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.

• Just remember before setting off to the casino with your doubling scheme that although you will win a little very often, when you loose you loose a LOT. Your expected return is still 0 on your even roulette wheel, and still negative on a wheel with one or two zeroes. – Tom Collinge Mar 11 '14 at 14:33
• In practice this strategy is foiled by casino bet limits. – MJD Mar 11 '14 at 14:51
• Congratulations, you just discovered the en.wikipedia.org/wiki/Martingale_(betting_system) While it will win in the long run, if there are no zer0s, it will win you only one betting unit. E.g. you will lose 1 + 2 + 4 + 8 (= 15) to win 16, or lose 31 to win 32, etc – Mawg Jan 11 '16 at 14:14

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: $$1 - \frac{1}{2}^{40} = 1 - \frac{1}{2}^{40}$$ • A careful reading of the Question will show you that winning (making a profit) does not depend on getting$40\$ reds in a row. – hardmath Apr 18 '18 at 16:25