Thu 4 February 2016

St Petersburg Paradox

Written by Hongjinn Park in Articles

A fair coin is flipped until a heads appears. You win $Y = 2^X$ dollars where $X$ is the number of flips to the first heads (including the heads). What is the expected value of this game?

$$ E[Y] = E[2^X] = \sum_{k=1}^\infty 2^k \left( \frac{1}{2} \right)^k = \sum_{k=1}^\infty 1 = \infty$$

How to make sense of this? Since there isn't an infinite amount of money, let's bound the payout to $2^{40}$ dollars which is more than a trillion.

Say that if it goes past $40$ tails in a row then he flees the country and you get $0$ dollars. So $Y = 2^X$ dollars if $X \le 40$ and $Y = 0$ if $X > 40$. Therefore

$$ E[Y] = \sum_{k=1}^{40} 2^k \left( \frac{1}{2} \right)^k + \sum_{k=41}^\infty 0 \cdot \left( \frac{1}{2} \right)^k= \sum_{k=1}^{40} 1 = 40 $$

If he doesn't flee the country and will pay you your trillion dollars then the expected value is $41$ dollars. Note that here $Y = 2^X$ dollars if $X \le 40$ and $Y = 2^{40}$ if $X > 40$.

$$ E[Y] = \sum_{k=1}^{40} 2^k \left( \frac{1}{2} \right)^k + \sum_{k=41}^{\infty} 2^{40} \left( \frac{1}{2} \right)^k = \sum_{k=1}^{40} 1 + 1 = 41 $$

since the geometric series $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... = 1$.



Articles

Personal notes I've written over the years.