Talk:St. Petersburg paradox
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Wiki Education Foundation-supported course assignment[edit]
This article was the subject of a Wiki Education Foundation-supported course assignment, between 27 August 2021 and 19 December 2021. Further details are available on the course page. Student editor(s): Yibeiiiii. Peer reviewers: EyeOfTheUniverse, 5740 Grant L, Jimyzhu.
Above undated message substituted from Template:Dashboard.wikiedu.org assignment by PrimeBOT (talk) 10:06, 17 January 2022 (UTC)
About "The St. Petersburg Game" heading.[edit]
The fair price to pay for the contestant to enter the game should not be based on the expected payoff ($), but on the expected number of tosses until a win. The expected number of tosses until a win is given by: E(x) = (1/2)*1 + (1/4)*2 + (1/8)*3 + ... = ∑x⁄2x (summing the infinite terms). This is not a divergent sum and it has the result 2. The win amount for scoring on the 2nd toss is 4 dollars, so a fair price to pay to enter the game should be 4 dollars. Anyone disagree?
--Alin Jonas (talk) 14:45, 19 December 2021 (UTC)
- As the number of tosses gets higher, the prize increases and the probability decreases. So you end up with an infinitely small chance of winning an infinitely large amount of money, right? So doesn't that balance out to unity? Or, since the prize money is infinite, and infinitely unlikely, wouldn't an infinitessimally small amount of money be the correct price to enter? Which is more or less what most people would pay, I suspect.
- If I'm supposed to come up with the $2 stake from my own pocket, I wouldn't play. There's no profit in the odds. Of course traditional gambling adds up to a loss, too, which is why I don't do that either. 84.71.64.87 (talk) 19:19, 17 April 2022 (UTC)
- Since you will always win at least $2 in this game, an infinitessimally small price, or anything less than $2 will be a no-brainer for the contestant...
- But I think that the lesson to be learned from this 'paradox' is that there is no logical basis for equating fair stake and expected value, or any other mathematically defined quantity generally. Also, when dealing with infinities and divergent sums the whole idea is too far into theoretical-land anyway. First of all, this example assumes the game continues forever, but we humans live only for a short while. If the game was set to never stop, then sure, the average prize sum for all the so far participated contentasts will increase in linear fasion with no bound, but what does that matter to any of the contestants? It has already been suggested that the median value could be a better predictor, that will mean (since half of the contestans will score tail on the first toss) that $2 is fair enough. But wait, what if the win prize for tailing on the first toss was an infinite amount of money, and zero for all the other cases, would that mean than an infinite amount of money was a fair price to enter the contest? Like I said, when dealing with infinities... But you could limit the value, to say that 50 million was the win amount for tails, and zero otherwise. Would then 25 million be a good price to pay? Well, if you had significantly more than 25 millions I guess you could go for that, but you can see why this line of reasoning is kind of ridiculous. Alin Jonas (talk) 19:57, 22 October 2022 (UTC)
The solution is obvious[edit]
If someone plays the game 64 times he will on average win:
$2 32 times = $64, $4 16 times =$64 $8 8 times = $64, $16 4 times = 64 , $32 2 times = 64, $64 once = $64 (plus one more go)
So he can expect just over 6*64 = 384 dollars.
Your expected winnings is (log base 2)+1 times the number of times you play. While it's true this tends to infinity if you keep playing forever, you would require a huge bankroll to get started. If they set the cost at $11 you would need to play around 1024 games to reach profit on average. So you would require around $10,000 to minimize the probability of ruin. And it would take days.
It's not a very good casino game, or much of a paradox. 2A00:23C6:3081:4501:418A:247D:1682:B772 (talk) 00:49, 24 July 2023 (UTC)