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I'm learning binomial distributions and I came across this problem:

Let r.v X be winning from a bet on a split in roulette and Y be be winnings from a bet on red color.

X = 17 (2/38 chance) and -1 (36/38 chance) Y = 1 (18/38 chance) and -1 (20/38 chance)

I calculated the variances for both: Var(X) = 0.4224 and Var(Y) = 0.9972

Now, if I were to play considerable amount of times, would I be better off only betting on a number or only betting on reds?

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It is not hard to verify that if you say bet a dollar a very small number of times, the low variance option gives you a better chance of coming out ahead. Now imagine playing a large number of times, so that the normal approximation is a good approximation of the total "winnings" (which are very likely to be negative). Then the higher variance option will give greater probability of non-negative winnings. I do not know how hard it would be to estimate the transition point. – André Nicolas Dec 13 '11 at 5:41
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I get a different value for the variance of $X$. $E(X^2)-(E(X))^2=\frac{289*2+1*36}{38}-\frac{1}{361}\approx 16.155$. Your expected return from many plays is the same in both cases, $\frac{1}{19}$ of your total wager. The variance will be much higher for the split.

If you play a considerable number of times, you are better off not playing.

Added: your chance of coming out ahead will depend upon the number of plays, but will generally be better with the split because of the higher variance. For example, if you play $16$ bets, you will be ahead on splits if you win at least once which is almost a 58% chance. Betting on red gives only a 10% chance to come out ahead. But if you lose on the splits you will lose 16. Only you can say which probability distribution is preferable-that is a psychology question, not a mathematics one.

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Oh right. Silly me. I understand but by choosing which one will maximize my chance of coming out ahead? I know the expectation is the same but the variances are different. Does this affect my decision? – user21128 Dec 13 '11 at 4:41

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