Wikipedia:Reference desk/Archives/Mathematics/2014 August 13

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August 13

Un-odd Rock Paper Scissors strategy

All of the version of RPS that I have seen have been with an odd number of states (2n+1) and beating n and losing to n. However I saw a t-shirt that apparently had RPS done with 4 states. I couldn't get the exact setup, but let's assume states A,B,C,D (A beats B &C, B beats C & D, C beats D and D beats A). What strategy would be superior here? I get the feeling that this may be a Markov chain problem, but I'd like help.Naraht (talk) 17:57, 13 August 2014 (UTC)

This sounds like a standard problem in game theory: Find the best (Nash equilibrium) mixed strategy for an imperfect-information game. (Not checking which of those are bluelinks — they should all be at least redirects and hopefully someone can help).

There's a classic text called The Compleat Strategyst: Being a Primer on the Theory of Games of Strategy. It is hosted by the Rand Corporation and you can view the complete text on-line; I won't give the deeplink in case that's somehow problematic, but you can trivially find it on Google.

It's not exactly up-to-date but I recommend reading it anyway; it's a lot of fun. It discusses a technique called the "pivot method", which I can't find that we have an article about. --Trovatore (talk) 18:32, 13 August 2014 (UTC)

[ec] This is a problem of finding a mixed-strategy Nash equilibrium. The payoff matrix is:

${\begin{array}{c|cccc}&A&B&C&D\\\hline A&0&1&1&-1\\B&-1&0&1&1\\C&-1&-1&0&1\\D&1&-1&-1&0\\\end{array}}$

Strategy C is dominated by B and so can be eliminated. The resulting table is:

${\begin{array}{c|ccc}&A&B&D\\\hline A&0&1&-1\\B&-1&0&1\\D&1&-1&0\\\end{array}}$

And this is the regular RPS table, so we know the equilibrium strategy is to choose A, B or D with probability 1/3 each. We can verify this is an equilibrium by showing that each of A,B and D has expected payoff 0 against it (if the solution wasn't obvious, we'd solve a system of linear equations where each supported strategy has payoff 0 against the equilibrium). -- Meni Rosenfeld (talk) 18:44, 13 August 2014 (UTC)

So the answer, is pretend it is normal RPS with C being thrown away. Thank you!Naraht (talk) 19:16, 13 August 2014 (UTC)

The following discussion is marked as answered. If you have a new comment, place it just below the box.

Yes, Meni's solution is very nice and is "compleat" for the problem as posed. I still recommend looking at The Compleat Strategyst, anyway. --Trovatore (talk) 19:18, 13 August 2014 (UTC)

It's not entirely clear what a "version of RPS" game is defined to be. Some basic requirements might be 1) It's a zero-sum gam, 2) the payoff values are 0, 1, or -1, 3) the best strategy for both players is to pick an option at random, 4) the average payoff with these strategies is 0, 5) the probability of there being a winner if both play follow the optimum strategy is greater than zero. Other possible rules might be 6) the payoff matrix is skew symmetric and 7) the payoff matrix is only 0 on the diagonal. Without 6 and 7 there is a game with two choices for each player, namely odds and evens. If you include 6 but not 7 then there is a game with four choices, but it has a 1 in 2 chance in ending with a tie instead of just 1 in 3 as in RPS. It seems pretty clear that with rules 1-7 there can be no game with and even number of options for each player. In such a game there are an odd number of payoff values in each row of the matrix, and since the nonzero values are either 1 or -1 the total for each row is either positive or negative. But if B'a strategy is to pick a choice at random then A has a better strategy in picking a row with a positive sum. The same argument shows that rules 1, 3, 4, 5 imply the row sums and column sums are all 0. --RDBury (talk) 14:03, 14 August 2014 (UTC)