Intuition says that if you play a fair game for a long time then on average you will be ahead half the time and behind half the time. Sounds reasonable doesn't it? But it couldn't be more wrong. The two most probable outcomes are that either you remain ahead or behind the whole time. Both are equally likely. Note that what I mean by a fair game is that you have an equal chance of winning and losing the same amount and the outcome of one game can in no way affect subsequent games.

So statistically if you get ahead or behind, you tend to stay there. It's not hard to show this mathematically but you can get a sense for why it's true with a simple example. Let's say the game involves tossing an unbiased coin and you win \(\$1\) on heads and lose \(\$1\) on tails. Now suppose you win on the first toss so you are up by one dollar. To go into negative territory you need to lose the next 2 games, anything else will keep you ahead. You have a \(3/4\) probability of staying ahead and only a \(1/4\) probability of falling behind in the next two games. With a larger lead the probability of falling behind becomes even less.

The probabilities can be made precise. If you play \(2n\) games, the probability that you are positive or flat for \(2k\) games is:

\[g_{2k}g_{2n-2k} = \frac{\binom{2k}{k}\binom{2n-2k}{n-k}}{2^{2n}}\]

where \(g_{2k}\) and \(g_{2n-2k}\) are the probabilities of being at zero after \(2k\) and \(2n-2k\) games respectively (see Getting Back to Zero). These probabilities are symmetric about \(k=n/2\) where there is a minimum. The maximum probabilities are at \(k=0\) and \(k=n\). The probabilities for \(n=10\) are shown below.

\(k\) | Probability |
---|---|

0 | 0.176197 |

1 | 0.092735 |

2 | 0.073643 |

3 | 0.065460 |

4 | 0.061684 |

5 | 0.060562 |

6 | 0.061684 |

7 | 0.065460 |

8 | 0.073643 |

9 | 0.092735 |

10 | 0.176197 |

The \(k=0\) probability means there is a \(17.6\) percent chance that you are either behind or at zero the whole time. The \(k=10\) probability means that there is a \(17.6\) percent chance that you are either ahead or at zero the whole time. There is only a \(6\) percent probability that you spend an equal amount of time ahead and behind.

For large \(n\) the probabilities can be approximated by the continuous arcsine distribution which has probability density function:

\[f(x) = \frac{1}{\pi\sqrt{x(1-x)}}\]

and the cumulative distribution:

\[F(x) = \frac{2}{\pi}\arcsin(\sqrt{x})\]

where the range of \(x\) is \(0 \leq x \leq 1\). The probability that \(k \leq xn\) is approximately equal to \(F(x)\). As the number of games goes to infinity the approximation becomes exact and is known as the Lévy arcsine law for Brownian motion.

© 2010-2012 Stefan Hollos and Richard Hollos

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