Chances that my Q high flush is beat?

Please check my math/logic.
I hold the Q of suit when the board (at the river) shows 4 others of that suit. What are the chances someone holds the A or K of suit? The board is unpaired so I don’t need to worry about a boat.

I think it is calculated as:
16 places the A could be in opponents hands(9 handed game) and 45 unseen cards. So 16/45 = .35 or 35.5%

Same for the K, 35.5%

Is it proper just to add those two and say there is a 71% chance someone holds either the A or the K?

What are the chances of all 8 players still being in the hand on the river?

(though yes, players with K or A of the flush suit will be much more likely to have stayed in the hand)

Yeah, when the flop is 3 suited cards, the A and K are likely to stay in just as I did since the bets were small enough to allow it.

Originally Posted by EricE

Yeah, when the flop is 3 suited cards, the A and K are likely to stay in just as I did since the bets were small enough to allow it.

True but what if they folded before the flop? For example, K2 os is a bad hand that is folded pre-flop.

Originally Posted by Eric

Originally Posted by EricE

Yeah, when the flop is 3 suited cards, the A and K are likely to stay in just as I did since the bets were small enough to allow it.

True but what if they folded before the flop? For example, K2 os is a bad hand that is folded pre-flop.

I agree that you take all these things into account but you have to know the base value before you can adjust. Is the 71% correct?
If it is, then I would not value my Q high so highly. I pushed, got called and lost. In general I don’t think pushing a Q high flush on the river is a situation where you will not be called by losing hands. J or T may call you (or even worse at the micro stakes). But if the chances of running into the A or K are that high then maybe I shouldn’t value my Q as the nuts. Hehe.

I think you hit it - the question isn't - "In a vacuum, what are the chance Opp has XYZ card". That's worthless info.
it's "If I push all my chips in here, what are the chances I get called by a worse hand than mine?"

And here the answer is very, very low.

Originally Posted by drmcboy

I think you hit it - the question isn't - "In a vacuum, what are the chance Opp has XYZ card". That's worthless info.
it's "If I push all my chips in here, what are the chances I get called by a worse hand than mine?"

And here the answer is very, very low.

Indeed. My push was strong enough that the J or T would likely not call. Whereas a bet of 20xBB or less could be called by worse hands.

Thanks all. That helps.

So…the same thing happened last night. 4 to a flush on the board and I held the Q. Guy before me pushes all in on the river for my whole stack. I fold and he shows 9 high flush. Hehe. Next time I’ll likely call…and I’ll likely be wrong. Hehe, my timing sucks.

Originally Posted by EricE

So…the same thing happened last night. 4 to a flush on the board and I held the Q. Guy before me pushes all in on the river for my whole stack. I fold and he shows 9 high flush. Hehe. Next time I’ll likely call…and I’ll likely be wrong. Hehe, my timing sucks.

I suggest that you fold the next time too :P

Here's how I think of it. I was sometimes in the situation of holding the second nut flush, the K of suit on a 4 flush board. Another guy went all-in or raised me all-in.

If you think
"I am holding the second nuts, I ONLY lose to ONE card, I will call" you are WRONG.

Though it is the second nuts and you only lose to ONE card, its not the same as holding the second nut FH... its a frigging 4 flush (non paired if we are talking for nuts) board! And someone went all-in (or bet huge) while YOU HOLD THE KING of that suit! Unless you have reasons to belive he is bluffing, calling means you actually think that he could have made this play with the queen or less...

I've lost some good pots in the past with K flush on a four flush board, then came to this realization.

16 places the A could be in opponents hands(9 handed game) and 45 unseen cards. So 16/45 = .35 or 35.5%

Same for the K, 35.5%

Is it proper just to add those two and say there is a 71% chance someone holds either the A or the K?

In both parts you're making the same mistake, when talking about two events and considering the "or" of them, you cannot add their odds. For example, if there's a 75% chance of rain today and a 50% chance of rain tomorrow, does it make sense that the chance of rain today or tomorrow is 125%? I sure hope that doesn't make sense...

What you have to do when confonted with an "or" in odds is inversion. It's based on the logical rule DeMorgan's Law, which states:
A | B <-> !(A & B)
Or in English, A or B is the same as not A and B.
So for our probability here, the chance of A or B happening is:
1 - (1 - P(A)) * (1 - P(B))
For the rain example, that's
1 - (1 - .75) * (1 - .5)
1 - .25 * .5
1 - .125
.875 - or 87.5%, not 125%.

So let's re-examine the odds here. It'd be possible to look at the odds of someone having the A then someone having the K, but it'd be easier to look at them both at the same time...why? Cause that's actually a case where we *can* add the odds.

Okay CrunchyNuts, you just said we couldn't do that in the case we're dealing with "or"s, and this would be "This guy's card is As or Ks", and that sure looks like an or to me. Why yes, it is. But the key difference here is that you're talking about *one event*, the card. Two possible outcomes of one event *are* added when or is involved, assuming there's no overlap in the outcomes. Going back to the rain example, if it's 75% chance to rain and 10% that it'll be sunny, the chance of rain or sun is 85%. An example of overlap is if you say it's 75% chance of rain and 85% chance of overcast...can't add those up 'cause it's overcast when it's raining.

So we can safely say the odds of the first of the 16 opp cards being As or Ks is 2/45. The next card, assuming the first wasn't As or Ks, is 2/44 - one card's gone already, and so forth for the rest of the 16.

At the end of the day though, we're looking at the "or" of all of these events, so we need to invert. We need to figure out the chances that the As or Ks do *not* come up, that's an "and" problem. It's not the first card, and it's not the second card, and it's not the third card, ect. Get your calculators handy:

Code:

43/45 * 42/44 * 41/43 * 40/42 * 39/41 * 38/40 * 37/39 * 36/38 * 35/37 * 34/37 * 33/35 * 32/34 * 31/33 * 30/32 * 29/31 * 28/30
(43 * 42 * 41 * 40 * 39 * 38 * 37 * 36 * 35 * 34 * 33 * 32 * 31 * 30 * 29 * 28) / (45 * 44 * 43 * 42 * 41 * 40 * 39 * 38 * 37 * 36 * 35 * 34 * 33 * 32 * 31 * 30)
Factor out a whole lot of X/X's...
(29 * 28) / (45 * 44)
812 / 1980
~0.41

So 41% chance that As and Ks do *not* so up, so 59% chance that they do.

Its not about math.

When someone re-raises you all in and you hold JJ do you think "he only has AA 1 in 221" and call?

Hand rarity does have something to do with the range of hands I put the opp on, yes.

But regardless, the man asked about odds, so odds he aught get.

Thanks Crunchy