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.
Best Answer
The following is very informal, since it does not take into account the (slim) possibility of multiple wins.
Note that if the player is not "up" in $35$ games, then she is down by $35$ dollars. Note also that if the player has only won once in $35$ games, she has lost in $34$, and therefore her net gain is $1$ dollar.
Should one be happy with a probability $0.607$ of winning $1$ dollar, and a $0.393$ probability of losing $35$ dollars? (Note again, it is not as bad as that, because of the possibility of multiple wins.)
But the logic of a basic expectation calculation is inexorable: if you play $n$ games, your expected net "gain" is $-\dfrac{2n}{38}$. On average, the more you play, the more you lose.
Remark: The expected loss per game is not terribly high. Contrast this with lotteries, that often return on average about $40$ cents per dollar bet.