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

Question

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?

Answer 1

Suppose $X_n$ denotes the color of the $n$th card in the shuffled deck.

Then note that the last card always denotes the end of a run. Other ends of runs are characterized by $X_n\ne X_{n+1}$ which indicates a run ending at $n$.

Note that $P(X_n\ne X_{n+1})=26/51$ (since once you fix a card, you can choose another card from remaining 51 out of which 26 will have a different color).

So summing up the indicators $X_n\ne X_{n+1}$ we get the number of runs -

$$\#\text{runs}=1+\sum_{n=1}^{51}\mathbb{I}_{X_n\ne X_{n+1}}.$$

So $$E(\#\text{runs})=1+\sum_{n=1}^{51}P(X_n\ne X_{n+1})=1+\sum_{n=1}^{51}26/51=27.$$

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