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In draw poker without the joker, what is the median pat hand, if we dont regard four of a kind as a pat hand? This should be straightforward, but i find it very difficult to calculate anyway...
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According to this chart,
http://www.poker1.com/mcu/tables/Table12.asp
the pat hands excluding four of a kind are:
Straight flush: 40 hands
Full house: 3744 hands
Flush: 5108 hands
Straight: 10200 hands
total = 19092, median = 9546
That puts us somewhere close to the top of the straights.
Each straight has 1020 different suits (4^5 - 4, we can't count the straight flush). So straights #1-1020 are all AKQTJ, which is where the median falls.
Bonus answer:
The 75th percentile is hand #4773, somewhere inside the flushes, flush #989 to be exact (maybe I'm off by one.)
Its identity is somewhat trickier to calculate, because the flushes have a more finely-ordered structure.
A-high flushes: 4 * (12C4 - 2) = 4 * 493 = 1980
So we're somewhere in this range.
AK-high flushes = 4 * (11C3 - 1) = 4 * 164 = 656 (#1-656)
AQ-high flushes = 4 * (10C3) = 480 (#657-1136)
So #989 is the 333rd AQ-high flush.
AQJ flushes = 4 * (9C2) = 144 (#657-800)
AQT flushes = 4 * (8C2) = 112 (#801-923)
AQ9 flushes = 4 * (7C2) = 84 (#913-996)
So #989 is the 77th AQ9-high flush.
AQ98x flushes = 4 * 6 = 24 (#913-936)
AQ97x flushes = 4 * 5 = 20 (#937-956)
AQ96x flushes = 4 * 4 = 16 (#957-972)
AQ95x flushes = 4 * 3 = 12 (#973-984)
AQ943 flushes = 4 (#985-#988)
AQ942 = flush #989-993, so this should be the 75th percentile pat hand.