Two players drawing balls from an urn until the first red ball is selected

My try:

$$\frac{\binom{3}{1}}{\binom{10}{1}} + \frac{\binom{7}{2} \binom{3}{1}}{\binom{10}{3}} + \frac{\binom{7}{4} \binom{3}{1}}{\binom{10}{5}}+ \frac{\binom{7}{6} \binom{3}{1}}{\binom{10}{7}} $$

Another approach:

$$\frac{3}{10} +\frac{7 \cdot 6 \cdot 3}{10 \cdot 9 \cdot 8} + \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6} + \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 3}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4} $$

What is the correct approach ?

Solution 1:

Your first approach:

$ \small \displaystyle \binom{3}{1} / \binom{10}{1} + \binom{7}{2} \cdot \binom{3}{1} / \binom{10}{3} + \binom{7}{4} \cdot \binom{3}{1} / \binom{10}{5} + $ $ \small \displaystyle \binom{7}{6} \cdot \binom{3}{1} / \binom{10}{7}$

The mistake:

Take the second term for example. The numerator is number of ways of choosing $2$ black balls and $1$ red ball in $3$ picks and the red ball can be in any place. But you want the red ball in the third pick.

So one of the ways to fix your work is as follows:

$ \small \displaystyle \binom{3}{1} / \binom{10}{1} + \frac 13 \cdot \binom{7}{2} \cdot \binom{3}{1} / \binom{10}{3} + \frac 15 \cdot \binom{7}{4} \cdot \binom{3}{1} / \binom{10}{5} + $ $ \small \displaystyle \frac 17 \cdot \binom{7}{6} \cdot \binom{3}{1} / \binom{10}{7}$

Solution 2:

Your second approach is correct.

To correct the first approach, you must multiply the probability that a red ball has not been received by one of the players in the first $k - 1$ rounds by the probability that player $A$ receives the ball during the $k$th round.

$$\Pr(A~\text{selects first red ball}) = \frac{\dbinom{3}{1}}{\dbinom{10}{1}} + \frac{\dbinom{7}{2}}{\dbinom{10}{2}} \cdot \frac{\dbinom{3}{1}}{\dbinom{8}{1}} + \frac{\dbinom{7}{4}}{\dbinom{10}{4}} \cdot \frac{\dbinom{3}{1}}{\dbinom{6}{1}} + \frac{\dbinom{7}{6}}{\dbinom{10}{6}} \cdot \frac{\dbinom{3}{1}}{\dbinom{4}{1}}$$

As Math Lover indicated in the comments, the problem with your first approach is that you did not ensure that player $A$ took the first red ball during the last round, just that the first red ball was taken by the time player $A$ made a pick in the $k$th round.