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Ok me and a buddy were just talking about it and I know this is rare but how rare is it, You have KQ hearts or even Kx hearts, player two has Ax hearts, how often do you both make a flush? A la the red head guy at the WSOP "i played perfect for two days." where he loses with his Kx to Ax suited. I don't care about how often you have the K and someone has the ace and there are four to your suit on the board. So What are the odds of both your kx and his Ax making a flush? from even both you being dealt the suited cards right down to it flopping or rivering or antyting inbetween the flush.

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Who cares? When it happens you're most likely gonna have to pay it off, sometimes you get cold-decked in poker. **** happens. It really doesn't matter what the odds of this particular happening are.

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Who cares? When it happens you're most likely gonna have to pay it off, sometimes you get cold-decked in poker. **** happens. It really doesn't matter what the odds of this particular happening are.
Ya i agree if you don't go broke in these spots your not playing good poker.
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Actually, if you do go broke in these spots you're not playing good poker.The best players know when to lay down a big hand when necessary.
You'll lose more money trying to avoid the A-high flush when you have K-high than you will by getting snapped by A-high on that rare occasion. Of course, if the board has 4 to the flush already, it's a different story.
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Yeah, let me know when you do that.
LOL....thank you...you read my mind...there is no way on a non paired/non straight flush board that you can get away from the second nut flush...I'm sorry, but ask any pro if they would lay it down and you will hear a resounding "NO"...this laydown can only be made in very specific tourney situations where you would not want to risk your whole stack on 2nd best...but if you aren't willing to risk going broke, why play in the first place????
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Actually, if you do go broke in these spots you're not playing good poker.The best players know when to lay down a big hand when necessary.
Nope. If you don't go broke in those spots you're playing scared - which isn't the same as playing good poker.
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LOL....thank you...you read my mind...there is no way on a non paired/non straight flush board that you can get away from the second nut flush...I'm sorry, but ask any pro if they would lay it down and you will hear a resounding "NO"...this laydown can only be made in very specific tourney situations where you would not want to risk your whole stack on 2nd best...but if you aren't willing to risk going broke, why play in the first place????
Pretty much.
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On the going broke question, it depends on the size of your stack don't you think? Also on how you think your opponent plays. :club:
Yes it depends, but there would have to be a combination of extremely tight play and extremely deep stacks.
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4 cards are already known... so it's basically the probability that 3 of the 9 cards left of the suit will hit the board out of the 48 that are left... sooo... probability majors get to it!

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4 cards are already known... so it's basically the probability that 3 of the 9 cards left of the suit will hit the board out of the 48 that are left... sooo... probability majors get to it!
No, that would be assuming that he has Axs everytime.
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No, that would be assuming that he has Axs everytime.
OP is asking if you have Kx and villian has Ax what are the odds both of you will hit the flushwe didnt assume anything
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Ok me and a buddy were just talking about it and I know this is rare but how rare is it, You have KQ hearts or even Kx hearts, player two has Ax hearts, how often do you both make a flush? A la the red head guy at the WSOP "i played perfect for two days." where he loses with his Kx to Ax suited.
Anyone know where I can find a video of this online?
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depending on if they raised/reraised preflop and/or how well u know the player you could have a good hunch that they had A something...and if they are still betting at a 4x or even 3x hearts board you could might entertain the feeling it was the Ah. Id think about that more than the math personally...

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heres how I think you would figure it out, first the odds fo being delt the hands at the same time, then the odds of floping a flush or a four flush(don't really care about back door flush because most likely going to fold the flop then) once you have that then you can figure your about 28% to both make a flush from there on with 7 or your suit remaining. The part that is hard is both being delt the hands. then of the 48 cards left 9 are your suit and 2 have to flop. And I have no idea how to do that math either.

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I don't see the relevance of knowing the odds of this particular scenario, but here are some of the realistic (...) scenarios if two players are dealt respectively Kxh and Axh:The chance of two players being dealt respectively Kxh and Axh:(((1/52)*(11/51)) * ((1/50)*(10/49))) * 100 = .00185% or 1/54145 (a)The chance of two players being dealt respectively Kxh and Axh and flopping a flush:((a/100) * ((9/48)*(8/47)*(7/46))) * 100 = 8.97E-6% or 1/11148713.33The chance of two hearts hitting on the flop if four hearts are missing from the deck:((9/48)*(8/47)*(39/46)*100) + ((9/48)*(39/47)*(8/46)*100) + ((39/48)*(9/47)*(8/46)*100) = 8.12% or 1/12.32 (b)The chance of two players being dealt respectively Kxh and Axh and flopping a flush draw and making a flush on the turn or river:((a/100) * (b/100) * (1-((38/45)*(37/44)))) * 100 = 4.35E-5% or 1/2300860.16Based on the principles above you can calculate the odds right down from the first card dealt to the river, but it would take a very long time. Also bear in mind that this calculation is based on two players, instead of e.g. 9 or 10 because that would also be an exhausting calculation.So the next time you're being dealt Kxh I don't think you have to worry about the reverse implied pot-odds of someone holding Axh.And BTW, I'm pretty sure that calculations are correct, but I may have entered something wrong on my calculator and I really don't feel like reading it through right now.[update] Due to editing the results are no longer 100% correct, but it is so close that it is irrelevant. The formulaes, however, are now correct.Regards,Mathias Munk Hansen

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