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A roulette wheel has 21 red, 21 black numbers and 4 zeros, A player places $20 bets on red numbers. If the roulette wheel spins a red number the player gets his $20 back and wins another $20. If the roulette does not spin a red number the player loses his bet. the player sits down with $40 and keeps playing until he has no money. How many spins can he expect to be able to play until he has no money?
This problem has got me in a bit of a pickle. I know I have to use the binomial theory but I am getting confused on what you would consider the number of successful trials.
Eg. to go bust after 2 spins you would have to land on a colour other then red twice. This probability is 23/44 and each event is independent.
For 4 spins there are 2 ways to go bust. For 6, there are 4 ways to go bust and so one and so forth.
I am not sure if I am heading the right path but the probability of going bust after 4 attempts say is:
Pr(x=4) = 2!/(2-4)!*4! * (21/44)^4!(23/44)^(4-2)...
Am I heading down the right path?