## Comment Solution (Score 1) 167

Let r, p and s be the probabilities we play rock, paper and scissors on a given game respectively. It follows that,

r + p + s = 1

where r, p and s are elements of the interval [0,1]

For our opponent we will use a similar notation, only with uppercase letters,

R + P + S = 1

where R is an element of [1/2, 1] and p and s are elements of [0, 1/2]

Rearranging, we find that,

s = 1 - r - p

S = 1 - R - P

Now the expected gains, G, we get from any one game is given by,

G = AW + 0D + (-A)L = A(W - L)

where A is the amount exchanged between players and W, D and L is the probability of winning, drawing and losing respectively.

The probability of winning on single game, W, is given by the sum of the probabilities of each of the winning configurations occurring is

W = rS + pR + sP

Similarly the probability of losing on a game is

L = Rs + Pr + Sp

and so the expected gains is given by

G = A( r - R + P - p + 3(pR - Pr) )

Now note that our opponent must play rock at least half the time, so whenever we play scissors, at minimum we will lose half of the time and so it will never improve our gains. Therefore we should never play scissors. In the worst case scenario, our opponent is aware that we will never play scissors and that playing rock more than he has to will never improve his gains. This means

R = 1/2

G = A( r + (p-1)/2 + P(1-3r) )

Assuming the opponent is aware of this, he will do what he can to minimize G. This would mean that if (1-3r) is negative then he would maximize P (the only variable he controls). Similarly if (1-3r) is positive, he will minimize P. Finally if (1-3r) is 0, he cannot affect G.

In all of these scenarios, if we set r = 1/3 (or at least as close as possible to it) we will maximize G. This means that we should play rock a third of the time and paper the rest of the time. Our expected gains per game is G = A/6 so if A = $100 we should only be willing to pay $16.66 or less if we wish to make any gains per game.