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Here’s the puzzle/problem:

Let’s presume we are best friends. I own the house next to you. I make a gaming proposition along the following lines: You throw a coin over and over again. So long as it comes up heads, you keep throwing. When it comes up tails you stop and I pay you $$$ — depending on how many heads you threw.

If you threw a tails to start, you get $0.

If you threw a heads then a tails, you get $1.

If you threw 2 heads then a tails, you get $2.

If you threw 3 heads, you get $4.

If you threw 4 heads, you get $8.

If you threw 5 heads, you get $16.

And on and on. The payoff doubles everytime you add an extra heads.

Since this game consists entirely of me giving you money (from 0 dollars to ?? dollars) — you will have to buy an “entry” —

How much will you pay me to play this game (once)?

Note: This is not a loophole question, you will know when you have the right answer.

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closed as off-topic by Dap, Cesareo, José Carlos Santos, clathratus, dantopa Apr 28 at 18:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – Dap, Cesareo, José Carlos Santos, clathratus, dantopa
If this question can be reworded to fit the rules in the help center, please edit the question.

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If you compute the expected value, each outcome after an initial tails contributes $\frac 14$ to the sum, so the sum diverges. This would say I should pay any price to play the game, which is counterintuitive as the chance of winning very much money is very small. The usual resolution is that you cannot pay me more than some amount of money, so we should cut off the sum at that point. When we do that, the expected value is $\frac14$ times the base $2$ log of the amount of money you can pay me. If you can pay me $\$1,000,000$ I should pay $\$5$ or so to play.

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Hint $\;$ Evaluate the expected value $E$ of the game. We know the game is fair if $E=$"Money Paid for entering the game".

$$E=\frac{1}{2}\cdot 0+\frac{1}{4}\cdot 1+\frac{1}{8}\cdot 2+\frac{1}{16}\cdot 4+\ldots=\sum^\infty_{i=0}\frac{1}{4}=\infty$$

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