October 18th, 2019 at 7:19:06 AM
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Can someone tell me what the probability in Texas Hold-em of two players having a flush on the same hand with both of them using both of their hole cards?

October 18th, 2019 at 8:02:34 AM
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This assumes heads-up play.

Note that all four hole cards must be the same suit. If, for example, one player has spades and the other has hearts, then the board needs three spades and three hearts for both of them to have flushes that use both hole cards, but that's six cards on the board.

Of the (52)C(4) sets of four hole cards, (13)C(4) x 4 are all the same suit; this is 2860 / 270,725.

In order to need to use both hole cards, exactly three board cards are the same suit as the hole cards. (Well, not exactly - if the board has a 2 in that suit, then both flushes will use both hole cards - but I'm ignoring that for now. I am also ignoring the possibility that either or both players have a straight flush.)

For each deal, there are (48)C(5) sets of board cards, of which (9)C(3) x (39)C(2) have exactly three cards of the hole card suit; this is 62,244 / 1,712,304.

The product is about 1 in 2604.

Note that all four hole cards must be the same suit. If, for example, one player has spades and the other has hearts, then the board needs three spades and three hearts for both of them to have flushes that use both hole cards, but that's six cards on the board.

Of the (52)C(4) sets of four hole cards, (13)C(4) x 4 are all the same suit; this is 2860 / 270,725.

In order to need to use both hole cards, exactly three board cards are the same suit as the hole cards. (Well, not exactly - if the board has a 2 in that suit, then both flushes will use both hole cards - but I'm ignoring that for now. I am also ignoring the possibility that either or both players have a straight flush.)

For each deal, there are (48)C(5) sets of board cards, of which (9)C(3) x (39)C(2) have exactly three cards of the hole card suit; this is 62,244 / 1,712,304.

The product is about 1 in 2604.

October 18th, 2019 at 10:39:45 AM
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Quote:ThatDonGuyFor each deal, there are (48)C(5) sets of board cards, of which (9)C(3) x (39)C(2) have exactly three cards of the hole card suit; this is 62,244

I agree with that.

I get 2,457 combinations with four suited cards on the board, and 35 with 5 suited cards on the board. This takes into account that with four suited cards on the board there is a 0.5 chance both player use both hole cards. With 5 suited cards on the board, that probability is 5/18.

The board combinations is 64736. The total of all combinations is 185144960. Dividing by combin(52,4)*combin(48,5) total combinations, the probability is about 1 in 2503.79.

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan

October 18th, 2019 at 11:32:02 AM
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These days, it all depends on where you play and who plays.Quote:kyjscrapCan someone tell me what the probability in Texas Hold-em of two players having a flush on the same hand with both of them using both of their hole cards?

I NEVER bet 5,3 (or 7,2) suited unless big blind forces me (NEVER have hit a flush that way either)

example: in 9 handed NL wsop.com tournament

FREEROLLS and free play the probability is about 1 in 66 rounds I have seen (greater than quads at 1 in 50 rounds - over a 15k sample size since last year and a half)

that explains all the 'flush chasers' that exist (play) there

It is free, one gets what the software gives and players bet what they see always winning.

wsop.com wants one to think they are a good player (winning with horrible hands) and deposit $$$ to play the many cash games they offer. they have no competition in Nevada - online.

I guess they do it for all the 'action'

most newbies (especially May and June of each year)

to that site have NO clue the freerolls are not fair (well known to those that play there)

done for 'action hands'

still, it really depends on who one plays against.

cash games with many tight players, doubt one would see that event

winsome johnny (not Win some johnny)