How Good is your Frush?

Brackchips sent me a HH where he forded a medium-sized frush (seriously, dude, you 6-bet shove like every other time you're on the button, then you fold a turned frush to one raise?) that turned out to be good, which set me thinking to what the exact "middle" flush is. For instance, A8 is the "middle" ace-high - there are 6 below it (A2,A3,A4,A5,A6,A7), and 5 above it (AK,AQ,AJ,AT,A9). It's the worst ace that beats most ace-high hands.

If you have a flush made up of four community cards and one hole card, it's easy to tell how your flush ranks: just count the cards of that suit that aren't on the board or in your hand, and see how many are above and how many are below the one in your hand. Example: The board is As Ts9s4h3s. You want to evaluate your Kh8s. There are three spades better than yours (K,Q,J) and five worse (7,6,5,4,2).

When your flush consists of two hole cards and three community cards, it's not so clear. You can't just count the cards not out, because there are more COMBINATIONS of suited hands that are ace-high than (for instance) three-high. In particular, there's only one card that goes with a three that makes your whole hand three-high (a deuce).

So let's say we're talking about a spade frush (if we're gonna talk about a frush, why not talk about the best kind?). You've got two spades in your hand, and three on the board, so there are eight spades that your opponent could have. In order to be clear, let's rank these spades frop highest card to lowest, with the highest being called "first spade", and the lowest "eighth spade". So if you have, say, Ts4s and the board is AcJs5s3s, the As would be the "first spade" and the 2s the "eighth spade". But if the board were AsKs5c2s, the Qs would be the "first spade" and the 3s would be the "eighth spade".

How many different two-card combinations can be made from eight cards? Well, he could have any of the eight spades as his first card, and then any of the seven remaining spades (after removing his first card) as his second card. That's 56 combinations. But then we divide by two, because we consider AsKs and KsAs to be the same hand. Which makes 28 combinations. Each of the eight cards is included in seven different combinations. How many of them have the first spade as the highest card? Clearly, seven. You can combine the first spade with any of the other seven spades, and it's always the highest card. How many have the second spade as the highest card? Six. The second spade is included in seven total combinations, and one of them is with the first spade. In all others it's the highest card. Continuing in this manner:

1st spade-high: 7 hands
2nd spade-high: 6
3rd spade-high: 5
4th spade-high: 4
5th spade-high: 3
6th spade-high: 2
7th spade-high: 1
______________

28 hands*

So if the highest spade in your hand is between the second and third spade (for instance, if you have JsTs on a Qs6s7c3s board), you beat most flushes. Otherwise, if your opponent's range is any and all flushes, you are behind.

-BRUECHIPS

*It's actually true in general that n*(n-1)/2 = 1+2+...+(n-1). To see why, note that you can use Gauss's method of doing long symmetric sums. That is, first add together the first and last term (1 + (n-1)), and the second and the next to last term (2+n-2), then the third and the third to last term (3+n-3), etc.. Note they all add to n. How many of these pairs do you have? Well, you clearly have n-1 total terms, so you must have (n-1)/2 pairs of terms. So the sum on the RHS is must be equal to n*(n-1)/2.