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A roulette wheel has $21$ red numbers, $21$ black numbers, and $4$ zeros. A player places $20$ dollar bets on the red numbers. If the roulette spins a red number, the player get his $20$ dollars back, and wins another $20$ dollars. If the roulette does not spin a red number, the player loses his bet. The player sits down with $40$ dollars, and keeps on playing until he has no more money. How many spins can he expect to be able to play until he has no more money?

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closed as off-topic by Namaste, JMoravitz, José Carlos Santos, Davide Giraudo, Siong Thye Goh Aug 17 '17 at 17:20

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Let's first calculate the expected winnings of each round.

He will win \$$20$ with probability $21/(21+21+4)$ and win -\$$20$ (that is, lose \$$20$) with proability $(21+4)/(21+21+4)$. Therefore his expected winnings per round is: $$ E:=20\left(\frac{21}{21+21+4}\right) - 20\left(\frac{21+4}{21+21+4}\right) = -\frac{80}{46} = -\frac{40}{23}. $$ Given that he starts with \$$40$, this means that he can expect to play $40/(-E)=23$ rounds.

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  • $\begingroup$ I think you'll find that if you don't round in the first step that you'll get an answer of exactly $23$. $\endgroup$ – JMoravitz Aug 17 '17 at 17:31
  • $\begingroup$ As an aside, this answer agrees with a direct calculation as well but skips the necessity of figuring out the pdf and evaluating the sum. $\endgroup$ – JMoravitz Aug 17 '17 at 17:33
  • $\begingroup$ @JMoravitz Indeed! $\endgroup$ – John Griffin Aug 17 '17 at 17:41

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