From the book The Green-Eyed Dragons and Other Mathematical Monsters
- Consider the following game. You flip a coin until you get a tails. The number of dollars you win equals the number og coins you end up flipping. (So if you immediately get a tails, you win one dollar; if you get one heads before a tails, you win two dollars, etc.) What is the expectation value of your winnings?
- Play the same game, except now let the number of dollars you win be equal to $ 2^{n-1} $, where n is the number of coins you end up flipping. What is the expectation value of your winnings now? Does your answer make sense?
Part 1:
\[E = \sum _{n=1} ^{\infty} \frac{n}{2^n} = 2\] \[1/2 + 2/4 + 3/8 + 4/16 + ... = 1 + 1/2 + 1/4 + ... = 2\]Part 2:
\[E = \sum _{n=1} ^{\infty} \frac{2^{n-1}}{2^n} = n/2\]In this case, the expectation over infinity trails is far from the real experiment (not infinity). How much money would a person be willing to put up for the opportunity to play these N games?