What is the answer to this Probability question?
Dear DannyM, I interpret “exactly 2 pairs” to mean that when you get two pairs the six cards do not contain more than two of any one kind, nor do they consist of three pairs. Notice that this is different from the probability of getting exactly two pairs by making a five-card poker hand from six dealt cards, since a possible third pair would never be part of a five-card hand. Now consider possible sequences that result in exactly two pairs. It helps to define notation for this. Let u be an unmatched card (i.e., it does not match any previously dealt card), and p be a card that forms a new pair. The possible successful sequences can be written as follows: s1 = {u, p, u, p, u, u}, s2 = {u, p, u, u, p, u}, s3 = {u, p, u, u, u, p}, s4 = {u, u, p, p, u, u}, s5 = {u, u, p, u, p, u}, s6 = {u, u, p, u, u, p}, s7 = {u, u, u, p, p, u}, s8 = {u, u, u, p, u, p}, and s9 = {u, u, u, u, p, p}. If you believe the cards are shuffled well so that each of the remaining cards is equally likely to be the n
