Imagine a game where you start with \$100 and toss a coin repeatedly. If it's heads - you lose \$1, if its tails - you double your money. Game ends when you lose all the money. Given infinite amount of tries is this game guaranteed to end?

I know that the game could end right at the start, with probability of 2^-100. I'm more interested in what happens when the number of tosses is very large (i.e. tending towards infinity)

  • $\begingroup$ You are very likely to win an infinite amount: the question would be more interesting if you started with $\$1$. $\endgroup$
    – Henry
    Nov 25, 2014 at 7:24
  • $\begingroup$ Empirically the probability of winning an infinite amount starting with $\$1$ is about $0.29564949$, and of winning an infinite amount starting with $\$100$ is about $1-2^{-99.1}$ $\endgroup$
    – Henry
    Nov 25, 2014 at 8:33

1 Answer 1


The game where one looses \$1 when the toss is heads and one wins \$2 when the toss is tails corresponds to a random walk with positive bias. These have positive probability to never hit \$0. The game where one looses \$1 when the toss is heads and one doubles one's fortune when the toss is tails, is at least as favorable as the first one at every step such that the fortune at least \$2, hence it never hits \$0 with positive probability.

Starting from some positive fortune \$$n$, the first game has probability $z^{n-1}$ to hit \$1, where $z$ deotes the positive root of $z^2+z=1$, hence the second game starting from \$100 has probability at most $z^{99}$ to hit \$0.

Numerically, $z=\frac12(\sqrt5-1)\approx0.618$ and the probability for the first game is $z^{99}\approx2\cdot10^{-21}$. The probability for the second game is smaller hence, for all practical purposes, one can consider that one never goes broke in the second game.

More refined estimates are that the probability $\theta_n$ to hit \$0 in the second game starting from some fortune \$$n$ is such that the sequence $(2^n\theta_n)$ increases from $1$ to some finite limit $c$ when $n\to\infty$, where $$1\lt c\leqslant\prod_{k\geqslant1}\frac1{1-\frac12z^k}\approx2.466.$$ In particular, the probability to get broke in the second game starting from \$$n$ when $n$ is large is best approximated by $$\frac{c}{2^n},$$ and, for $n=100$, $$7.9\cdot10^{-31}\leqslant\theta_{100}\leqslant2.0\cdot10^{-30}.$$

  • $\begingroup$ Could you possibly direct me to the source of the values for $z$? Thanks. $\endgroup$
    – alonso s
    Nov 25, 2014 at 9:05
  • $\begingroup$ @alonsos You mean, why $z^2+z=1$? This is classical Markov chains theory and rather peripheral to the question at hand, hence I suggest that you ask this as a separate question (that is, if this was not already asked ten times on the site)? $\endgroup$
    – Did
    Nov 25, 2014 at 9:08
  • $\begingroup$ If the game goes on for an infinite amount of time, aren't you guaranteed to hit zero? $\endgroup$
    – robson
    Dec 31, 2022 at 18:48

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