probability in the casino I have a question regarding probability.
Lets assume that roulette has just red and black, no green and an equal amount of both.
Lets also assume we start each game with a bet of 1 dollar.
We also have enough dollars to be able to double our bet up to 40 times.
Now an example game is we bet our 1 dollar on black.
If we lose we bet 2 dollars on black, the total expenditure is now 3 dollars.
If we lose again we bet 4 dollars on black, the total expenditure is now 7 dollars.
If we win we get 8 dollars back. 1 of which is a profit.
Remember we can lose 40 times in a row before we cannot double up any longer. 
The odds of getting 2 reds in a row is 25% so getting 40 reds in a row would be very low. 
My question is what is the probability of making a profit after one entire game ( a game is where you keep doubling until you can or you profit)
and after 1000 games.
 A: After one game, you make a dollar unless you lose $40$ times in a row, which happens $2^{-40}$ of the time, about once in $10^{12}$ tries.  After $1000$ games, you have to win them all, so you will lose approximately $1000 \cdot 2^{-40}$ of the time, or about once in a billion tries.  This is not exact-it is actually smaller because some of the probability comes from two losses.  The chance of that is greater than zero, but very small.
A: The probability of you losing after an entire game is $\frac{1}{2}^{40}$, because you have to get 40 reds in a row to lose, and the probability of getting a red is $\frac{1}{2}$. The probability of making a profit is then $1-\frac{1}{2}^{40}$.
Starting with 1\$, the (maximum) amount of money you could lose after an entire game is $2^{40}-1$(lost 1\$ after round one, 3\$ after round two, 7\$ after round three...), if you win once, you earn 1\$. So playing this game you have a chance of $\frac{1}{2}^{40}$ to lose $2^{40}-1$\$, and a chance of $1-\frac{1}{2}^{40}$ to win 1\$, which means that if you repeated the full game forever (restocking to be able to play 40 rounds again), you would neither gain money nor lose money, because $\frac{1}{2}^{40}\times (2^{40}-1) = (1-\frac{1}{2}^{40})\times 1$, as worked out below:
Lose: $\frac{1}{2}^{40}\times (2^{40}-1) = \frac{1}{2}^{40}\times 2^{40} - \frac{1}{2}^{40} = 1 - \frac{1}{2}^{40}$
Win: $(1-\frac{1}{2}^{40})\times 1 = 1-\frac{1}{2}^{40}$
Lose = Win: $1 - \frac{1}{2}^{40} = 1 - \frac{1}{2}^{40}$
