1

I was in my Algebra 2 class today, and we're learning basic probability. Talking with my friend, he proposed this carnival game. The game begins with two dollars in the bowl, and every turn you flip a coin. If the result is heads, the carnival doubles the money in the bowl, and if tails, you take the money out of the bowl. The question is, what is the expected value you would get from this game?

Starting from the first flip, you have a $1/2$ chance of getting 2 dollars. Multiplying these together, you get 1. Measuring the next flip, you have a $1/4$ change of getting 4 dollars, which multiply to 1 as well. Continuing this, you end up with the infinite sequence $1 + 1 + 1 + 1...$

How is this possible? Would you really statistically pay infinite money to play a game where you have a chance $50$% chance of getting $2?

Kyranstar
  • 113
  • 2
    This seems relevant: math.stackexchange.com/questions/495053/st-petersburg-paradox – jdc May 13 '14 at 23:21
  • Real life is actually worse than the St. Petersburg paradox. Recall that "80% of everything is bad". But if you don't play, you die, much much sooner. – nomen May 13 '14 at 23:35

1 Answers1

4

This is known as the St. Petersburg paradox, and there are many proposed methods to resolve it.

One is that the carnival worker doesn't have infinite money, and if you truncate the payout at a given point the expected value falls dramatically (as the logarithm of the carny's money). Another is that if you think of it in terms of the usefulness you get out of the money, this is not simply a linear function of how much money you get. The difference between \$1 and \$2 prizes is "larger", in terms of what you can do with it, than the difference between a \$1,000,001 and a \$1,000,002 prize.

pokes
  • 131