# Expected value for a roulette wheel problem?

A roulette wheel is divided into 38 pockets. 18 are numbered red pockets, 18 are numbered black pockets and 2 are green pockets. You can place bets on specific red or black pockets with odds of 35:1. That is, if you are correct you win 35 dollars (plus the dollar you bet), and if you are incorrect, you lose the dollar you bet. A ball is spun into the wheel, and we will assume it lands in all pockets with equal probability. Your friend has developed what he describes as a “foolproof system” for winning. It involves him betting on $1 on 7 red again and again. You try to discourage him, but he is insistent. What are his expected winnings after 36 spins of the wheel? Any help would be appreciated. I know it's the probability of red x value of red + probability of not red x value of not red, but don't know how to extend that onto 36 spins. ## 1 Answer The probability of getting a$7$red is$\frac{1}{38}$. Thus, the expectation of one game is: $$\mathbb{E} = 35\cdot\frac{1}{38} -1\cdot\frac{37}{38} = -\frac{2}{38}$$ Now, realize that what happens in one game does not influence any subsequent games. Thus, after$36$spins, your friend will lose \$ $\dfrac{72}{38}$.

Your friend seems to be suffering from Gambler's Fallacy.