Fallacy number one. Yes, the chances of losing on one specific bet are small. But keep in mind that you're only winning one dollar on each bet. The number of runs needed to break the bank (or even win a significant sum) is very large. And "small" * "very large" yields a number that is large enough for discomfort. Better to keep your day job.

Fallacy number two. Casinos have betting limits. Typically, the difference between the maximum and minimum bets are enough for 6 or 7 doubling runs. This is why it doesn't work. A run of 7 consecutive blacks happens once out of 64 runs.

Assume you have a stockpile of \$1023 (enough for 10 straight losses) and the wheel includes 0 & 00, so that your odds of winning are 18/38.

It's easy to figure the probability of losing, and that half of the expected value equation:

p = (1-(18/38))^10 = .001631 ev(l) = (1+2+4+8+...+512)*.001631 = 1023*.001631 = -1.668551

But, if the probability of NOT losing is 1-p, and the value of the win is \$1, the other half of the expected value equation would be:

1*.998369 = .998369

Thus:

ev = .998369 - 1.668551 = -.6701

This is your dollar expectation every time you begin your doubling series, ending with either your win of \$1 or your ruin. And it is about negative 67 cents.

On each bet, your rate of return is:

r = 18/38 - 20/38 = -0.05263 dollars

If you just bet on red 10 times for a dollar each time, you'd expect to gain 10r = -0.5263 dollars (i.e. lose about 53 cents).

Now of course you won't always play the whole 10 games, because you might win and cash out early. In fact, your chance of playing game number k is only (20/38)^(k-1), or for k > 1, a little better than 1/2^(k-1). But if you do play the game, you'll wager 2^(k-1) dollars on it, so its contribution to the overall expectation (amount won or lost times the probability of that happening) is just as high as the first game's, in fact it gets higher as k increases.

Since the later (possible) spins of the wheel actually weigh in against you a little heavier than the first, it's no surprise that your strategy is somewhat worse than 10 straight spins at \$1 each, i.e., the expected value is

E = -.6701 = 12.7 r < 10r

as you computed.

The strategy of doubling the amount of losing bets, ad infinitum, until one wins, is called the martingale betting strategy and was first proposed by a Frenchman a long time ago. Some claim that this lead to the name "martingale process" for certain stocastic processes but I've also heard other stories. This strategy does guarantee gains if you start with an infinite bankroll but then again, it's hard to invent a losing strategy assuming you start with an infinite amount of money. Also, if you already have an infinite amount of money, how can you win more?

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