# Probability of winning in roulette

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?

• 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? Feb 27, 2013 at 21:14
• If it's Russian rouletter, then you better not play it.
– mez
Feb 27, 2013 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.
The probability of winning n/N games at win probability p is: $$f(n) = p^n(1-p)^{N-n}\frac{N!}{n!(N-n)!}$$
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. • Is it because it is higher than probability of not making profit? Jan 5, 2016 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. Jan 5, 2016 at 15:52