This is a discussion on Combinatorics question suited combinations? within the online poker forums, in the Cash Games section; So, I know that there are 4 combinations of any 2 specific suited cards. For example, 4 combos of QKs, 4 combos of AKs, etc. 


#1




Combinatorics question suited combinations?
So, I know that there are 4 combinations of any 2 specific suited cards. For example, 4 combos of QKs, 4 combos of AKs, etc. Where I run into problems with finding the number of combinations of suited cards in a villain's range are situations where villain plays suited kings, queens, or God forbid, any two suited cards. If villain plays only suited aces, I think that he would have 48 combos of any suited ace, and only 12 that match the suit in question on the board, unless there are 2 double suits by the turn in which case it would be 24, right? What if a looser player, whether it be from late position or just a bad player from EP, opens any suited ace, king, or queen? So, removing the Ace of that suit that we already accounted for when counting suited aces, would there be 44 combos of suited kings, with 11 of them matching the suit in question on the board or 22 on double suited turn boards? Then, when estimating suited queens, and removing the previously accounted for suited aces and kings, would there be 40 combos with 10 that match the board's suit? I'm really bad with figuring this out when it comes to removing the previously accounted for suited hands. Can anyone give me a formula for this? If not a formula, even just answering how many combos of suited cards a villain could have if he plays only suited aces, suited aces+kings, suited aces, kings, and queens, and the dreaded any two suited cards would help me out a lot. Maybe there's an easy way to figure this out that I can't put my finger on. Please help!

#2




Just glancing at your post it seems you have it right. There are countless different ways to approach the question.
There are 1326 combinations of holdem hands and 312 (12/51 of the total) are suited. This is 78 for each suit which can also be determined by 13 ranks × 12 and then divided by 2 as we don't distinguish J5 and 5J. 45 of those 78 hands have a high card of J or lower (you can do the same thing here and go 10 × 9 / 2) and the other 33 have at least one A K or Q. The 33 can be calculated just adding 12 11 and 10 together which I think is what you did your post except you did for all suits as opposed to one specifically. 
#4




re: Poker & Combinatorics question suited combinations?
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#5




I'm trying to figure this myself. but more when the flush comes in, how many combos that "have it"
normally to figure out how many combos of something it is "m choose n" or in the case of suits there are 13 spades and we want all 2 card combinations which would be "13 choose 2" Let "i" stand for "factorial". The equation for "M choose N" is mi divided by (n)i x (mn)i however, in the case where the spades have come in (say on the river) there are actually only 10 unseen spades to choose from. so it would be expressed as "10 choose 2" I'm not going to bore you with showing my math, but if you apply the equation for "M choose N" to "10 choose 3" the answer (I believe) is 45 combos. If you just want to know "how many combos of 2 spades" preflop, without there being any dead spades it would be "13 choose 2" and the answer would be 78. 