Probability of being up in roulette A player bets $\$1$ on a single number in a standard US roulette, that is, 38 possible numbers ($\frac{1}{38}$ chance of a win each game). A win pays 35 times the stake plus the stake returned, otherwise the stake is lost.
So, the expected loss per game is $\left(\frac{1}{38}\right)(35) + \left(37/38\right)(-1) = -\frac{2}{38}$ dollars,
and in 36 games $36\left(-\frac{2}{38}\right) = -1.89$ dollars.
But, the player is up within 35 games if he wins a single game, thus the probability of being up in 35 games is $1 - \left(\frac{37}{38}\right)^{35} = 0.607$. And even in 26 games, the probability of being up is still slightly greater than half.
This is perhaps surprising as it seems to suggest you can win at roulette if you play often enough. I'm assuming that that this result is offset by a very high variance, but wouldn't that also imply you could win big by winning multiple times? Can someone with a better statistics brain shed some light onto this problem, and extend my analyse? Thanks.


This is perhaps surprising as it seems to suggest you can win at
  roulette if you play often enough.

The most optimal way of "winning" at roulette is bold play. Bet your entire fortune or the amount desired to "end up," whichever is less.
However, in other sub-fair games when $p > 1/2$ then timid play is optimal.
