I was reading something a few days ago where somebody mentions a bet regarding putting 25 random cards into five pat poker hands (that is, straights, flushes, full houses, quads). As it was written, it sounded like a near certainty that it was possible to do so. Of course, I dealt out two samples and yes, I was able to do it both times.

Yes, I know two times out of two doesn't mean certainty, and I can think of counterexamples (A267JQ of two suits, 3489TK of the other two suits, plus an additional ace), but is there some better argument of why this should work more often than not?

Regarding your example:
AAAKK (full house)
267JQ (flush)
267JQ (flush)
3489T (flush)
3489T (flush)

You should have left a 5 in there instead of adding the ace.

One of the reasons that this works so often is that flushes are value-independant, and 4-ups give you a 'free card' either of which will give you a lot of breathing space, and you're guaranteed to see at least two potential flushes.

Whoops! Quite right. I did a little more searching and someone coded up a way of doing this and had 98 hits out of 100.

http://www.math.fau.edu/Locke/Course...ving/F04Up.htm

One heuristic to consider is the expected number of ways to produce 5 pat hands. If you deal out 5 hands from separate decks, then the probability that each is a pat hand is about 0.76% ~ 1/132, so the probability that all 5 are pat hands is about 1/132^5 ~ 1/(4 x 10^10). When all 5 hands are deal from the same deck, the probability is different, but it's not obvious whether it should be higher or lower, so for a ballpark estimate, we can assume that this is the probability.

The number of essentially different ways of choosing 5 poker hands from 25 is (25 multichoose 5,5,5,5,5)/5! = 25!/5!^6 ~ 5.2 x 10^12. The extra 5! in the denominator is that we don't care about the order of the hands. Otherwise, successful divisions would come in groups of 120. So, the average number of essentially different ways to produce 5 pat hands is about (5.2 x 10^12) / (4.0 x 10^10) ~ 130.

If the average value were under 1, it would be a meaningful upper bound (nonrigorous because we chose hands from different decks) for the probability, and a better estimate for the probability might be the probability of a count of 0 from a Poisson distribution with that mean. The probability of a count of 0 with a Poisson distribution with mean 130 is e^-130. By this heuristic, we'd estimate that the probability that you can't divide the hands into pat hands is low.