[Muted]

As others have mentioned this is indeed not true and is referred to as the Martingale betting system [0]. In the case of casino games Martingale still has negative EV. Your EV for any round n is (1-q^n)B-q^n(B^n-1) (chance of winning within n rounds times amount won minus chance of losing n rounds times amount lost) [1] which is B(1-(2q)^n) with B the initial betting amount and q the chance of losing. Since in any round your chance of losing is always slightly greater than 1/2, you will have a negative EV in any round you play, there is no number of rounds for which the EV becomes positive.

The standard deviation of this is fairly large though so that the chance of this happening is very small but nevertheless it isn't zero and what you lose is more than you can expect to win. As sleepychu mentioned, the amount you bet goes up exponentially ($1M after 20 rounds).

Aside: It is sometimes said that in order to do this "double when you lose" strategy, you need to e.g. "stick to red for the entire run". This is false. If I bet red-red-red-...-red-red I am just as likely to win/lose as when I bet red-black-odd-odd-even-...-red

[0] https://en.wikipedia.org/wiki/Martingale_%28betting_system%2...

[1] https://en.wikipedia.org/wiki/Expected_value