Percentage of A, K, Q high flops

Ace high: ~21.7%
King high: ~18.4%
Queen high: ~15.2%
Jack high: ~12.4%
Ten high: ~9.9%
Nine high: ~7.6%
Eight high: ~5.6%
Seven high: ~3.9%
Six high: ~2.6%
Five high: ~1.5%
Four high:~0.74%
Three high:~0.23%

Enjoyed this thread, but to me, a better question would be how often does someone else have an ace when I have one?

All cards flop with exactly the same frequency. Aces will be the highest card on 100% of the flops in which they occur. That’s why the percentage of ace high flops is higher than for all other cards. Not all flops with a king will be king high flops - any AKx flop would be such a flop. Kings do not flop less often than aces, but flops with a king are king high flops less often than flops with an ace are ace high ones. And likewise for all lower cards.

Different, but somewhat related question that I always thought had an interesting answer. Suppose you are dealt a five card hand from a one suited deck. That is a 13 card deck with only one card of each rank. Straights donÂ’t count. What hand beats exactly 50% of all possible hands? IÂ’ll give anyone interested a chance to think about this one, but the answer might be counterintuitive (at least it was for me when I first saw and solved this problem).

I've never solved a question like this, but I'll give it a try!

Possible combinations: C(13,5) = 1287

A high: C(12,4) / 1287 = 38.5% -> Not enough to beat 50%
K high: C(11,4) / 1287 = 25.6% (64.1% cumulative) -> Enough to beat 50%

The highest card in the hand is K high.

Then I did the same for the next couple of cards:

Spoiler

Show

We now have KJ9

With two cards left: C(8,2) = 21 combinations

We now want to find the hand that beats 10 combinations and loses to 10 combinations (i.e hand #11).

Final hand: KJ972

I'm not sure I got it exactly right, but hopefully pretty close 😀

edit: based on some more brute-forcing I'm gonna revise my answer to KJ987. Still not sure it's right, but whatever 😃

Close and your math is good but I think you counted combos a bit wrong. There are 1287 total. We are looking for a target hand that beats 643 and loses to 643. A high gives 495. KQxxx gives 10C3 or 120 for 615 total. We need 28 more. KJxxx gives 9C3 or 84 — too many. Target must be KJxxx. KJTxx gives 8C2 or 28 — exactly what we needed. All KJTxx beats us so our hand is KJ987.

Obviously the idea is similar. All cards are equally likely but A high is more probable than any lower hand

Yes, that was the method I used in my second attempt. I definitely learned something!