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If you can bet $1$ dollar and win with a probability of $\dfrac{1}{38}$ in a game of roulette. What is the probability that you will make a profit (i.e. $> 105$ dollars) if you currently have $105$ dollars, and thus can make $105$ bets on the wheel?

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How much do you win? Usually you would get back $36$ when you win. You imply that you will bet each dollar once and not bet any of the winnings. Is that correct? – Ross Millikan Feb 27 '13 at 21:14
If it's Russian rouletter, then you better not play it. – mez Feb 27 '13 at 21:18

2 Answers 2

Hint: if my assumptions are correct, how many wins do you need to make a profit? It is easier to calculate the probability of $0,1,2,3$ as required and calculate the chance that you lose money. The answer is surprising.

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The probability of winning n/N games at win probability p is: $$f(n) = p^n(1-p)^{N-n}\frac{N!}{n!(N-n)!}$$

Your numbers for these would be 105 and 1/38 for N and p respectively.

If you assume that you win \$36 each time you win, you make a profit if you win 3 or more games. Your chance of making a profit is $1-F(0)-F(1)-F(2)=.524$ or 52.4%.

That actually does make sense but ask if you need to know why.

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