# What is the expected number of runs of same color in a standard deck of cards?

Standard deck has $52$ cards, $26$ Red and $26$ Black. A run is a maximum contiguous block of cards, which has the same color.

Eg.

• $(R,B,R,B,...,R,B)$ has $52$ runs.
• $(R,R,R,...,R,B,B,B,...,B)$ has $2$ runs.

What is the expected number of runs in a shuffled deck of cards?

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Let $X_n$ be the color of the n'th card. For n<52, the n'th card is the end of a run if and only if $X_n\not=X_{n+1}$ and the last card in the pack is always the end of a run. So, the total number of runs is $$N=\sum_{n=1}^{51}1_{\{X_n\not=X_{n+1}\}}+1.$$ The expected number of runs is $$\mathbb{E}[N]=\sum_{n=1}^{51}\mathbb{P}(X_n\not=X_{n+1})+1.$$ Whatever colour the n'th card is, there are 51 remaining cards in the deck of which 26 of them are a different colour from $X_n$. So, $\mathbb{P}(X_n\not=X_{n+1})=26/51$, giving $$\mathbb{E}[N]=51 (26/51) + 1=27.$$
More generally, a deck with $2k$ cards, $k$ of each color, has expected number of runs $k+1$. – Michael Lugo Aug 19 '10 at 2:44