you should not play rock with probability more than 50% as player 1 and not play scissors at all.

Also, to be more precise, the strategy for player 1 is to play rock with probability 1/2, paper with probability 1/6 and scissors with probability 1/3.

## Randomize 2/3 paper, 1/6 rock, 1/6 scissors (1, Insightful)

## Anonymous Coward | about 10 months ago | (#46671093)

Over the long term, the strategy must converge to stable, therefore true random can be the only optimized strategy.

50% of time opponent must play R. The remaining 50% of the time they can equally choose R,P,S.

R -> P = 1/2 + 1/3 * 1/2

S -> R = 1/3 * 1/2

P -> S = 1/3 * 1/2

For a guaranteed win, roll a die: 1-4 => P, 5 => R, 6 => S.

By how much? Consider you are random as above and opponent is fixed wlog at 100% R. You win 2/3 of the time and lose 1/6 of the time.

The expected payoff to play is 2/3, the expected cost is 1/6. Payoff - cost = 1/2, so the most you should be willing to pay to play a $100 game is $150.

## Two Games (3, Insightful)

## inhuman_4 (1294516) | about 10 months ago | (#46671095)

The key (if you RFTA) is that whether or not your opponent plays rock is determined by a coin toss. So really you are playing a compound game. You are playing a coin toss and rock paper scissors (RPS). Since the coin toss determines your opponents move, you can think of it as playing 50% coin toss and 50% RPS. The RPS is a subgame of the coin toss.

Since the coin toss is the dominate game, you play with win that first. But instead of heads/tails, it is paper/other. The answer to the coin toss is a 50/50 guess of heads/tails, so the answer to the paper/other is 50% paper, 50% other.

The "other" is the RPS game. And since the answer to the RPS game is 1/3 rock, 1/3 paper, 1/3 scissors, we know what the solution to the other 50% of the game is.

So the equations are:choice = (Coin Toss) + (RPS) so: paper = 1/2 + 1/3, rock = 0 + 1/3, scissors = 0 + 1/3. Or paper = 4/6, rock = 1/6, scissors = 1/6.

## Re:Two Games (2)

## MtHuurne (602934) | about 10 months ago | (#46671201)

The opponent could respond to this by playing scissors on all non-forced-rock turns. If the opponent plays rock, you win 4/6 of the time and lose 1/6 of the time, but if the opponent plays scissors you lose 4/6 of the time and win 1/6 of the time, so overall you'd be even.

## Re:Two Games (1)

## inhuman_4 (1294516) | about 10 months ago | (#46671485)

1/2 scissors x 4/6 paper = 2/6 = 1/3 victory for the opponent. 1/2 scissors x 1/6 scissors is 1/12 tie. And 1/2 scissors x 1/6 rock is 1/12 lose. So the "all scissors" strategy only nets him 1/3 victory not 4/6.

## Re:Two Games (1)

## MtHuurne (602934) | about 10 months ago | (#46672397)

I listed the chances in the context of the opponent move ("if the opponent plays rock"). The chance of playing rock or playing scissors is 1/2 each (the coin toss), so if you list it as overall chances you get 1/3 win and 1/12 loss (same as you wrote) due to the opponent playing scissors and also 1/3 loss and 1/12 win due to the opponent playing rock; the expected result result is still 0.

## Re:Two Games (1)

## mestar (121800) | about 10 months ago | (#46671685)

How the fck would you manage to lose 4/6 of the time to an opponent who must play rock at least 50% of the time?

## Re:Two Games (2)

## Reaper9889 (602058) | about 10 months ago | (#46671387)

There is a flaw in your reasoning. You do not know that your oppoent flipped so you can not condition on it like you do here (you can not play paper all the time if he "flips" rock because you do not know his coin flip). If you think about it you should NEVER play scissors. In the best case for you he plays rock 50% and paper 50% and you get 0 in expectation and clearly you got an advantage so 0 is not good.

The optimal strategy is to play 1/3 rock, 2/3 paper. It gives at least 1/6 against anything he could play. He can similarly ensure that you can not get more than 1/6 a game by playing rock with probability 1/2, paper with probability 1/6 and scissors with probability 1/3. Your strategy would get less than 1/6 against that (more precisely, you get 1/6 if you play either rock or paper and you lose 1/3 if you play scissors. Therefore you get 1/6*5/6-1/3*1/6=1/12 which is less than the 1/6 you get for playing 1/3 rock and 2/3 paper).

See my above post for a indepth analysis.

