Note that I had to use P( X_1 ) to find the probability that 2 specific players do not have a pair. Similarly, I'd need to use P( X_2 | X_1 ) to find the probability that 3 specific players do not have a pair. So let me define some more terms.
Let Y_1 = X_1, and then Y_i = X_i | Y_i-1 for i > 1. Then the probability of n specific players not having a pair would be P( Y_1 ) * P( Y_2) * ... * P( Y_n )
Let d_i = # of cards in the deck for a given player i, i.e. d_1 = 48, d_2 = 44, and so on. d_i = 48 - 4*(i-1) Let a_i = # of cards with no pairs remaining for a given player i (assuming all previous players also have no pairs), i.e. a_1 = 0, a_2 = 4, a_3 = 8, and so on. a_i = 4*(i-1) Let b_1 = # of cards whose pair remains in the deck under the same conditions, i.e. b_i + a_i = d_i
Then for 1 < n < 12 P( Y_n ) = [ (4 C 0 * b_n C 4) / (d_n C 4) ] * [ (b_n / b_n) * (b_n - 2 / b_n - 1 ) * ( b_n - 4 / b_n - 2) * ( b_n - 6 / b_n -3 ) ] + [ (a_n C 1 * b_n C 3) / (d_n C 4) ] * [ (b_n / b_n) * (b_n - 2 / b_n - 1 ) * ( b_n - 4 / b_n - 2) ] + [ (a_n C 2 * b_n C 2) / (d_n C 4) ] * [ (b_n / b_n) * (b_n - 2 / b_n - 1 ) ] + [ (a_n C 3 * b_n C 1) / (d_n C 4) ] * [ (b_n / b_n) ] + [ (a_n C 4 * b_n C 0) / (d_n C 4) ] * 1
and P( Y_12 ) = 1 because no cards with pairs will remain in the deck after the 11th player doesn't have a pair.
In particular if one wishes to find the probability that no player has a pair, you would need to calculate P(Y) = P( Y_1 ) * P( Y_2 ) * P( Y_3 ) * P( Y_4 ) * P( Y_5 ) * P( Y_6 ) * P( Y_7 ) * P( Y_8 ) * P( Y_9 ) * P( Y_10) * P( Y_11 ) * P( Y_12 )
Then to get the probability that at least 1 player has a pair, you would calculate 1 - P(Y)