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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.

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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    $\begingroup$ This is known as St Petersburg paradox.en.m.wikipedia.org/wiki/St._Petersburg_paradox $\endgroup$ – Tojrah Apr 22 at 17:49
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    $\begingroup$ All the answers (until now) seem to tackle the mathematical / probabilistic angle of the question. I really hope that the "correct" answers requires to take into account the bits about "best friends" and "owning house next to you". $\endgroup$ – Neo Apr 23 at 8:34
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    $\begingroup$ "How much will you pay me to play this game (once)?" I think it would be more clear to phrase this as, "How much would you be willing to pay me to play this game (once)?" Unless you have phrased it very intentionally. $\endgroup$ – jpmc26 Apr 23 at 19:48
  • $\begingroup$ The last line indicates that the asker intends this to be a math exercise, but many of the answers treat it as a subjective question rather than a puzzle. $\endgroup$ – noedne Apr 24 at 14:40
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    $\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful. $\endgroup$ – Rubio May 7 at 4:51

12 Answers 12

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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

$$\frac12\log\frac{w-p}w+\sum_{k=0}^{\infty}\frac1{2^{k+2}}\frac{a}{a+2^k}\log\frac{w-p+2^k}w=0.$$

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.

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  • $\begingroup$ While I feel like this is rather sound, I do feel like you've introduced bias at the end of the spectrum (a billionaire playing a 12 bucks game does not seem realistic). Adding a source for evidence where the value of the dollar grows slower and making it more realistic would make it an even stronger answer. Or are you saying the range of 2-13 would be narrower based on that relation? $\endgroup$ – PascalVKooten Apr 23 at 12:41
  • $\begingroup$ What do you mean by "bias" and by "realistic"? You should expect the "value" of this thing to grow at most logarithmically with your (and your counterparty's) wealth, because the contribution to your expected return in dollars is 25 cents from the case where you win \$1, 25 cents from the case where you win \$2, 25 cents from the case where you win \$4, etc. -- each doubling brings in a constant amount of extra expected money. $\endgroup$ – Gareth McCaughan Apr 23 at 14:48
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    $\begingroup$ In reality a billionaire might well be totally uninterested in playing this game at all, because presumably they got to be a billionaire by doing something and doing more of it probably brings them a much bigger return on time taken than playing this game could. $\endgroup$ – Gareth McCaughan Apr 23 at 14:49
  • $\begingroup$ Isn't this analysis kind of arbitrary and subjective? It doesn't really seem like the solution to a puzzle. $\endgroup$ – noedne Apr 24 at 14:42
  • $\begingroup$ I certainly don't imagine it's what OP had in mind. (Perhaps they wanted "an infinite amount" or something like that.) There are plenty of tweaks you could make -- different utility function, different way of handling the fact that very large payoffs probably never actually happen -- but I would expect the numbers to come out in something like the same sort of range for any plausible way of doing it. $\endgroup$ – Gareth McCaughan Apr 24 at 15:30
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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.

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    $\begingroup$ As a small addition to this (already good and complete) answer, I think it might be worth point out that for the paradox, you don't need the utility of money to be constant; you just need the utility of an unbounded amount of money to be unbounded. That is, no matter how much money you already have, there is always an amount of money that makes you 1% happier. Then simply replace each successive doubling of money with whatever number doubles the utility of the previous amount of money. $\endgroup$ – Mees de Vries Apr 22 at 21:39
  • $\begingroup$ For most people, they simply never reach the state where this understanding makes sense. However, if you have 100 million, an extra $10 million is basically not going to do anything close to improve your happiness by 10%. In fact, the first thought might be "Damn, how did I miss 10 mil? Do I need to fire my accountant(s) now?" or "Damn, how do I prevent getting taxed on this?" $\endgroup$ – Nelson Apr 24 at 4:41
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I would pay

Zero, because I dont need to convince you anything. You are offering me to toss and have a fixed value to give me. Why would I pay anything?

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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.

