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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%

Thanks in advance

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    $\begingroup$ Probability doesn't work that way. The wheel doesn't remember what past spins yielded....each spin is exactly like all the others (trusting that this is a fair and honest wheel of course). $\endgroup$ – lulu Dec 24 '16 at 14:04
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Presumably the roulette wheel has no memory. So it does not know if the past 10 rolls were green, red or whatever. So the chance of getting a green is always the same (1/15 in your case).

Note that the fact that the probability of 11 greens in a row is low means that the ensemble ("11 greens in a row") is unlikely, not that any of the 11 rolls ("2nd roll is green", "10th roll is green", ...) is any different from the others.

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).

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  • $\begingroup$ @Ralph J, our edits conflicted, I used your suggestions anyway, thank you $\endgroup$ – Rolazaro Azeveires Dec 24 '16 at 14:20

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