**OR: Guaranteed way to fool gullible people into losing lots of money**
This "paradox" is quite simple to straighten out. Yes, with this method, your probability of *not* winning $1 halves every step (i.e. approaches zero exponentially). However, at the same time, the amount of money necessary to be able to continue after a loss *doubles* after every step.

Let's put it like this: the "profitability" of a game is the average amount you win each round. Let's call this G (for goodness). It is equal to the probability of winning P, times the amount to be won W, minus the probability of losing Q, times the amount to be lost L.

P*W - Q*L = G

In the case of a single coin toss, it's simple

½*1 - ½*1 = 0

i.e. you will, on average, break even.
What about 2 coin tosses in series, with the second being conducted only if the first is lost, with twice the amount being bet?

¾*1 - ¼*3 = 0

So it's not really any better. What about n coin tosses in series?

(2^{n}-1)/2^{n} * 1 - 1/2^{n} * (2^{n}-1) = 0

Again no difference. The increasingly tiny chance of losing is exactly cancelled out by the increasingly huge amount to be lost, while the amount to be won doesn't change. And that's the

*theoretical*,

asymptotical result. In reality, you will eventually (and sooner than you think) end up in a situation where you cannot double the bet because all your money is gone and nobody is stupid enough to loan you any more. And then you're

fucked quite thoroughly.

Basically, the belief that one "cannot lose" with this doubling strategy is based on a lack of understanding of how mathematical limits work and/or the skewed perception all too typical for gambling addicts where one only looks at the potential for winning and ignores the consequences of losing.