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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

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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Is it because it is higher than probability of not making profit? – SarpSTA Jan 5 at 15:46
This makes sense because you are more than likely (52.4%) to make a profit but the average profit if you win will be smaller than the average lose if you lose. It is analigous to rolling a die under the following conditions: if you get a 1, you lose \$100 dollars but other wise you get \$1. – kaine Jan 5 at 15:52
So it doesn't make sense because it boosts up the house edge. – SarpSTA Jan 5 at 15:53
@SarpSTA the house always has an edge so it is never monetarily smart to play roulette. Only gamble if: 1) the expected fun is worth much more than the expected losses, 2) the law of diminishing returns is reversed for some reason, 3) you are playing strip poker or similar. In my experience, none of these are true at a casino past the first bet (which has novelty). – kaine Jan 5 at 16:00

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