## Re:Two Games (1)

## inhuman_4 (1294516) | about 10 months ago | (#46671833)

If I play paper 4/6 of the time, than I should expect 1/2 of my paper to align with his rock. So 4/6 * 1/2 = 2/6 = 1/3. So I should expect to win 1/3 of the time, plus my winnings on the other combinations. That means 1/3 is the lower bound.

If you play 1/3 rock and 2/3 paper, his response will be 1/2 paper and 1/2 rock. So you are going to get 2/3 * 1/2 = 1/3 for your paper. But your 1/3 rock will never win because he will never play scissors either. But his 1/2 paper will meet your 1/3 rock, giving him 1/2 * 1/3 = 1/6 win. Putting you head by only 1/6.

This is where the two games key comes in. You and I both recognize that 2/3 paper is the right move because 1/2 of his moves will be rock. But by playing the other half as regular RPS with a win/tie/loss of 1/1/1 you can expect the win/loss to cancel out, leaving you with your 1/3 lower bound advantage.

## Re:Two Games (1)

## Reaper9889 (602058) | about 10 months ago | (#46671947)

Ok lets see: you play 2/3 paper (I shorten the fraction I hope that is ok :) ) and 1/6 scissors and 1/6 rock. You play against the strategy 1/2 rock, 1/3 scissors, 1/6 paper. Fast version: lets look at a random round in which you play rock: In those you win 1/3 against his scissors and lose 1/6 against paper, thus you get 1/6 on avg.

Next, random round in which you play paper: In those you get 1/2 against his rock, and lose 1/3 against his scissors, i.e. again you gain 1/6.

Next, random round in which you play scissors: In those you get 1/6 against his paper and lose 1/2 against his rock, i.e. you LOSE 1/3.

On avg you play rock 2/3 of the time and get 1/6 in those rounds, scissors in 1/6 of the time and LOSE 1/3 and paper 1/6 of the time and get 1/6. Thus, on avg. 2/3*1/6+1/6*(-1/3)+1/6*1/6=1/12. This is below your lower bound so there is something wrong with it. (the reason is that you lose whenever a bit on avg. whenever you do not play paper).

My strategies, played against each other gives 1/6. Thus, you can not say that yours is better always. I can argue that against ANY strategy mine gets 1/6. You can not get better than 1/12 (because you get that against mine strategy for the other player). Thus, yours can not be optimal sorry :(

## Re:Two Games (1)

## Reaper9889 (602058) | about 10 months ago | (#46671979)

Note, I say I can argue that my strategy gives 1/6 on avg. in the sense that I did argue that in a looong post above :/

## 2/3 Paper 1/3 Rock (0)

## Anonymous Coward | about 10 months ago | (#46671125)

The payoff is 16 2/3.

The defender should do 1/3 paper 2/3 scissors. (When not forced to play rock.)

This is just a standard mini max problem. The payoff matrix looks like:

0 -50 50

100 50 0

-100 0 -50

(With row and columns being in RPS order.)

## Re:The bonus question2/3 Paper 1/3 Rock (0)

## Anonymous Coward | about 10 months ago | (#46671335)

For the bonus question you can set up a similar payoff matrix. For the case where the restricted player picks rock on the first round the value of the second round will be zero. When rock is not picked in the first round by the resticted player the value of the second game will be 50 to the unrestricted player.

So the payoff matrix for the first round that includes the value of the second round looks like:

0 0 100

50 50 0

-50 100 50

(Again rows and columns are in RPS order.)

The payoff to the unrestricted player is 33 1/3 for both rounds combined.

The unrestricted player should use 2/3 paper 1/3 rock in the first around. In the second round either just paper or 1/3 RPS depending on whether or not the restricted player must play rock.

The resticted player should player should play 2/3 rock 1/3 scissors in the first round.

There is a program to calculate these results at:

http://wolff.to/bruno/strategy... [wolff.to]

## Huh? (0)

## Anonymous Coward | about 10 months ago | (#46671155)

Nobody seems to be asking what the definition of "50% of the time" is ?

Over what interval/time frame?

If the opponent throws scissors or paper on his first move is he then required to throw rock on his second move?

## Re:Huh? (1)

## Seraphim1982 (813899) | about 10 months ago | (#46672535)

Nobody seems to be asking what the definition of "50% of the time" is ?

Over what interval/time frame?

If the opponent throws scissors or paper on his first move is he then required to throw rock on his second move?

Probably because they read the link which describes how "50% of the time" is determined.

## No actual advantage? (3, Insightful)

## MtHuurne (602934) | about 10 months ago | (#46671157)

First, make sure you read TFA, since it explains what the summary doesn't: how the 50% is determined and how the opponent can play in the non-forced turns.

