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How to win at rock-paper-scissors

bbc.com

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Re: How to win at rock-paper-scissors

#41

Post by Dn_Ab »

valdiorn wrote: post #9

"The game of rock-paper-scissors exhibits collective cyclic motions which cannot be understood by the Nash equilibrium concept." This is complete bullshit. It is well understood that if your opponent is not playing an optimum strategy, you can gain an advantage by shifting away from the Nash equillibrium yourself. Example: Image a player that always picks "Paper". By shifting away from a random choice (Nash optimum)…

You're right that they're incorrect to say RPS can't be understood in terms of the Nash Equilibration concept. Possibly they mistook RPS's lack of a Pure Equilibrium to mean NE is inapplicable. Instead, it has a Mixed Nash Equilibrium.

The advice in the article - bias towards previous opponent win is just an under-specification of how to play a no regret strategy (alter your distribution to match opponent play, which is more or less what your example describes). If you've already installed a random number generator in your brain then playing optimally (in the sense that it will exploit weakness and shift to random as necessary) can be very easily done via a no-regret algorithm such as randomized weighted majority.

This class of algorithm is proven to converge on the NE for zero-sum games while being much quicker than linear programming (this actually has practical consequences for poker bots - see counterfactual regret).

Re: How to win at rock-paper-scissors

#42

Post by jerf »

jakobe wrote: post #6

TLDR: Players are more likely to pick the move that just won. To exploit that, the best strategy would probably be to always pick the move that wasn't played. Should give you an edge until your opponent notices your pattern :)

My wife and I play RPS to determine who does things like change a diaper when we're out and about and stuff. We tie a lot. Far, far more than chance would dictate. We can tie for 10 in a row, quite easily, and before you jump up about how this can happen by chance, we tie in sequences a lot, not just 1/3^10 times. We're trying to second guess each other.

Since humans are bad at random, "just play randomly" doesn't really work; humans don't have access to "random" to play that way. So you often do get into the sorts of strategies you mention, to compensate for this.

Besides, it's more fun this way.

Re: How to win at rock-paper-scissors

#43

Post by im3w1l »

valdiorn wrote: post #9

"The game of rock-paper-scissors exhibits collective cyclic motions which cannot be understood by the Nash equilibrium concept." This is complete bullshit. It is well understood that if your opponent is not playing an optimum strategy, you can gain an advantage by shifting away from the Nash equillibrium yourself. Example: Image a player that always picks "Paper". By shifting away from a random choice (Nash optimum)…

The only reason that sharks can pretend to be fish is that there is real fish out there, that make it an expected win to deviate from nash equilibrium. The fish are not a property of the game, but rather a property of the playerbase.

You can still use game theory to analyze the game though, if you "include" the player model into the game model.

"Rock-paper-fool goes like this. First a coin if flipped without you seeing If tails, you face a fool npc, who will play according to..., if head, you face a (homo economicus) player"

You could then try to analyze the equilibria of this game.

Re: How to win at rock-paper-scissors

#44

Post by Monkeyget »

andyhmltn wrote: post #40

Is it actually a 1/3 chance of winning though? Is that including a draw as a win? I've tried coding it up here: https://gist.github.com/andyhmltn/5c8c5734fe894db5aa02 Baring in mind it was a rushed 5 minute bodge-job so it could be wrong but there seems to be a massive skew towards draws After more tests with this code it seems there's an 11% chance of winning if it's random. Baring in mind javascripts RNG

There is an equiprobability of winning/losing/draw. Your code is erroneous. You do a +1 to the random number you generate giving you 1, 2, 3. You then use that number to access an array with indexes [0, 1, 2]. You end with an undefined which screws your results.

Re: How to win at rock-paper-scissors

#45

Post by bradleysmith »

jerf wrote: post #42
jakobe wrote: post #6

TLDR: Players are more likely to pick the move that just won. To exploit that, the best strategy would probably be to always pick the move that wasn't played. Should give you an edge until your opponent notices your pattern :)

My wife and I play RPS to determine who does things like change a diaper when we're out and about and stuff. We tie a lot . Far, far more than chance would dictate. We can tie for 10 in a row, quite easily, and before you jump up about how this can happen by chance, we tie in sequences a lot , not just 1/3^10 times. We're trying to second guess each other. Since humans are bad at random, "just play randomly" doesn't…

I played for years with a close friend in high school to decide who was driving or whatever, or just pass time. The better we knew each other (ie the longer we were playing) the more ties.

I believe there's a natural intuition for reading the expected movements of the other player, and have seen this reproduced (if anecdotally by only testing between myself and her, and not recording results) hundreds of times. Like you, 10-12 streaks of ties were not uncommon, certainly less common than statistics would seem to dictate.

somewhat relevant, there's a competitive league: http://www.usarps.com/

I always wanted to see it in vegas: https://www.youtube.com/watch?v=htX4T20t6lU

EDIT: clarity

Re: How to win at rock-paper-scissors

#46

Post by avaku »

danieltillett wrote: post #33
avaku wrote: post #20

>> What are your odds of winning rock-paper-scissors? Simple - one in three. At least, that's what chance predicts. I thought the chance of winning with no prior is 1/2... Otherwise, okay, you win with 1/3, your opponent wins with 1/3, and where is the other 1/3? :) I know what the article means, but they phrase is wrongly.

The odds are 1/3 win, 1/3 loss, 1/3 tie :)

Sorry I didn't know the rules properly :) Which combination is a tie? In my version someone always wins :) ha ha

Re: How to win at rock-paper-scissors

#47

Post by andyhmltn »

Monkeyget wrote: post #44
andyhmltn wrote: post #40

Is it actually a 1/3 chance of winning though? Is that including a draw as a win? I've tried coding it up here: https://gist.github.com/andyhmltn/5c8c5734fe894db5aa02 Baring in mind it was a rushed 5 minute bodge-job so it could be wrong but there seems to be a massive skew towards draws After more tests with this code it seems there's an 11% chance of winning if it's random. Baring in mind javascripts RNG

There is an equiprobability of winning/losing/draw. Your code is erroneous. You do a +1 to the random number you generate giving you 1, 2, 3. You then use that number to access an array with indexes [0, 1, 2]. You end with an undefined which screws your results.

Well if that was the case, then one result would always have 0. The indexes are used to find what something wins against
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