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45 members have voted

ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5755
June 7th, 2023 at 6:50:43 PM permalink
Quote: Wizard

This one is probably not "easy," so all due apologies for putting it here.

Consider a game of rock-paper-scissors where:

• If rock beats scissors, scissors pays rock \$1
• If scissors beats paper, paper pays scissors \$2
• If paper beats rock, rock pays paper \$3
• No money changes hands on a tie

Assume two logicians are playing. What is the optimal strategy for each?

Let p be the probability of playing paper, and s the probability of playing scissors; (1 - p - s) is the probability of playing rock

Expected values of the strategy, if the opponent plays:
Rock: 3p - s
Paper: 2s - 3(1 - p - s) = 5s + 3p - 3
Scissors: (1 - p - s) - 2p = 1 - s - 3p

Assuming an equilibrium (i.e. all EVs = 0):
5s + 3p - 3 = 3p - s -> s = 1/2
1 - s - 3p = 3p - s -> p = 1/6
Check:
Rock = 3p - s = 0
Paper = 5s + 3p - 3 = 0
Scissors = 1 - s - 3p = 0

Each player rolls a 6-sided die, and plays scissors on 1-3, rock on 4-5, and paper on 6

Wizard
Joined: Oct 14, 2009
• Posts: 25471
June 7th, 2023 at 7:40:34 PM permalink
Quote: ThatDonGuy

Let p be the probability of playing paper, and s the probability of playing scissors; (1 - p - s) is the probability of playing rock

Expected values of the strategy, if the opponent plays:
Rock: 3p - s
Paper: 2s - 3(1 - p - s) = 5s + 3p - 3
Scissors: (1 - p - s) - 2p = 1 - s - 3p

Assuming an equilibrium (i.e. all EVs = 0):
5s + 3p - 3 = 3p - s -> s = 1/2
1 - s - 3p = 3p - s -> p = 1/6
Check:
Rock = 3p - s = 0
Paper = 5s + 3p - 3 = 0
Scissors = 1 - s - 3p = 0

Each player rolls a 6-sided die, and plays scissors on 1-3, rock on 4-5, and paper on 6

I agree! Not only the correct answer, but the solution was very simply and elegantly stated.

That one is beer worthy!
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5755
Thanks for this post from:
June 8th, 2023 at 5:57:27 AM permalink
I was a little surprised that this even had a solution, so I tried to see if there was a general solution, where Rock receives R if it wins, Paper receives P, and Scissors receives S, and in this case, there is:

The expected value based on each of the opponent's plays is:
Rock: q P - (1 - p - q) R = p R + q (P + R) - R
Paper: (1 - p - q) S - p P = S - p (P + S) - q S
Scissors: p R - q S
Assuming a Nash Equilibrium, EV(Rock) = EV(Scissors) when:
p R + q (P + R) - R = p R - q S
q = R / (R + P + S)
EV(Paper) = EV(Scissors) when:
S - p (P + S) - q S = p R - q S
p = S / (R + P + S)
1 - q - p = P / (R + P + S)

Check:
Rock: p R + q (P + R) - R = RS / (R + P + S) + R (P + R) / (R + P + S) - R = 0
Paper: S - p (P + S) - q S = S - (P + S) S / (R + P + S) - RS / (R + P + S) = 0
Scissors: RS / (R + P + S) - SR / (R + P + S) = 0

Play Rock, Paper, Scissors in the proportion S, R, P
i.e. play each in proportion to what the move that this move beats would collect if it was the winning move