Tjyylli

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.

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.

OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that.

The simplest somewhat-plausible way to handle #1 is to suppose that the value to you of your material assets is roughly proportional to the logarithm of their total value or, equivalently, that the value to you of your "next" dollar is roughly inversely proportional to your current net wealth. (There's some evidence that it actually grows a bit slower than that, but it'll do).

The way I'll handle #2 is to suppose that the probability that you pay up when I win $$X$ in the game is roughly proportional to $frac a{a+X}$ where $a$ is a fairly large number; so when $X$ is small you almost certainly pay and when $X$ is very large you almost certainly don't. (We should expect $a$ to be of the same order of magnitude as your net wealth in dollars; it's the size of payoff at which I think it's equally likely that you will or won't pay up.)

Then, if my net wealth before playing the game is $$w$ and your pay-or-don't parameter is $$a$, the "correct" price $$p$ is such that

$$frac12logfrac{w-p}w+sum_{k=0}^{infty}frac1{2^{k+2}}frac{a}{a+2^k}logfrac{w-p+2^k}w=0.$$

This isn't the sort of thing we should expect to be able to solve analytically; so far as I know there is no nice closed form for that sum. But we can do it numerically. Here are some specific results. (I cannot guarantee that I haven't messed up the calculations, though the numbers seem plausible enough to me.)

• Suppose your $a$ parameter is ($)1000000, so that your probability of paying up has reduced to 1/2 by the time you owe me a megabuck. And suppose my own net wealth happens also to be one megabuck. Then I should be willing to pay about $4.87.


• Suppose my net wealth remains at $1M but now $a$ is $$10^9$; even once you owe me a billion dollars you might well pay up. Then the amount I'm willing to pay increases to about ... $5.46.


• Suppose you remain a billionaire but I am much poorer, having only $1000 to my name. Then the amount I'm willing to pay goes down to $2.98. (It goes down because the poorer I am, the faster the value-to-me of a dollar decreases as I get more.)


• Suppose we're both poor, so I'm at $1000 and your $a$ parameter is the same. Then I am willing to pay about $2.39.


• Suppose I am a billionaire and you have infinitely much money. Then I am willing to pay about $12.80.

So I'm fairly comfortable answering the original question as follows: I am willing to pay somewhere between about $2 and about $13, depending on how wealthy we both are and how much I trust you to pay up even if doing so is painful for you.

This gambling problem is the famous St. Petersburg paradox. It is a paradox because

the wager has infinite expected value regardless of the amount paid to play. A rational agent should pay
any amount of money to play this game, and over many rounds of the game, they can expect to make a profit.

The one issue with this theoretical result is that it requires no upper limit on the possible winnings - if you make it through enough coin flips, you can win more money than the combined wealth of everyone on the planet. If we limit the lottery to a maximum payout of the world GDP (~$55 trillion), the expected value of the lottery is

around $45.

This analysis also assumes that each dollar won has equal utility, which is a measure of what a dollar means to you personally. Utility of money tends to flatten out for large amounts - a $10 million, for example, is a life-changing amount of money, but the next $10 million won't have as big an impact on your life as the first $10 million. So, $20 million is not really "worth" twice as much as $10 million. The "infinite wager" solution to the St. Petersburg paradox assumes constant utility of money, even though that's not a realistic model of how people behave.

I would pay you

All my money, and then be sad when I lost it all on the first flip

because

Let the amount I pay you be represented as C. After paying C, you have a 1/2 chance of winning 0, a 1/4 chance of winning 1, a 1/8 chance of winning 2, a 1/16 chance of winning 4 etc. Your expected payout would be $$-C + sum_{i=0}^infty frac{1}{2^{i+2}}2^i = -C + sum_{i=0}^infty frac{1}{4} = -C + infty$$ Theoretically if you played forever and ever, you would profit no matter how much you paid per game. Realistically, you would probably not profit no matter how much you paid.

There are several nuances to this question. First of all, it asks how much you are willing to pay, not what price is fair. Second, you have to understand, that even if a game is fair, that does not mean that It is reasonable to play it.

For example, if someone offers me a one in a million chance to win a million dollars for 1, I will take it. It seems reasonable. But if someone offered me a one in a million chance to win a billion dollars for 1000, I would not. What’s the difference? I am willing to lose a dollar for a small chance at success, but I am not willing to lose a thousand dollars for that same small chance, even if the reward is adjusted accordingly.

You have to factor in how much money you are willing to lose if the game does not go your way, especially if you only play once.

Since there is a 50% chance to lose your whole wager, the question becomes how much is the reward, but also how much are you willing to lose for it? If the reward was a million dollars, I would not wager 500,000 for a 50% chance to lose it, even if that is a “fair” game.

Realistically, I know the chance of getting a long streak of heads in a row is very low. Even though in 100 coin flips it’s not uncommon to have several streaks of 5 to 7 heads or tails in a row, playing the game only once means that the law of large numbers doesn’t apply.

For these reasons, I would not wager more than $1 for this game, because I don’t expect to see more than 3 heads in a row.

I will pay you

One dollar

to play this game, because

The amount you pay me has no bearing on what I pay you.

If there was a single throw there would be 50-50 chance of getting my dollar back.

But the game can go on, so there is better than 50-50 of getting my dollar back.

Your bankroll is the limit — your house against my $1.

To play once I would pay:

50 cents, since it is a doubling game and one heads is a dollar.