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I would not pay anything. I would not play. I would encourage you to not play. Are you doing okay? I'm willing to help you out of you need help. I would offer you a hug.

You are my best friend, and you live right next to me. Any outcome of this game that would be monetarily meaningful to either of us would also most likely be highly damaging to our friendship. Having a very good friend who lives next door is worth far more to me than whatever money I might manage to extract from you over a game like this if I did win big, and if I did not, that too would be damaging. Better to instead pay attention to the person who is my best friend. Why would you make an offer like this that is so potentially self-destructive?

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$16, Consider what else we know:

  1. You are my best friend
  2. The currency you're using is US dollars
  3. You own a home (namely the one next to mine)

What do these things imply?

  1. Because I'm your best friend, I won't want you to have to move away by you having to sell your house in order to pay me.
  2. We're in the USA
  3. The median price of a home in the USA is around $200,000

Conclusions: I'm probably not going to hold you to your agreement if you owe me more than half of your home's worth, because we're friends. So \$2^16 is \$65,536 is where we'll probably stop if I keep throwing heads. Therefore the expectation value is really 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1=16 so \$16, \$17 if I'm a jerk who is willing to take $131,072 from his best friend.

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  • $\begingroup$ This is the only answer I see that actually takes into account both the house and the friendship. I think winning a third of your best friend's net worth is a bit much though :) $\endgroup$ – JollyJoker Apr 24 at 8:50
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    $\begingroup$ @JollyJoker Please note that I'm not assuming that the value of his home is his net worth. We don't know anything else, and depending on the affluence of the neighborhood, it could be much less than a third of his home's value. I'm trying to put an upper bound on the value. $\endgroup$ – Mathaddict Apr 24 at 16:51
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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.

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  • $\begingroup$ Even at only investing 1000 dollars (everything you have), you realize you only profit from this in 0.09% of the cases. The gain to you, even if you would win, is not going to be worth losing everything. You might want to risk 1000 dollars when you have a billion, but putting everything on the line for a theoretical expected value of infinity is horrible. I suggest you investigate this further. $\endgroup$ – PascalVKooten Apr 23 at 12:33
  • $\begingroup$ @PascalVKooten The Wikipedia page on the St. Petersburg paradox has an interesting discussion about this which takes into account the utility of money. A rich person can safely wager more than a poor one, because they won't become bankrupt if they lose. For very poor individuals, though, it's actually the correct move to go into debt to play this game! $\endgroup$ – Nuclear Wang Apr 23 at 18:50
  • $\begingroup$ @NuclearWang Yes, I will give you that, it would hold if you have like 1 dollar or something. But already at 1000$ I would argue that this does not hold. You'd need a lot of consecutive heads for this to be worth it. A one-shot 0.09% chance of getting even (or better) is not going to be worth the money in this situation. $\endgroup$ – PascalVKooten Apr 23 at 18:56
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What I will pay:

$14.29

updated after fixing a bug in my code, old wrong value was

$14.54 Gosh that's not much of a difference

Because:

That is one penny less than I expect to get paid out on average per game, so that means that if I offer that price to the neighbor and the neighbor accepts, then I have a better than 50% chance of getting more money back, even if only by a penny.

Why do I think that?

Well, I am not 100% up on my statistics, so I wrote a program to simulate playing the game and then ran that simulation about 200,000,000,000 times. Averaging the payout each time. My code could totally be off as I coded it naively in JS(I don't know R or python for NumPy and didn't have time to learn enough to make this simulation). In fact I realize that just my heads or tails code, if not exactly 50%, could mess up my calculations. But that's my best shot.

