Easy math puzzles
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45 members have voted
June 7th, 2023 at 6:50:43 PM
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Quote: WizardThis 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?
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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
June 7th, 2023 at 7:40:34 PM
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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
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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
June 8th, 2023 at 5:57:27 AM
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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
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