If you play using a deterministic algorithm, for example always play paper, the opponent can figure it out and beat you on all the non-forced turns. At best you'll get an even result.

If you play using a random algorithm, the opponent can figure out the frequencies you're using and compensate for that. For example, if you decide to play paper 50% of the time and rock and scissors 25% of the time, you'd win against an opponent playing rock 50% of the time and paper and scissors 25% of the time. However, if the opponent decides to play rock 50% of the time and scissors the other 50%, the result is even again. If the opponent would be forced to play rock more than 50% of the time, there is no room to compensate and you would win consistently with 100% paper. I think that with 50% rock, there is enough room to respond to any frequency distribution you can come up with, although I have no proof for that.

You could change your algorithms during play, but if there isn't any algorithm that results in an advantage when playing it consistently, gaining an advantage from changing your algorithm would depend on how well your opponent responds to your changes. In other words, you're playing mind games. I don't think the 50% rock restriction is going to be of any help here.

## Re:No actual advantage? (2)

## Reaper9889 (602058) | about 10 months ago | (#46671349)

You can get an advantage. The important point is to notice that you should not play scissors ever. You can only get 0 in expectation IF he plays paper 50% and rock 50% and he gets an advantage otherwise and 0 is not good for you :/ See my above post for further details (spoiler: The optimal choice for you is 1/3 rock, 2/3 paper).

## Easy win, isn't it? (0)

## Anonymous Coward | about 10 months ago | (#46672001)

You play all paper, all the way. You beat him in all the forced turns, giving you an even score. However it will cost him a few lost points on the non-forced turns to figure out your all paper strategy and switch to an all scissor strategy. At that point, haven't you already won? Opponent will only win 50% of the remainder of the games but is already behind, so will never win. For extra points, you can switch to rock at random intervals for 1 turn only, after opponent picks an all-scissor strategy on the non-forced turns. Those will give you a 50/50 chance of a draw (on the forced turns) or a win against scissors (on the non forced turns).

51 to 49 is a win as good as 99 to 1 in RPS.

## Re:Easy win, isn't it? (0)

## Anonymous Coward | about 10 months ago | (#46672411)

Not true: You win and lose $100 for each game of RPS in this model. So the goal is to maximize your winning.

## Re:No actual advantage? (1)

## Dr. Spork (142693) | about 10 months ago | (#46672055)

First, there is some uncertainty about what ratio of your opponent's throws will be paper. Let's call that ratio x. This means that we can define the likelyhood of his throws this way:

HeDoesRock = .5, HeDoesPaper = x, HeDoesScissors = .5 - x

Now you need to figure out the optimal winning ratio of your throwing either rock or paper.

YouWinIfRock = HeDoesScissors = (.5 - x); YouLoseIfRock = HeDoesPaper = x .5; YouLoseIfPaper = HeDoesScissors = .5 - x

YouWinIfPaper = HeDoesRock =

Winning Margin = FreqRock[(.5 - x) - x] + FreqPaper[.5 - (.5 - x)] = FreqRock(.5 - 2x) + FreqPaper(x)

Now take the first derivative of winning margin and set it to zero to find the maximum: .5FreqRock - 2FreqRock(x) +FreqPaper(x) = -2FreqRock +FreqPaper = 0

d/dx

So 2(FreqRock) = FreqPaper, so you should throw paper twice as often as you throw rock. And surprisingly, you can get this result without ever needing to know x, the optimal ratio of your opponent throwing paper. Is that right?

## Re:No actual advantage? (1)

## Reaper9889 (602058) | about 10 months ago | (#46672133)

This is right yes :) Well, assuming that you know that he is playing 50% rock (and not more - that is btw. right - he would worse off if he played rock with even higher probability). Also, to be truly formal about it you should argue that it is a maximum and not a minimum you found :)

## Throw rock (1)

## FuzzNugget (2840687) | about 10 months ago | (#46671185)

## Go for paper (0)

## Anonymous Coward | about 10 months ago | (#46671191)

If I play rock all the time, I tie 50% of the time, lose 25% and win 25%. Perfect tie overall.

If I play scissors all the time, I lose 50% of the time, win 25% and tie 25%. Lose two out of three.

If I play paper all the time, I win 50% of the time, tie 25% and lose 25%. Win twice as much as lose.

Go for paper.

