Why is the expected value infinite? It seems fairly obvious that you will not make much money on such a game as we all know you can't flip a coin many times without tails showing up. To make infinite money in such a game you would need effectively "infinite luck" no?

If you get 2 dollars on the first head, 4 dollars on the second, etc., then the probability of receiving each dollar amount multiplied by the dollar amount is 1 dollar.

1/2 * 2 + 1/4 * 4 + 1/8 * 8 ... = 1 + 1 + 1 ... = ∞

Of course, the proof only works if dollars continue to have linearly increasing value to you as you keep accruing them. If their value to you logarithmically increases, the expected value is not infinite.

> as we all know you can't flip a coin many times without tails showing up.

We don't all know that. There's an ever-decreasing but nonzero probability that you will keep getting heads.

The probability that the next N flips will all be heads is small and gets smaller the larger N is.

But the probability that the next flip is heads is .5 no matter how many of the previous flips were heads :)

> There's an ever-decreasing but nonzero probability that you will keep getting heads.

But with enough flips, the odds getting heads on all of them comes close enough to zero that they are effectively zero.

"Effectively zero" times infinity is infinity. Infinity is like... really big.

Oh, I wish I could find it, but there's an excellent video by a mathematician explaining why an infinity doesn't work in your favor here. IIRC, it's because as the number of throws approach infinity, the odds also approach being infinitesimal, and you'll still lose.

This paradox is famously a case where the expected value of the game is infinite.

Consider the chance of winning at least 100 times. It's 1 over 2^100, and you win at least 2^100. So the small chance and the big reward "balance out", kind of.

But the reward is not limited there. Once you reach each and every "balance point", the next step is 50% chance for doubling the reward. If you currently have X, the value of continuing to play is 1.5X. This is independent of how big X is, or how unlikely you have made it this far (sorta a reverse Gambler's Fallacy). And it's why the expected value is infinite.

it's an example of how expected value doesn't tell the full story

https://en.wikipedia.org/wiki/Risk_of_ruin

It seems that the scenario Cowen put forth and the St. Petersburg paradox are not quite the same? In St. Petersburg (at least as Wikipedia describes it) you walk away with the accumulated winnings, whereas in the example put forth if you lose you walk away with nothing.

I was wondering because it would seem with Cowen's example the expected value is just $1 (ignoring the 51% edge). I suppose the 51% edge ends up making the expected value infinite, so the intuition lines up even though the exact details don't.

I see it as a clever and extreme variation of the same underlying fallacy: a) the favored outcome is slightly more probable (51%) b) the expected value of losing could be expressed as negative infinity

That SBF answered this (rhetorical) question that confidently and quick with not necessarily but what followed was simply something like look on the bright side is effectively imho parroting but at least he handled the verb eloquently. It would certainly take me like 20-30 seconds to process it.