# Probability of the Second Marble

### Solution 1

Assume the bag at hand contains $n\gt 0$ marbles. There are $n+1$ possibilities for the amount of black marbles, ranging from $0$ to $n;$ all $n+1$ are said to be equiprobable. Let $N_k$ denote the event that there were $k$ black marbles, to start with; let $B_1$ denote the event that the first drawn marble was black, $B_2$ is the event that the second marble was black.

Thus, the probability that the bag at hand carries $i$ black marbles is $\displaystyle P(N_i)=\frac{1}{n+1},$ for any $i=0,\ldots,n,$ and the probability that a randomly drawn marble from that bag is black is $\displaystyle P(B_1|N_i)=\frac{i}{n}.$ Then, by the Bayes theorem, the probability that the bag from which a black marble was drawn had $k$ black marbles is

\displaystyle\begin{align} P(N_k|B_1)&=\frac{P(B_1|N_k)P(N_k)}{\sum_{i=0}^{n}P(B_1|N_i)P(N_i)}=\frac{\displaystyle \frac{k}{n(n+1)}}{\displaystyle \sum_{i=0}^n\frac{i}{n(n+1)}}=\frac{k}{\sum_{i=0}^ni}\\ &=\frac{2k}{n(n+1)}. \end{align}

Now, if the event $N_k$ took place, the probability of the second marble being black is $\displaystyle \frac{k-1}{n-1}$ and the total probability of the second black marble is

\displaystyle\begin{align}P(B_2)&=\sum_{k=1}^n\frac{2k}{n(n+1)}\cdot\frac{k-1}{n-1}=\frac{2}{n(n+1)(n-1)}\sum_{k=1}^nk(k-1)\\ &=\frac{2}{n(n+1)(n-1)}\sum_{k=1}^n(k^2-k)\\ &=\frac{2}{n(n+1)(n-1)}\left(\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}\right)\\ &=\frac{1}{n-1}\left(\frac{2n+1}{3}-1\right)=\frac{1}{n-1}\cdot\frac{2(n-1)}{3}\\ &=\frac{2}{3}.\end{align}

Note that the probability is independent of $n.$

### Solution 2

There is a shortcut for evaluating $P(N_k|B_1).$ Since we do not know how many black marbles there were in the bag at hand, we should consider all $n+1$ possible configurations. The total amount of black marbles for all eventualities is $\displaystyle \sum_{i=0}^ni=\frac{n(n+1)}{2},$ making the probability of getting a black marble at all equal $\displaystyle \frac{2}{n(n+1)}$ and the probability of getting one from a bag with $k$ marbles $\displaystyle P(N_k|B_1)=\frac{2k}{n(n+1)}.$ From here we proceed as in the first solution.

### Solution 3

Let, $D_i$: the event that the $i^{th}$ draw is black, $B_k$: the event that there are $k$ black marbles out of a total of $n$.

$\displaystyle P(D_1)&=\sum_{k=0}^n P(D_1|B_k)P(B_k) =\frac{1}{n+1}\cdot\sum_{k=0}^n\frac{k}{n} \\ P(D_2\cap D_1)&=\sum_{k=0}^n P(D_2\cap D_1|B_k)P(B_k)=\frac{1}{n+1}\cdot\sum_{k=0}^n \frac{k}{n}\cdot\frac{k-1}{n-1}. \\ P(D_2|D_1)&=\frac{P(D_2\cap D_1)}{P(D_1)}=\frac{1}{n-1}\cdot\frac{\sum_{k=0}^n k(k-1)}{\sum_{k=0}^n k}\\ &=\frac{2}{3}.$

The answer brings to mind the Monty Hall problem. First grab two marbles blindfolded and then apply $n=2$ on those two. After the first marble is black, we know there are two black marbles or one, but not zero. The first one could've been the one in $BW$ or either one in $BB.$ So $P(B_1B_2)=\displaystyle \frac{2}{3}.$

### Acknowledgment

This is a slight modification of problem 318 from J. G. McLoughlin et all, Jim Totten's Problems of the Week, World Scientific, 2013. Solution 2 comes from the book.

Solution 3 is by Amit Itagi.