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Solution

Answer: 100

Let \(X\) be the number of colour switches in the full sequence of \(200\) draws. A switch can only happen between consecutive draws, so there are \[ 199 \] possible places where a colour change might occur. It helps to look at each gap separately. For the gap between draw \(i\) and draw \(i+1\), define \[ I_i = \begin{cases} 1, & \text{if the colour changes between draws } i \text{ and } i+1,\\ 0, & \text{otherwise.} \end{cases} \] Then the total number of switches is \[ X = I_1 + I_2 + \cdots + I_{199}. \] So the expected number of switches is \[ \mathbb{E}[X] = \mathbb{E}[I_1]+\mathbb{E}[I_2]+\cdots+\mathbb{E}[I_{199}]. \] Since each \(\mathbb{E}[I_i]\) is just the probability that the two consecutive balls have different colours, we only need to find that probability. Take any two consecutive positions in the random ordering. Because all orderings of the \(100\) red and \(100\) blue balls are equally likely, the probability that the two positions are different colours is \[ P(\text{different}) = P(RB)+P(BR). \] Now \[ P(RB)=\frac{100}{200}\cdot\frac{100}{199}, \] because the first ball is red with probability \(100/200\), and then there are still \(100\) blue balls among the remaining \(199\) balls. Similarly, \[ P(BR)=\frac{100}{200}\cdot\frac{100}{199}. \] So \[ P(\text{different}) = \frac{100}{200}\cdot\frac{100}{199} + \frac{100}{200}\cdot\frac{100}{199} = 2\cdot \frac{100}{200}\cdot\frac{100}{199}. \] Since \(\frac{100}{200}=\frac12\), this becomes \[ P(\text{different})=\frac{100}{199}. \] Therefore, \[ \mathbb{E}[I_i] = \frac{100}{199} \] for every \(i\), and hence \[ \mathbb{E}[X] = 199 \cdot \frac{100}{199} = 100. \] So the expected number of colour changes is \( 100. \)