## Profit if the opponent is human and visible (0)

## Anonymous Coward | about 10 months ago | (#46671299)

If the opponent is human and both players can see each other, the opponent won't be able to completely hide his foreknowledge of when he must throw rock. Learn his body language and tells, then throw paper whenever he inevitably gives away his move. If you're good enough to successfully predict, better than chance, when he must throw rock, then over a large enough number of games you can throw randomly the rest of the time and still make a profit.

## Simple. (0)

## iCEBaLM (34905) | about 10 months ago | (#46671303)

If your opponent must throw rock 50% of the time, then you throw paper 100% of the time.

You will win AT LEAST 51% of the time, because you get the 50% gifted to you, and the other non-0% of the times that your opponent throws paper will cause a rematch.

## Progressive Betting (0)

## cosm (1072588) | about 10 months ago | (#46671377)

## I'm the author -- video solution coming soon (1)

## Alex Sutherland (3604875) | about 10 months ago | (#46671531)

## Re:I'm the author -- video solution coming soon (1)

## Alex Sutherland (3604875) | about 10 months ago | (#46671613)

## Re:I'm the author -- video solution coming soon (1)

## Alex Sutherland (3604875) | about 10 months ago | (#46671641)

I'll give a few other hints. Your optimal strategy never plays scissors. Your optimal strategy plays rock 50% of the time. Your optimal strategy does profit on average.

ugh, I meant your optimal strategy plays rock 50% of the time. sorry.

## Re:I'm the author -- video solution coming soon (1)

## Alex Sutherland (3604875) | about 10 months ago | (#46671673)

## Re:I'm the author -- video solution coming soon (1)

## viperidaenz (2515578) | about 10 months ago | (#46671687)

Never playing scissors and playing rock 50% means you play paper 50%.

That's not a hint, that's an answer.

## Re:I'm the author -- video solution coming soon (1)

## Reaper9889 (602058) | about 10 months ago | (#46671737)

"you" are not the guy playing rock 50% of the time. "you" are the guy beating on the poor guy playing rock 50% of the time. The optimal choice is to play rock 1/3 and paper 2/3 (his is to play rock 1/2 and paper 1/6 and scissors 1/3).

## Depends on the stakes (0)

## Anonymous Coward | about 10 months ago | (#46671581)

Personally, I always throw 'gun' if there is sufficient cash involved.

## Bonus question (2/3 paper 1/3 rock is opt) (1)

## Reaper9889 (602058) | about 10 months ago | (#46671671)

Since I already explained the optimal solution to the basic question mentioned in the summery lets solve the bonus question too (my solution also matches the solution given in comments on the article side so it should be good (and said to be correct by the author) - note currently no answer with a high score is correct - mine has 1).

The bonus question is that you play two rounds, and your oppoent must play atleast rock once. So, if he plays something not rock in the first round he must play rock in the second and loss (you just play paper). If he plays rock in the first he can play 1/3 all in the second (which leads to a draw like normal). Thus, if he plays rock first it is like normal RPS (because he get 0 in the next). Otherwise you get one free win (for the second round).

Thus, we can model the first game of the bonus question as (where the numbers is the number of rounds he wins on avg given the choice in round 1):

R P S

R 0 -1 1

P 0 -1 -2

S -2 0 -1

Where you pick columns and him rows. We see that rock dominates paper for the row player. We get

R P S

R 0 -1 1

S -2 0 -1

For the column player, the choice of rock now dominates scissors. We get

R P

R 0 -1

S -2 0

Playing rock 1/3 and paper 2/3 for the collumn player gives -2/3 wins on avg. Similarly, the row player can get -2/3 wins on avg by playing rock 2/3 and scissors 1/3.

## What should opponent do on non-forced turns? (1)

## Dr. Spork (142693) | about 10 months ago | (#46671769)

To figure out what you should do, first assume your opponent is rational, and will make good choices whenever he is able. Since he knows that you will play a paper-heavy strategy to counter his rock-heavy strategy, it would not be rational to

voluntarilychoose more rocks. That could only make things worse.But if he tried to exploit your paper-heavy strategy by throwing scissors on turns when he gets a choice, you'd have a perfect strategy against this: All rock. On forced rock, you get a redo, and on non-forced rock, he does scissors and you win. So on first pass, I think that your opponent should favor paper when he can choose. It's not like you'll

everdo scissors. That's auto-lose half the time - basically a complete surrender of your advantage. So paper is a safe move for your opponent.The problem is that if he did 50% rock and 50% paper, then all-paper will be your perfect counter. He won't let you do that, so he'll have to throw in some scissors. Just how many? It looks now like you will both simultaneously have to determine optimal strategies in order to answer that question, and this will require derivatives of two sets of optimal-choice equations - so that you can solve for the two maxima. Sounds like a fun problem!

