Reading an old article by Richard Burke from Poker Player Online : Median flops. He is talking about unpaired flops and the median flop. He goes on to state that the median unpaired flop is a q93. He gets there by saying no of flops with at least q8x equals 9024 which is less then his math target of half of 18304 which is 9152. He then states each additional rank adds 64 flops. So the median unpaired flop equals q93. I understand this whole article except how he is narrowing down his over under math wise for the flop values. He gives an example of taking out a,k,q leaving only 7680. I get that part. Any help is greatly appreciated.

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Not sure what you are asking.

There are clearly C(13,3) = 286 unpaired flops where only the ranks matter. Median can be said to occur at either flop number 143 or 144. In an ordering of three unpaired ranks, we'd see that flop number 143 is Q94 and flop number 144 is Q93 (where flop #1 is AKQ and flop #286 is 432). It's not hard to confirm these numbers using math or a spreadsheet.

Burke seems to be using the fact that there are 4^3 = 64 actual flops for each flop category where suits are taken into account. The he counts in chunks of 64 until he reaches the middle (or as close as he can come).

Some people count "up" to find a median and some people count "down" to find a median. Of course, it doesn't really matter and only is relevant if you have an even number of items which we do here.

Not sure if that helps.

Could you link the article?

https://www.dropbox.com/s/pk6hpxftip...Flops.pdf?dl=0

Helped a lot thanks. Linked the article which isn't web available anymore also.

This is a cool little fun fact, probably doesn't help that much as a player, but cool nonetheless.

Well, it helps that JTs makes the friggin nuts way more than an average hand...