How much money?

[liniarc]

This is a question I came up with and asked my friends who couldn't give me a satisfactory answer.

A mysterious person tells you to pick any positive integer, x.
He picks a random whole number smaller than x.
If you guess it, he will give you that $x^2.

What number should you choose for x, why?

Assume his choice is truly random.
Assume the quantity of money you receive will not cause inflation or any economic problem in any way.
You may only pick once.

[notzeb]

I pick x = 1000, because my utility curve starts flattening out around a million dollars.

[Laguana]

And what if he charges you $x to enter?

Personally I'd be inclined to pick x ~= 2, but I'm not much of a gambler.

[t1mm01994]

Let's see. EV = 1/(x-1) * sum 1 to x-1 i^2.
My gut tells me this is rising over the naturals, and handwaviness gets me there, too, but neither are accepted proofs. So, take it away!

[Magnanimous]

Expected payoff, if I'm interpreting the question correctly. So yes, the EV does approach infinity.

[jaap]

Expected payoff, if I'm interpreting the question correctly. So yes, the EV does approach infinity.

I interpret the problem differently from you and t1mm0. The payout is the square of x, not the square of the number he chooses.

If you decide to play the game with x=2, then he is certain to choose 1, which you are certain to guess, and then you win $ x^2 = $4.
If you decide to play the game with x=3, then he choose 1 or 2, so you have 50% probability of guessing correctly. If you win you get $9, if you lose 0, so the expected payout is $ 4.50.
If you decide to play the game with x=4, then he choose 1, 2, or 3, so you have 33% probability of guessing correctly. If you win you get $16, if you lose 0, so the expected payout is 16/3 = $ 5.33.
...

In general, the expected payout for game x is x^2 / (x-1) = x + x/(x-1). This is still obviously increasing if x increases.

If the entry fee is x, then the expected payout is just x/(x-1). This is decreasing with x, so you should choose the smallest valid x possible, i.e. x=2 for a payout of $2. I assume you are not allowed to choose x=1 because then he has no numbers to pick from.

[mfb]

For large numbers, it does not matter whether you get the square of x or the square of the number you pick. In the latter case, you pick x-1 and get (x-1)^2 (with probability 1/(x-1)) which is nearly as good as x^2.

As notzeb already mentioned, the expectation value is not the interesting thing here - nearly nobody would prefer a 1/900 chance to get 1 billion (and 899/900 to get 0) over a 100% chance to get 1 million.
With sqrt as utility function, all x are (nearly) the same. If it is growing slower, smaller x are better, if it is growing faster, larger x are better. If it has some strange shape, the general case is to optimize f(x^2)/(x-1) or f((x-1)^2)/(x-1).

[liniarc]

The problem we ran in to is similar to that of mfb. In order to maximize your expected payoff, you would want to pick the largest number possible(infinity), however, in doing that, you're making it unlikely that you'll get any money at all. And at what point would it be better to not worry about getting more money.

Also, the payout is not of the number he chooses, it's of the number you picked initially.
Simply speaking, you have a 1/x chance of getting x^2 dollars

[Moonbeam]

Simply speaking, you have a 1/x chance of getting x^2 dollars

..... shouldn't that be 1/(x-1) chance ??? (assuming that "a random whole number smaller than x" cannot be 0).

[Tnarg]

This is very much like the TV show deal or no deal. It’s all to do with the difference between mean and medium and what is a life changing amount. Of course if you play these game lots then you play for the mean and pick a high number. But if you’re only playing this once then things are different. Let’s say you get to play this game once and one $ in the game is worth 1,000,000,000 US$ now I don’t know about you but I would pick 2.

It’s kind of like insurance in a way, insurance is bet weighted in favour of the insurance company which is why I don’t insure my phone it’s a bad bet I have loads of small items I don’t insure so over a large number of bets the medium ~= the mean. But I do insure my house I only have one of them so I 'play' for the best medium.

Edit:
I guess I have not anser the questoion I guess I would be happy with any value where the pay out is below the point where I see money stops having linner worth, Which for me would be about the amount I have left on my morgage. so any number below that would be for me. But I dont think this is a maths question. it has a lot of maths in you need to get to grips with to understand the problem the the final anser is more about the person being asked.

[rigwarl]

This is a question I came up with and asked my friends who couldn't give me a satisfactory answer.

Really any answer the person wants is valid. Your EV is higher if you choose a higher number but so is your risk. Personally, as a comfortably employed but not rich person, I would choose 1000$ and invest the million if I hit that 1/1000 chance of winning.

[xkcd follower]

The best bet would be to bet about 3 dollars. Thus, you have a 50% chance of winning if the better is not bias and choose to try and decrease your odds more. The result would be you winning at least something over nothing.

[ibgdude]

The best bet would be to bet about 3 dollars. Thus, you have a 50% chance of winning if the better is not bias and choose to try and decrease your odds more. The result would be you winning at least something over nothing.

I guess this is just a matter of preference, but I'd rather have a 1/1000 chance of winning a million dollars than a 1/2 chance of winning $9. Why do you think you prefer the latter?

[Adacore]

I suspect it's a combination of risk aversion and monetary utility. It's really quite an interesting subject (and, as Tnarg said, the sole reason shows like Deal Or No Deal make good television). The fact there is no stake involved, merely an opportunity cost of losing out on the guaranteed $4 if you say '2' means it is less complex than some similar problems, though - I feel like most people would be prepared to take quite low odds of winning a large amount of money if the stake is small and constant, whereas if the stake increases then they would be less inclined to do so.

For example, if instead you had to pay the man $x, for a 1/(x-1) chance of winning (2x)^2$, I think people would go for considerably lower odds, because - while the expected value is higher - they stand to lose more if the long-shot bet doesn't come off.

I would probably choose a number around 1000 (payout $1000000, expected value ~$1001) for the original scenario, but would probably stick closer to 100 (payout ~$40000, expected value ~$304) for my suggested game.