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jackson24

  • 3 years ago

Q.If you bet on “red” at roulette, you have chance 18/38 of winning. (There will be more on roulette later in the course; for now, just treat it as a generic gambling game.) Suppose you make a sequence of independent bets on “red” at roulette, with the decision that you will stop playing once you’ve won 5 times. What is the chance that after 15 bets you are still playing?

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  1. CarlosGP
    • 3 years ago
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    You are still playing if in 15 bets you have won zero times or once or twice or three times or four times. That means:\[P(x <5)=\sum_{k=0}^{4}\left(\begin{matrix}15 \\ k\end{matrix}\right)\left( \frac{ 18 }{ 38 } \right)^k \left(\frac{ 20 }{38 }\right)^{15-k}\]

  2. CarlosGP
    • 3 years ago
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    Substracting P(x=5) from 1 gives you the probability of obtaining less than 5 but also more than 5 in 15 bets in case you went on playing. In any case, you have to substract from one the probability P(x>=5), which leads to the same result that I have proposed.

  3. Khlara
    • 3 years ago
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    0.0874

  4. CarlosGP
    • 3 years ago
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    @Khlara, that´s correct

  5. perl
    • 3 years ago
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    on calculator binomcdf(15, 18/38, 4)

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