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A gambler decides to keep betting on red at roulette, and stop as soon as she has won a total of 5 bets.

a. What is the probability she has to make exactly 8 bets before stopping?

b. What is the probability she has to make at least 9 bets?

For a, I was thinking you could just use the binomial distribution to get ${8\choose 5}(18/38)^5(20/38)^3$. Apparently this is wrong, why?

For b, I thought you could do 1 - P(she makes exactly 5 bets) - P(she makes 6 bets) - P(she makes 7 bets) - P(she makes 8 bets). Apparently this is wrong too. Not sure why I'm not understanding this.

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  • $\begingroup$ a. think of it this way. Let X="number of trials to the 5th success" ... does that suggest anything? $\endgroup$ – Glen_b -Reinstate Monica Sep 5 '17 at 6:19
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I think you should use a negative binomial distribution since your random variable$X$ is how many trials you need to make 5 successes. Not the other way, you have 8 trials then what is the probability of 5 successes.

For the negative binomial distribution:

$P(X=x)=\dbinom{x-1}{r-1} (1-p)^{x-r} p^r\\ P(X=8)=\dbinom{8-1}{5-1} (1-\frac{18}{38})^{8-5} \frac{18}{38}^5$

suppose a winning probability of a roulette is 18/38.

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Think of it this way:

a. Obviously, $8th$ bet is going to be the $5th$ win. Right? Now you can fix that and look for combinations of $4$ wins, $3$ losses out of the remaining $7$ bets; multiply each combination with the $5th$ win probability, i.e. $18/38$; and add each combination. This will be exactly equal to what @Deep North told you.

Example combinations: $WWWWLLLW$, $WWWLWLLW$, $WWWLLWLW$, $WWWLLLWW$, $WWLWLLWW$, $WWLLWLWW$, $WWLLLWWW$, and so on. Each trial is independent.

Can you do it for b. now?

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