5 ms·

For anyone wondering about the verb to St. Petersburg paradox: The game is a simple 50/50 for a win or loss. You begin with one unit (say: $1). Every round yo

by dav_Oz 3y ago

For anyone wondering about the verb to St. Petersburg paradox: The game is a simple 50/50 for a win or loss. You begin with one unit (say: $1). Every round you don't lose the stake doubles. The game terminates if you lose. How much should you pay (expected value) to enter such game? Given the bank - offering you the gamble - has infinite resources the expected value is infinite. Well of course one way [0] (another more nuanced way is to question the simple axiom of maximizing the expected value [1]) to resolve the paradox is to limit the resources realistically, analogous to Martingale [2]. Given the world's GDP at ~ $100 trillion (or about N ~ 46 rounds) the expected value (beginning with $1) would be only about $(N+1)/2 so in this case $23,5. [0]https://en.wikipedia.org/wiki/St._Petersburg_paradox#Finite_St._Petersburg_lotteries https://en.wikipedia.org/wiki/St._Petersburg_paradox#Finite_... [Note: the numerical examples begin with $2] [1]https://plato.stanford.edu/entries/paradox-stpetersburg/ https://plato.stanford.edu/entries/paradox-stpetersburg/ [2]https://en.m.wikipedia.org/wiki/Martingale_(betting_system) https://en.m.wikipedia.org/wiki/Martingale_(betting_system)

saberience 3y ago
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?
- bitshiftfaced 3y ago
  > 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.
  - Kon-Peki 3y ago
    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 :)
  - JohnFen 3y ago
    > 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.
    - nwallin 3y ago
      "Effectively zero" times infinity is infinity. Infinity is like... really big.
      - JohnFen 3y ago
        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.
        traek 3y ago
        This paradox is famously a case where the expected value of the game is infinite.
- rytill 3y ago
  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.
- phunehehe0 3y ago
  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.
- tomjakubowski 3y ago
  it's an example of how expected value doesn't tell the full story https://en.wikipedia.org/wiki/Risk_of_ruin https://en.wikipedia.org/wiki/Risk_of_ruin
mindslight 3y ago
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). The 51% edge ends up making the expected value infinite, so the intuition lines up even though the exact details don't.
- dav_Oz 3y ago
  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.

by dav_Oz 3y ago

saberience 3y ago

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?

bitshiftfaced 3y ago
> 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.
- Kon-Peki 3y ago
  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 :)
- JohnFen 3y ago
  > 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.
  - nwallin 3y ago
    "Effectively zero" times infinity is infinity. Infinity is like... really big.
    - JohnFen 3y ago
      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.
      - traek 3y ago
        This paradox is famously a case where the expected value of the game is infinite.
rytill 3y ago
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.
phunehehe0 3y ago
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.
tomjakubowski 3y ago
it's an example of how expected value doesn't tell the full story https://en.wikipedia.org/wiki/Risk_of_ruin https://en.wikipedia.org/wiki/Risk_of_ruin

mindslight 3y ago

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). The 51% edge ends up making the expected value infinite, so the intuition lines up even though the exact details don't.

dav_Oz 3y ago
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.