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I was wondering what the odds would be if I guessed the same 5 numbers on roulette for 17 spins. (38 diff outcomes on a zero and double zero wheel) I think it is 13% but I don't know if I did that right I don't know if that was the percent to guess ALL 5 numbers or just 1 out of the 5 numbers. because you only need to hit one to win. I don't know if I am wording this correctly so if you have questions I can explain myself

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You mean that you choose 5 different numbers of the 38 that are on the roulette and don't change them for 17 spins? –  Georgey May 2 '13 at 10:48
    
"I was wondering what the odds wwould be..." The odds of what? of winning 17 times? of winning at least once? of never winning at all? Why not show us how you got 13%, and we'll see what we can say about your reasoning. –  Gerry Myerson May 2 '13 at 12:53

1 Answer 1

It is often useful when trying to calculate the probability of an event happening at least once to calculate instead the probability of the event not happening at all (let's call that probability $q$) and then taking $p=1-q$ to get the probability of the event happening at least once.

The probability of not winning on any particular spin is $\frac{38-5}{38}$. The probability of not winning 17 spins in a row is $(\frac{38-5}{38})^{17}$ since they are independent.

Thus, the probability of winning at least once is $1-(\frac{38-5}{38})^{17}\approx9.087\%$

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