# Clustering And Distribution in An Unfair Coin Flip [closed]

A coin is biased such that lands heads up 40% of the time and tails the other 60% of the time.

You flip this coin eight times and three of those flips land heads up.

If the greatest clustering is defined as three heads in a row anywhere in the series of flips and the least clustering is an equal number of flips between those three heads, what is the expected clustering of those three heads in the series of flips?

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## closed as off-topic by Yes, Michael Albanese, T. Bongers, user91500, studiosusJul 10 '14 at 5:25

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If exactly 3 of the eight coins are heads, then we don't need to worry about the bias of the coin flip. The other 5 coins flipped up tails.

In order to assign an expected value to clustering we need a numerical measure. Let's assign it the count of tails between the first and last head. Thus the greatest cluster has a value of 0, the least a value of 5.

There's a total of $\frac{8!}{5!3!} = 56$ possible arrangements of the 5 tails and 3 heads.

To count the number of arrangements of clusters of size $n$, count the permutations of $n$ tails and the single head between the end heads, and multiply by the count of permutations of the remaining $5-n$ tails and the cluster itself. That's $(n+1)(6-n)$.

Hence $P(N=n) = \dfrac{(n+1)(6-n)}{56}$

$$E[N] = \sum_{n=0}^5 n \cdot P(N=n) = \frac{1}{56} \sum_{n=0}^5 n(n+1)(6-n)$$

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Thanks for answering. –  Gabe Apr 24 '14 at 5:14