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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?

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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
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1 Answer 1

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?

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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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