100 Balls: Change of Color

You have an urn with 100 balls (50 black and 50 white). They are drawn one by one without replacement until the urn is empty. A ``color change'' occurs whenever two consecutive draws differ in color. Find the expected number of color changes.

Answer

Label the consecutive pairs of draws as $i=1,\dots,99$ . For each pair $$i$$ , define the indicator variable $$X_i=1$$ if the colors differ and $$0$$ otherwise. Then the total number of color changes is

X=X_1 + X_2 + \dots + X_{99}.

By linearity of expectation,

\mathbb{E}[X] = \sum_{i=1}^{99} \mathbb{E}[X_i] = 99\,\mathbb{E}[X_1].

Because of symmetry (see Comment 1 below), the probability that a given pair $$(i,i+1)$$ is a color change is the same for all $$i$$ . For one pair, the first draw is black with probability $\tfrac{50}{100}=\tfrac12$ and then the second draw is white with probability $\tfrac{50}{99}$ . Or the first draw is white ( $\tfrac{50}{100}=\tfrac12$ ) and the second is black ( $\tfrac{50}{99}$ ). Hence,

\mathbb{P}(\text{color change in a given pair}) = \tfrac12\cdot\tfrac{50}{99} \;+\; \tfrac12\cdot\tfrac{50}{99} = \tfrac{50}{99}.

Thus, $\mathbb{E}[X_1]=\tfrac{50}{99}$ . Multiplying by 99 yields

\mathbb{E}[X] = 99 \times \tfrac{50}{99} = \boxed{50}.

Notes and comments

Comment 1: My intuition for why all $$X_i$$ 's have the same distribution is due to the fact that we are talking about calculating the EV without any information at all w.r.t. the ordering. In that scenario it makes sense why.

Comment 2: Notice the similarily of this question with the two questions before.