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View Full Version : Probability of an opponent having a given hand in a range

DavidC
10-16-2005, 08:32 AM
Let's assume that I can narrow down an opponent to having either KK, AQ, or QQ, and that none of these cards are in our hand. Additionally, none of those cards are on the board, and we're on the turn.

Possible ways to get the hands:
KK: 12 ways
AQ: 16 ways
QQ: 12 ways

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Firstly, is this table correct?

Secondly, is KK equally likely as QQ, or is KK slightly more likely, because QQ and AQ overlap?

Dave G.
10-16-2005, 08:47 AM
There's only 6 ways to get KK and QQ, and 16 ways to get AQ.

KK and QQ are equally likely. There are still 4 Qs to choose from until one of the Qs is accounted for. You can consider it a possibility with AQ, but they don't affect each others probability until you actually place that Q in someones hand or on the board.

DavidC
10-16-2005, 09:40 AM
[ QUOTE ]
There's only 6 ways to get KK and QQ, and 16 ways to get AQ.

KK and QQ are equally likely. There are still 4 Qs to choose from until one of the Qs is accounted for. You can consider it a possibility with AQ, but they don't affect each others probability until you actually place that Q in someones hand or on the board.

[/ QUOTE ]

Thanks. /images/graemlins/blush.gif

DavidC
10-16-2005, 09:53 AM
Mathematically, I'd like an explanation here, I'll try it first:

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Any time we're drawing from different pools of cards (to make AQ), we just multiple the cards in one pool by the cards in the other, right?

When we're drawing from the same pool of cards, we're doing this:

TC*X/CT!, such that:

TC = Total Cards
CT = Cards Taken
X = (TC-1)(TC-2)... (TC-CT)

So in this case of KK with no K's exposed, it's:

4(3) / 2(1) = 6

But if we wanted to know what the chances were of getting three kings in our hand in pineapple or something, we'd do:

4(3)(2) / (3)(2)(1) = 4 possibilities...

This would make sense, as you'd be excluding one card only, since there's four suits... you'd pick one card out each time, leaving you with 4 unique hands.

However, I have no idea if this is called "pick" notation or "choose" notation...

It's been a while since I've done any of this in school. It was fun stuff, though.

Obviously I need a refresher.

I'm curious...

When would someone use the opposite of whatever we're using now (pick/choose)?

Would we be doing that if we were putting cards back into the deck and redrawing them or something (which would be possible in a weird home game of draw poker with lots of players, or something like that...)?

Edit: and do you find the # of hands for KK using pick or choose notation?

AaronBrown
10-16-2005, 11:07 AM
This is correct, but there is a hidden assumption if you want to say the conditional probabilities are 30% for each pair and 40% for AQ. That would be true if you dealt two cards until you got one of these three hands, shuffling after each deal. But in a Poker game it might be true that, say, KK would be played the observed way 100% of the time, QQ 90% and AQ 80%. That would affect your conditional probabilities.

The function you describe, C(4,2) = 4*3/(2*1), is called the "choose" funtion (read "4 choose 2"). It's COMBIN(4,2) in Excel.

I think by pick you mean a situation where order matters. For example, in five card stud, getting A in the hole with Q exposed is a different hand than Q in the hole and A exposed. So although it's two cards like Hold'em, there are more starting possibilities. There are 13 pairs in both, but 13*12 = 156 non-pairs in five card stud, each of which can be suited or not, versus 156/2 = 78 non-pairs in Hold'em, each of which can be suited or not.