AIME Problem 3

Let's represent the socks in the drawer as $AABBCCDDEE$ .

Monday: After drawing one sock , there are $9$ socks remaining of which $1$ matches, so there is $\frac{8}{9}$ chance that we make it to Tuesday without a match.

At this stage the remaining socks are (without loss of generality) $ABCCDDEE$ .

Tuesday: The following draws are possible (where $AB$ means $A$ and $B$ in either order):

• $AB$ : $\frac{1}{28}$
• $AC, AD, AE, BC, BD, BE$ : $\frac{2}{28}$ each
• $CC, DD, EE$ : $\frac{1}{28}$ each
• $CD, CE, DE$ : $\frac{4}{28}$ each

So if we make it to Wednesday without a pair being found, the possible states with their probabilities are (again, sorting the result without loss of generality)

• $CCDDEE$ : $\frac{1}{28}$
• $BCDDEE$ : $\frac{12}{28}$
• $ABCDDE$ : $\frac{12}{28}$

Wednesday :

• Case 1 $CCDDEE$ : $P(match) = \frac{3}{15}$
• Case 2 $BCDDEE$ : $P(match) = \frac{2}{15}$
• Case 3 $ABCDEE$ : $P(match) = \frac{1}{15}$

Combining all these cases , we have

$\begin{aligned} P &= \frac{8}{9} ( \frac{1}{28} \frac{3}{15} + \frac{12}{28} \frac{2}{15} + \frac{12}{28} \frac{1}{15} ) \\ &= \boxed{\frac{26}{315}} \end{aligned}$