# Roulette Probability. Single spin or the past ones?

To make the calculations easier we'll just use a roulette with 15 number (0-14)

A roulette has 15 numbers (0-14) 0 is green 1-7 is red 8-15 is black

Green wins x14 of your bet Red and Black win x2 of your bet. The bet can be placed on either of the three colours.

So my question is: What's the best strategy to profit. My thinking was to bet on green after green hasn't rolled for 10 spins or more as then the probability is more than 50% because the chance of getting 11 not greens in a row is <50%

Actually, you know that the chance of not having 10 greens is more than 50% $$P(10)=(14/15)^{10}\approx50.2\%$$ but the chance of having 11 is less than 50% $$P(11)=P(10)*(14/15)=(14/15)^{11}\approx46.8\%$$ precisely because the 11th roll has a chance just like any other of not being a green (14/15).