LogFAQs > #938648480

That's similar to another estimation method I came up with while in the shower.

Imagine dealing out both cards of a pair randomly to any of the 12 people. The # of ways to deal the pair to different people is 11*12, whereas the number of ways to deal out the pair at all is 12*12, so the odds of any particular pair being dealt to 2 different players is 11/12. Then just do this for all pairs by taking (11/12)^24, which is about a 12.4% chance of no player obtaining a pair.

As calculated, this is not an exact solution because it neglects the constraint that each player must be given 4 cards. (If you really do this randomly, sometimes you reach valid arrangements, but sometimes you get arrangements where player A has 5 cards, player B has 3 cards, etc.) However this would be close if the relative frequency of pairless arrangements and pair arrangements for the subset of evenly distributed cases is similar to the same frequency for the distribution as a whole, which I would intuitively guess to be true for numbers large enough where the probability distribution is dominated by cases close to uniform distribution of cards.

On that line of thought, I think this estimation method represents a lower bound on the true value since the odds of obtaining a pair should increase if you consider cases where cards begin to cluster in the hands of one player*, so this analysis should be introducing cases where you are more likely to obtain pairs than you would have been in just the 4-cards-to-each-player cases. However I don't quickly have a way to provide a fully rigorous proof of that statement.

*In the extreme limit of the subset where 25 or more cards go to 1 player there is a 100% chance of a player having a pair, and it's intuitively obvious that the odds are quite high in the cases where one player gets, say, 12-16 cards.
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Oh woops. Putting Advokaiser in my sig like this until I think of something more clever