Paper On, Dude!Math brain teasers require computations to solve.
The All-Portland Rock-Scissors-Paper League follows the canonical rules: at a given signal, both contestants reveal their choices. Rock beats scissors, scissors beats paper, and paper beats rock. However, for tournament play, the stakes are altered. Losers pay winners 3 chips if the latter wins with "rock", 2 points with "scissors", and 1 point with "paper". Each minute, everyone must find a new opponent, play one "throw", pay off, and get ready for the next switch.
The players with the most chips at the end of the evening get the big prizes: first place is a bag of Portland cement, second is a hedge trimmer, and third place is a choice of stationery. Also, any chips you've won are redeemable for drinks at the bar.
Knowing nothing about the other players' preference, mind games, etc., what's your best strategy (mixture of rock, paper, and scissors) for maintaining your balance of chips throughout the evening?
Hint(1) The answer is not 1:1:1; this loses to people who throw nothing but rocks.
(2) This is a "zero-sum" game: what one player wins, the other loses.
(3) The game is "symmetric": everyone is playing with the same information, the same options, and the same risks.
(4) The game is "open": there is no hidden information, such as secret agreements or rules that only one player knows.
AnswerRandomly throw paper 1/2 the time, rock 1/3 of the time, and scissors the remaining 1/6.
To avoid dredging through matrix math, I'll present a couple of basic principles of game theory.
(1) Playing the best strategy possible allows you to expose your overall strategy to your opponent without affecting your expected payoff. No matter what your opponent does in response, you'll still expect at least a given amount in the long run.
(2) Since the game is symmetric and open, every player will have the same best strategy -- and the same expected payoff if they use it.
(3) Since the game is zero-sum, that expected payoff must be zero.
(4) Given the payoff structure, the best grand (overall) strategy will involve all three choices.
With these principles, we can now look for a strategy that gives an expected payoff of zero, regardless of what the opponent does. Granted, if we *knew* the opponent was playing some other strategy, we could change to get an advantage -- but getting the best stable payoff includes the assumption that the opponent will take advantage of any exposed weakness.
So, we're looking to mix our "throws" in some useful proportion that will give us a stable, zero payoff. First, let's compare against someone throwing nothing but rocks. If we throw rock, it's a tie (no payoff). If we throw scissors, we lose 3. If we throw paper, we win 1. This yields the equation
E = 0*R -3*S +1*P
E = expected payoff
R = proportion of rocks
S = proportion of scissors
P = proportion of paper
Against an all-paper opponent:
E = -1*R + 2*S -0*P
Against someone running with scissors all the time:
E = 3*R + 0*S -1*P
We know that E=0; we can now solve the three equations in three variables. We get a stable proportion, rather than a fixed answer. P = 3*S; R = 2*S. Normalizing to a total probability of 1, we get
S = 1/6, R = 1/3, P = 1/2.
I realize that it's a bit counterintuitive that the choice with the best potential gain (rock) is not the largest proportion of the optimal mix. Rather, note that what *beats* rock gets the highest share, and what beats scissors gets the next highest.
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