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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.

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  • $\begingroup$ 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. $\endgroup$ – Tom Collinge Mar 11 '14 at 14:33
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    $\begingroup$ In practice this strategy is foiled by casino bet limits. $\endgroup$ – MJD Mar 11 '14 at 14:51
  • $\begingroup$ 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 $\endgroup$ – Mawg Jan 11 '16 at 14:14
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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.

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

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  • $\begingroup$ A careful reading of the Question will show you that winning (making a profit) does not depend on getting $40$ reds in a row. $\endgroup$ – hardmath Apr 18 '18 at 16:25
  • $\begingroup$ @hardmath Oops, thanks for pointing that out, I had given the probability of losing instead of that of winning. $\endgroup$ – alvitawa Apr 19 '18 at 19:43

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