## 2/3 paper 1/3 rock (0)

## Anonymous Coward | about 10 months ago | (#46671857)

The best strategy is to randomly choose paper 2/3 of the time and randomly choose rock 1/3 of the time. You will be expected to win at least 2/3*1/2 = 1/3 of the time and tie 1/3*1/2 = 1/6 of the time when the opponent randomly chooses rock. When free to choose, the best option for the opponent is then to always choose scissors which results in you winning 1/3*1/2 = 1/6 of the time and losing 2/3*1/2 = 1/3 of the time. So, overall you will be expected to win 1/2 the time, tie 1/6 of the time, and lose 1/3 of the time.

This is a typical game theory type problem. One simply needs to optimize a max-min problem. If the probability of the player choosing rock, paper, scissors is x,y,z, respectively, and the probability of the opponent choosing is a,b,c, then one optimizes (Wins - Losses):

max_{x,y,z}min_{a,b,c} (b-c)*x + (c-a)*y + (a-b)*z subject to x + y + z = 1, x,y,z >= 0, a + b + c = 1, a >= 1/2, b,c >= 0.

There are no solutions in the interior since those points boil down to the original rock paper scissors game. So, one only needs to check the boundary, when b = 1/2 and, thus, c = 1/2 - a. If you fix x,y,z then the extreme points occur at a = 0 and a = 1/2 giving a Win-Loss of x/2 - y/2 when y >= (x+z)/2 and y/2 - z/2 otherwise . Again the interesting behavior only happens at the boundary and, so, setting y = (x+z)/2 gives a Win-Loss of x/4 - z/4 which is maximized when z = 0, and, therefore 1 = x + y = x + x/2 =(3/2)*x, yielding x = 2/3, y = 1/3, and a Win-Loss of 1/6.

## Re:2/3 paper 1/3 rock (0)

## Anonymous Coward | about 10 months ago | (#46672021)

(Corrected for dyslexia)

The best strategy is to randomly choose paper 2/3 of the time and randomly choose rock 1/3 of the time. You will be expected to win at least 2/3*1/2 = 1/3 of the time and tie 1/3*1/2 = 1/6 of the time when the opponent randomly chooses rock. When free to choose, the best option for the opponent is then to always choose scissors which results in you winning 1/3*1/2 = 1/6 of the time and losing 2/3*1/2 = 1/3 of the time. So, overall you will be expected to win 1/2 the time, tie 1/6 of the time, and lose 1/3 of the time.

This is a typical game theory type problem. One simply needs to optimize a max-min problem. If the probability of the player choosing rock, paper, scissors is x,y,z, respectively, and the probability of the opponent choosing is a,b,c, then one optimizes (Wins - Losses):

max_{x,y,z}min_{a,b,c} (c-b)*x + (a-c)*y + (b-a)*z subject to x + y + z = 1, x,y,z >= 0, a + b + c = 1, a >= 1/2, b,c >= 0.

There are no solutions in the interior since those points boil down to the original rock paper scissors game. So, one only needs to check the boundary, when a = 1/2 and, thus, c = 1/2 - b. If you fix x,y,z then the extreme points occur at b = 1/2 and b = 0 giving a Win-Loss of y/2 - x/2 when x >= (y+z)/2 and x/2 - z/2 otherwise . Again the interesting behavior only happens at the boundary and, so, setting x = (y+z)/2 gives a Win-Loss of y/4 - z/4 which is maximized when z = 0, and, therefore 1 = x + y = y + y/2 =(3/2)*y, yielding y = 2/3, x = 1/3, and a Win-Loss of 1/6.

## A video with the solution and an explanation (1)

## arsheive (609065) | about 10 months ago | (#46671889)

## How many games are we playing? (1)

## TheRealMindChild (743925) | about 10 months ago | (#46671907)

## Not much of a brain teaser. (0)

## Anonymous Coward | about 10 months ago | (#46672131)

Always play paper. Worst case scenario is a perfect stalemate. Over the long term it's not possible for you to lose, only tie.

## Well, you could... (0)

## Anonymous Coward | about 10 months ago | (#46672137)

Do your own statistics homework.

## 50 percent of the time (1)

## kruach aum (1934852) | about 10 months ago | (#46672341)

doesn't include a consideration of my lifespan. If I only live 100 years my opponent can play scissors every single game and play rock continuously after I am dead and the restriction would still be satisfied. In other words, "having to play rock 50 percent of the time" doesn't give you any relevant information about your opponent's behaviour.