Probability of paired combinations There are 8 friends: Alfred, Alex, Billy, Barbara, Connie, Chris, David and Donna.
The friends are paired up at random. What is the probability that at least two pairs will share the same first initial?
I believe the probability of having all four pairs share the same first initial is:
$(4/28)*(3/15)*(2/6)$ = .0095
where the numerator is the number of remaining possible pairs who share a first initial, and the denominator is the number of possible pairs left (8/6/4 choose 2).
How can you determine the probability of at least 2 pairs having the same initial?
 A: Here's a less advanced approach than whuber's:
The total number of possible "pairings" (ways to divide the 8 friends into four pairs) is
$$
\frac{\binom{8}{2} \binom{6}{2} \binom{4}{2}}{4!} = 105.
$$
This is because there are $\binom{8}{2}$ ways to choose the first pair, then $\binom{6}{2}$ ways to choose the second pair, then $\binom{4}{2}$ ways to choose the third pair, but because we don't distinguish between the "first pair", "second pair", "third pair" and "fourth pair", we divide by the $4!$ ways of labelling the four pairs as "first", "second", "third" and "fourth".
The number of "pairings" with the same initials in all 4 pairs is clearly 1.
There are no "pairings" with the same initials in exactly 3 pairs, because the last pair would then also have the same initials.
The number of "pairings" with the same initials in exactly 2 pairs is
$$
\binom{4}{2} \times 2 = 12,
$$
because there are $\binom{4}{2}$ ways of choosing the "matched" initials from $\{A,B,C,D\}$, and then 2 possibilities for the unmatched pairs.
The probability that at least two pairs have the same initial is the number of "pairings" in which at least two pairs have the same initial divided by the total number of possible "pairings", which, from the above, is
$$
\frac{1 + 12}{105} = \frac{13}{105} \approx 0.124.
$$
