An EV paradox (cross-posted in probability)

[zac777]

I posted this in the probability forum, but it hasn't received much traffic so I thought I'd try it over here. I'm curious to hear how everyone would handle this situation.

Suppose you have the opportunity to participate in the following lottery based on coin flips:

One fair coin is flipped repeatedly until tails comes up for the first time. The payout is made according to how many times the coin was flipped. If tails came up on the Kth flip, than each participant wins 2^K dollars. That is, if the first flip is tails, you win $2. If tails doesn't come up until the 4th flip, you win 2^4 dollars = $16. If tails doesn't come up until the 10th flip, you win 2^10 dollars = $1024.

What's the most you would be willing to pay in order to participate in this lottery, and why?

- Zac

[mmbt0ne]

You should sell everything you own, and put every last penny on this game, because the expected result is infinite. Just another example of why you can't play solely on mathematical formulas.

[Louie Landale]

If lingering memories of Calculus are correct, this sequence approaches $2. But the minimum you can win is $2 and you can easily win more than that. And I guess that's the paradox.

$2.

- Louie

[EjnarPik]

Doesn't the sequence approach $4 (from below).

So anything below $4 should be fine.

Ejnar Pik, Southern-Docks.

[well]

The sequence approaches infinity.
EV = 1/2 * 2 + 1/4 * 4 + 1/8 * 8 + ... = 1 + 1 + ... = Infinity.
Look at the replies to the cross-post at the probability forum.

Regards.

[EjnarPik]

Again, I am not that good at math, but it seems wrong to add the sums. If heads comes, for instance, the third time, you only get the 8 dollars for the third time, not the 2 and 4 dollars from the first two flips.

Ejnar Pik, Southern-Docks.

[well]

For a discrete random variable X, the expectation value of f(X) is defined by

EV[f(X)] = Sum [f(x)P(X=x)], where the summation is taken over all possible outcomes of X.

We're interested in the payoff: f(x)=2^x
P(X=x) clearly equals 2^-x.

So EV[f(X)] = 2^1 * 2^-1 + 2^2 * 2^-2 + ... = 1 + 1 + ... = Infinity.

I hope this makes it clear for you.

[EjnarPik]

It doesn't, but thanks anyway. After contemplating a little, and making some strange calculations, I think you're right.

Ejnar Pik, Southern-Docks.