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  • $\begingroup$ Here's my try of this algorithm in javascript : jsfiddle.net/15kwo2gf Average result found is around 5 or 6$ $\endgroup$ – F3L1X79 Apr 25 at 12:11
  • $\begingroup$ Thanks, I will put my code into a fiddle and post the link when I can. Among other things, I am using a different heads or tails but not very different. $\endgroup$ – Elliot_The_Curious Apr 25 at 13:24
  • $\begingroup$ If needed, I updated the code a bit: jsfiddle.net/s2b951nr/1 Obviously, the results tend to go higher the higher you input the number of tries but I limited it to 20000000 tries for the safety of my browser. It should be fun to use the existing population on earth (7.7 billion) for the tests. $\endgroup$ – F3L1X79 Apr 25 at 13:56
  • $\begingroup$ Here is the fiddle: link full disclosure: It includes a mistake I found thanks to having to look over my code again to see where our code is different(Thanks!). I left the mistake in, but I will run another simulation without the mistake and see what difference it makes. $\endgroup$ – Elliot_The_Curious Apr 25 at 14:26
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One dollar. It's a nice round amount.

This answer was previously deleted on the alleged basis that it "doesn't answer the question", with no opportunity for me to object to that claim, which I find quite rude. The question asks "What would you be willing to pay". This answers that question. The OP does not ask "What is the best bet, according to this objective metric", it simply asks what the reader would pay. If no metric is presented by which to evaluate amounts, then any arbitrary number is an acceptable answer.

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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.

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    $\begingroup$ Why not pay one penny instead of a whole dollar? Alternatively, if the game was offered for $1.01, you would reject the offer? $\endgroup$ – Nuclear Wang Apr 22 at 19:01
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    $\begingroup$ @NuclearWang please don't reveal the hidden answer. In answer to your question, because in the land of this puzzle I don't know if a penny exists. But I do know that the amount I stated exists. $\endgroup$ – Weather Vane Apr 22 at 19:03
  • $\begingroup$ Your answer to @NuclearWang ignores his/her second question. Replace 1.01 by 2 if you like it better, the point of the question is the same. $\endgroup$ – Arnaud Mortier Apr 22 at 19:27
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    $\begingroup$ "Note, this is not a loophole question". $\endgroup$ – SpqrTiang Apr 22 at 20:57
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    $\begingroup$ You're taking "How much will I pay" too literal. $\endgroup$ – PascalVKooten Apr 23 at 12:36
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Maybe I'm making a mistake in this reasoning, but I believe that the logical answer is:

I would pay 1 $

Explanation

Let $n$ be number of throws, following a risk-adjusted cost approach, the probability corrected money $M$ you win from playing is equal to the average of the prize (2 to the power of n consecutive throws) multiplied by the probability of getting that prize (n consecutive throws):

$ M = (2^n - 1) \times (1/2)^n \Leftrightarrow $

$ M = \frac{2^n - 1}{2^n}$

$M_{n \rightarrow \infty} = 1$, and therefore the average money earned by playing this game will tend to 1 $ (remember that this game has not cumulative earnings, in opposition to the St. Petersburg paradox)

Because I am playing with my best friend, I don't want him to lose money (but also not me) and, therefore, I will pay him the average prize that I will win so that the net loss/earnings between us is zero.

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  • $\begingroup$ Your equation doesn't seem to match the game description. According to your math, at n=1, M=0.5. According to the game description, at n=1, M=1. $\endgroup$ – glibdud Apr 24 at 12:51
  • $\begingroup$ @glibdud It does, but maybe I did not explain myself correctly. From a "risk analysis" point of view, I will have a "balanced" earning of 0.5 for 1 throw. For 2 throws, my "balanced earnings" will be 0.75, and so on. It's like saying that if I play once, my average earning will be 50 cent. I will edit my answer to try to make myself more clear. $\endgroup$ – cinico Apr 24 at 12:59
  • $\begingroup$ The other answers are not assuming cumulative payoff. If you have a set of mutually exclusive events, each with their own payoff, then your expected payoff is the sum of the probability of each event times their own respective payoff, even knowing that only one will occur. Your answer only works if the friend says "I will only pay you if you specifically get n heads then a tails, but not any other combination". Even though only one payoff will occur, the expected payoff is still infinity. $\endgroup$ – Bridgeburners May 7 at 19:21
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To play once I would pay:

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

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