# Roulette probability

In a roulette game, a player bet only on the Red color (we are given that the probability of Red is 18/38). The player will stop playing after at least one winning or losing 11 rounds in a row. On the first game he bets \$1, in the 2nd game he bets \$2, and in the $i$th game he bets $2^{i-1}$. If he won, he will get the money that he bet on, else he lose the money he bet on. (meaning that if he won/lost a game he can start a new round and bet from \$1 again) 1. How much he will earn if he wins the$i$th game? 2. What is the expected value of the number of rounds? - The house always wins. See: amazon.com/The-Eudaemonic-Pie-Thomas-Bass/dp/0595142362/… – Fred Kline Jun 21 '13 at 0:04 Which course is this homework for? – Did Jun 21 '13 at 17:07 add comment ## 1 Answer Hint: try calculating the outcome of winning the first, second, third, fourth games. The pattern should be clear, then prove it. For the expected value, you need the chance he loses$11$games in a row. What is that? - i edited the question, i hope now it's more clear, thanks ross! – guestCS Jun 20 '13 at 22:19 @guestCS: it was clear before. Have you been able to make any progress given the hint? – Ross Millikan Jun 20 '13 at 22:31 with the first question yes, the 2nd no progress – guestCS Jun 20 '13 at 22:39 If your probability for winning is$p$, the probability for losing is$q = 1-p$. Independent random events(such as your roulette example or throwing a dice), allow you to multiply the probabilities. E.g. scoring 6 twice in a row with a dice has a probability of one out of six, times one out of six:$\frac{1}{6}*\frac{1}{6}=\frac{1}{36}\$. –  Fab Jun 20 '13 at 22:48
the chance he lost a bet is 20/38, if he loses 11 rounds in a row that's 11*(20/38) so the expected value is ~5.798? –  guestCS Jun 20 '13 at 22:58
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