## Solution to the coin tossing problem

Submitted by plusadmin on January 1, 1998The chance that any fixed set of *k*+1 tosses gives *k* heads followed by a tail is:

In *n* tosses there are *n*-*k* possible starting points of such sequences. Therefore the expected number of wins is:

Using probability theory, one finds that the actual number of wins has what is called an *approximate Poisson distribution*. That is to say, the probability that you win exactly *r* pennies is given approximately by the *Poisson probability*:

Your average number of wins equals *lambda*. If *lambda* is big, then the number of wins will generally be big also, and if *lambda* is small then there will be only a small number of wins (perhaps 0).

The threshold between big and small occurs when *lambda* is a 'reasonable' number, say the value 1. When *lambda* = 1, then *k* is very close to log_{2}(*n*). In this threshold case the Poisson probabilities are given in the following table.

Substantially longer runs than log_{2}(*n*) are exceedingly unlikely, while substantially shorter runs are commonplace.

## Acknowledgements

This solution was written by Geoffrey Grimmett. You may also be interested in his article "What a coincidence!" elsewhere in this issue.