# Petersburgh Paradox

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• #21629
pacopasa
Participant

I’m not much of a mathematician but I do like to go to Atlantic City, so this is my take on the paradox. I suspect this approach was previously taken by someone else and I would appreciate any feedback.
It appears to me that whatever fee one decides to pay, the odds of losing that fee(investment) is 50%.
What are the odds of breaking even?
If your fee is \$2, the odds are 50%
If your fee is \$4, the odds are 25%
If your fee is \$8, the odds are 12.5%
If your fee is \$16, the odds are 6.25%
If your fee is \$32, the odds are 3.125%
If your fee is \$64, the odds are 1.5% (rounded)
If your fee is \$128, the odds are .75%.
Therefore the odds of losing \$128 is approx 99%, very bad bet.

Since the odds of losing one’s fee is never in one’s favor a reasonable person would not play the game at \$2. If you don’t play there is a 100% chance of keeping \$2, if you play there is only a 50% chance of keeping \$2.
Playing at a higher fee is more unreasonable….
…. on the other hand if you pay only \$1 you still have a 50% chance of losing that dollar but you also have a 50% chance of DOUBLING your investment. The less you pay still keeps you at 50/50 but you can triple, quadruple… your investment.
So I think a reasonable person could play for less than \$2.
I think the utility theory comes into play only when you change the starting amount of the game. As a middle class cautious gambler I would be comfortable paying a \$20 fee for a game that started at \$20 and a \$50 fee for a game that started at \$100.
I would not be comfortable paying a \$100 fee for a game that started at \$100 or \$200.

#21630
pacopasa
Participant

If the game had a four sided die instead of a two sided coin and three sides were winners, one a loser, then at a \$2 game

a \$2 fee has a 75% chance of breaking even
a \$4 fee has a 56% chance of breaking even
an \$8 fee has a 42% chance of breaking even
a \$16 fee has a 32% chance of breaking even.

A reasonable person would pay \$4,
If I had one drink I would pay \$8
If I had two drinks I would pay \$16
If I had three drinks I would pay \$32
If I had four drinks I would pass out and not play.

The lesson: don’t drink or have 4 drinks.

#21631
bob.murphy.ancap
Participant

Hi pacopasa,

I’m not exactly following what you are saying here. If you have a specific question can you try rephrasing?

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