Re: Five pat hands in 25 cards... sucker bet?
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.