St Petersburg "paradox" solved
September 22nd, 2021 at 4:18:06 PM
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The St Petersberg game is based on repeated coin flips. The payoff is $(2^n) where n is the number of flips. The game ends with the first tails.
So the EV is $2/2 + $4/4 + $8/8 + $16/16 + ... = $infinity
The variance is also infinite.
The question: As a function of your bankroll B, what is the max price P you should willing to pay to play the game?
Solution:
The max is where CE=0, which is where the log of your bankroll equals the expectation of the log of your post-game bankroll.
lg(B) = lg(B-P+2)/2 + lg(B-P+4)/4 + lg(B-P+8)/8 + ...
(the base of the logarithm doesn't matter here because it just multiplies both sides by the same constant, so I'm using base 2 for convenience)
B = 2^( lg(B-P+2)/2 + lg(B-P+4)/4 + lg(B-P+8)/8 + ...)
B = (B-P+2)^(1/2) * (B-P+4)^(1/4) * (B-P+8)^(1/8) * ...
Solve it numerically by plugging in various values of B-P
B-P=5 -> B=9.97
B-P=10 -> B=15.47
B-P=20 -> B=26.09
B-P=40 -> B=46.81
B-P=80 -> B=87.62
B-P=1000 -> B=1010.97
This means when your bankroll is $1010.97 you should be willing to pay up to $10.97 to play the St Petersburg game.
So the EV is $2/2 + $4/4 + $8/8 + $16/16 + ... = $infinity
The variance is also infinite.
The question: As a function of your bankroll B, what is the max price P you should willing to pay to play the game?
Solution:
The max is where CE=0, which is where the log of your bankroll equals the expectation of the log of your post-game bankroll.
lg(B) = lg(B-P+2)/2 + lg(B-P+4)/4 + lg(B-P+8)/8 + ...
(the base of the logarithm doesn't matter here because it just multiplies both sides by the same constant, so I'm using base 2 for convenience)
B = 2^( lg(B-P+2)/2 + lg(B-P+4)/4 + lg(B-P+8)/8 + ...)
B = (B-P+2)^(1/2) * (B-P+4)^(1/4) * (B-P+8)^(1/8) * ...
Solve it numerically by plugging in various values of B-P
B-P=5 -> B=9.97
B-P=10 -> B=15.47
B-P=20 -> B=26.09
B-P=40 -> B=46.81
B-P=80 -> B=87.62
B-P=1000 -> B=1010.97
This means when your bankroll is $1010.97 you should be willing to pay up to $10.97 to play the St Petersburg game.
September 22nd, 2021 at 4:30:17 PM
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Quote: scroogeThe St Petersberg game is based on repeated coin flips. The payoff is $(2^n) where n is the number of flips. The game ends with the first tails.
So the EV is $2/2 + $4/4 + $8/8 + $16/16 + ... = $infinity
The variance is also infinite.
The question: As a function of your bankroll B, what is the max price P you should willing to pay to play the game?
Solution:link to original post
The max is where CE=0, which is where the log of your bankroll equals the expectation of the log of your post-game bankroll.
lg(B) = lg(B-P+2)/2 + lg(B-P+4)/4 + lg(B-P+8)/8 + ...
(the base of the logarithm doesn't matter here because it just multiplies both sides by the same constant, so I'm using base 2 for convenience)
B = 2^( lg(B-P+2)/2 + lg(B-P+4)/4 + lg(B-P+8)/8 + ...)
B = (B-P+2)^(1/2) * (B-P+4)^(1/4) * (B-P+8)^(1/8) * ...
Solve it numerically by plugging in various values of B-P
B-P=5 -> B=9.97
B-P=10 -> B=15.47
B-P=20 -> B=26.09
B-P=40 -> B=46.81
B-P=80 -> B=87.62
B-P=1000 -> B=1010.97
This means when your bankroll is $1010.97 you should be willing to pay up to $10.97 to play the St Petersburg game.
An initial observation. It’s trivial to modify the payoffs of the paradox so that the log of the payout series is infinite. What should you pay then?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
September 22nd, 2021 at 4:46:52 PM
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If the payoffs are 2^(2n) instead of 2^n then that infinite product diverges. I think that means no matter how large your bankroll is, you can go all in. Even the worst case you still get $4 back. If the payoff is 0 for n=1 and 2^(2n) for n>1, that's hard.
September 22nd, 2021 at 8:21:27 PM
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I show the player should be indifferent to paying 4.36019% of his wealth to play.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
September 22nd, 2021 at 9:26:28 PM
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Quote: Wizardlink to original postI show the player should be indifferent to paying 4.36019% of his wealth to play.
Small hint:
As bankroll increases, the percentage of bankroll you're willing to pay should decrease.
September 23rd, 2021 at 2:15:53 AM
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Quote: scroogelink to original postIf the payoffs are 2^(2n) instead of 2^n then that infinite product diverges. I think that means no matter how large your bankroll is, you can go all in. Even the worst case you still get $4 back. If the payoff is 0 for n=1 and 2^(2n) for n>1, that's hard.
As to that, you could have $0 returned if flip 1 is tails but then keep everything else the same. Series should still diverge.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
September 23rd, 2021 at 2:57:52 AM
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When you go all in and there's a possibility of zero payoff, the first term in the infinite product will be zero.
September 23rd, 2021 at 6:40:46 PM
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Quote: scroogeSmall hint:
link to original postAs bankroll increases, the percentage of bankroll you're willing to pay should decrease.
You're absolutely right. I stand corrected.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
