Hi, I'm trying to work out the number of open ended straight draws in all possible 5 card boards. So 4 consecutive ranks, where the 5th card does not make a straight or a flush. The 5th can be paired with any of the four. This means A2345 and TJQKA being single ended are not counted. So I'm thinking there is 4*4*4*4*8 possible combinations of these open ended straight draws, from 23456 to 9TJQK, but not sure how to calculate the 5th card flush and straight exclusions and remaining combinations. Any help is greatly appreciated.

Not sure the relevance, but there's a couple of stat guy's who may appease your (2 part) request .. but I'll tell you it will be something significantly lower than 2,598,960.

Not sure on your terminology either .. I think you mean to eliminate A234 and JQKA as 'single' ended, but 2345 and TJQK are considered open-ended, which makes for 9 combos, not 8.

There are 270,725 different combinations of 4 cards in a 52 card deck, but you only want the ones that make 4 in a row. So I believe that leaves us with 2304 qualifying 4-card Boards.

Now the easy part .. you need to calculate the possibilities that a straight card doesn't come .. which is only 8 of the remaining 48 cards, so 40 'good' cards. So that gives us 92,160 possible 5-card Boards that have 4 connecting cards but not the 5th.

Now the hard part (?) .. we still need to eliminate the flushes too. There are 36 possible Boards that are 4-flushes, four from each set of 9. We have already eliminated 2 of those cards since they would make a straight flush, so we need to eliminate the 7 other possible flushes as well. That gives us 252 Boards to eliminate from our total above, dropping us down to 91,908 5-card Boards that have 4-in-a-row but aren't 5-card straight or flush Boards.

1) There are 256 combinations when you have 4-cards-in-a-row (4*4*4*4)
2) There are 9 possible combinations of 4-in-a-row in a card deck .. 2345 to TJQK, giving us 2304 4-card-in-a-row Boards 'going to the River'
3) There are 40 non-straight cards that can be the 5th card, giving us 92,160 5-card non-straight Boards.
4) There are 36 possible 4-flush Boards that each have 7 more 'bad' cards, so eliminate 252 flush Boards
5) 92160 - 252 leaves 91,908 5-card Boards that have 4-in-a-row but aren't a 5-card straight or flush.

Hope I'm right, I'm sure someone else will chime in if I'm not .. GL

PS .. The word 'draw' is key here, you would (hopefully obviously) have a much higher amount of 4-card Boards if you don't care what order the 5 cards come out in, but you still ultimately end up with the same final number of 5-card Boards that meet the criteria.

I agree with answer20.

Assuming double-gutters don't count, there are 9 rank combos that make an open-ender.

There are 4^4 suit combos for flop+turn, then (4*3 + 7*4) suit combos for the river. But there are 4*7 flush combos to subtract.

So it's 9[(4^4)(4*3+7*4) - 4*7] = 91908