In an urn containing $n$ white balls and $n$ black balls, we proceed to draw balls without replacement, until the urn is completely empty. The draws are equally likely at each draw. For every integer $k$ between $1$ and $2n$, we say that the integer $k$ is an equality index if, after drawing the first $k$ balls without replacement, there remain as many black balls as white balls in the urn. We note that the integer $2n$ is always an equality index. We denote by $M_n$ the random variable counting the number of equality indices $k$ between $1$ and $2n$. By using for example the events $B_i$: ``the integer $i$ is an equality index'', show that the variable $M_n$ has finite expectation equal to: $$\mathbb{E}(M_n) = \sum_{i=0}^{n-1} \frac{\binom{2i}{i} \cdot \binom{2n-2i}{n-i}}{\binom{2n}{n}}.$$
In an urn containing $n$ white balls and $n$ black balls, we proceed to draw balls without replacement, until the urn is completely empty. The draws are equally likely at each draw. For every integer $k$ between $1$ and $2n$, we say that the integer $k$ is an equality index if, after drawing the first $k$ balls without replacement, there remain as many black balls as white balls in the urn. We note that the integer $2n$ is always an equality index. We denote by $M_n$ the random variable counting the number of equality indices $k$ between $1$ and $2n$.
By using for example the events $B_i$: ``the integer $i$ is an equality index'', show that the variable $M_n$ has finite expectation equal to:
$$\mathbb{E}(M_n) = \sum_{i=0}^{n-1} \frac{\binom{2i}{i} \cdot \binom{2n-2i}{n-i}}{\binom{2n}{n}}.$$