[elitegimp]

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Say you sit down at the wheel and agree to bet $1 38 times on the same number.

Given 0 and 00, Roulette pays 35-1 on a single number bet and you have a 1/38 shot of hitting your number each spin. So after 38 spins, you will on average have $36.

After 38 spins, which is more likely?

(a) You have more than $36.
(b) You have less than $36 (0 dollars).
(c) You are as likely to have 0 dollars as more than $36.

My gut says c. Can anyone confirm?

- Jason

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Here's my way of doing it -- the probability that you win on any given roll is 1/38, or 2.632% of the time. In order to end with $0, you need to be wrong all 38 times -- this happens (1-1/38)^38 = (37/38)^38 = 0.3630. So there is a 36.3% chance you lose all 38 rolls.

In order to end with $36, you need to be right exactly once, and wrong the other 37 times. This happens 38*(1/38)*(37/38)^37 = 0.3728 or 37.28% of the time.

Recap:
You go broke 36.3% of the time
You end with exactly $36 37.28% of the time.
Therefore you end with more than $36 1-.363-.3728=0.2642 or 26.42% of the time.

So you are most likely to have exactly $36, but you are more likely to have $0 than more than $36. This makes sense -- you are more likely to lose 38 times than you are to win at least twice.

Note: the 38 in
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38*(1/38)*(37/38)^37 = 0.3728 or 37.28% of the time.

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is the number of possible rolls for you to be right on. There is a (1/38)*(37/38)^37 chance that you are right on a specific roll (i.e. the third roll), but you just want the chance that any of the 38 are correct.