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.
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.
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: $1 - \frac{1}{2}^{40} = 1 - \frac{1}{2}^{40}$