Pai Gow Poker Low Hand Question...

What are the odds of you 'copying' the house/banker in your low hand when you have a pair in your low hand. Say you have King/King/x/x/x with 4/4 up and the House/Banker has 6/6/x/x/x with 4/4 up. I would have to think it's pretty low since there are only 2 cards left in the deck and the House/Banker has to not only have them in their hand, but has to play them in their 'low' hand. Thanks.

Ok edited because I answereed the wrong question. This is the probability that the dealer will have a pair for their low hand not considering that you already have a pair.

Dealer low hand pair	Probability
2	0.022299
3	0.021033
4	0.019793
5	0.018611
6	0.01752
7	0.020697
8	0.019207
9	0.017774
10	0.016424
j	0.014342
q	0.012299
k	0.010313
a	0.010929

Miplet -

I think you're calculating for simple poker, not Pai Gow. And even for simple poker, I think those figures are too high.

Hey miplet, can you please post the details of your calculation? I'm curious as to how you arrived at those numbers. Thanks!

Ooops I see what I did now. I did the odds that the dealer will have a pair in the low hand. Silly miplet. Back to the drawing board.
Some more stuff:
Given that you have a pair of dueces and 5 non dueces, the odds that the dealer has the other two dueces are about 0.0202899 How often they will put it for their low hand is harder to figure out.

Quote: miplet

Given that you have a pair of dueces and 5 non dueces, the odds that the dealer has the other two dueces are about 0.0202899 How often they will put it for their low hand is harder to figure out.

I still think that's high.

Unless the only thing you calculated was the odds of the dealer having two dueces in his 7 card hand, when only two duesces are available.

The player doesn't need to have both of the other deuces. Two players can have one deuce each, and other players, no matter how many there are, do not have a deuce.

I.E. Without knowing anything other than two deuces are accounted for, then I think that a 2% chance of the dealer having both remaining deuces might be right.

There are 46 cards left in the deck. The odds that the dealer has both of your cards in a remaining pair is 7 x 6 x 44 x 43 x 42 x 41 x 40 / 46 x 45 x 44 x 43 x 42 x 41 x 40 = 42 / (46 x 45) = 2.0290% [Edited for my correction]

But the odds that the dealer will play it is their low hand is very difficult to determine. The dealer will play the matching pair (MP) in the lower hand under the following conditions according to AC Trump way.

- In a two pair hand, if the MP is the lower pair and the MP is not superceded by having a King in their hand (Medium/Low pair) or an Ace in their hand (Hi / Low, Medium / Medium and below). If the hand is a straight, flush, or straight flush, the two pair don't care rule applies (split according to the rules)
- In a straight, flush, or straight flush, with a matching pair, MP is played in the front.
- In a full house, the MP is the smaller pair, unless the MP is 2s and the remaining cards is an A-K
- Four of a kind and a pair, the MP is the matching pair.
- Five aces and a pair of Kings, and Kings is the MP.

So then you have to look at the odds of the dealer getting these hands and compare it to the hand that you have. This is too complicated for me, but maybe Goat or MB or even the Wiz might know this one.

The 0.0202899 figure that miplet came up with is the correct probability of the dealer having the other pair that you have. As an Excel formula, it is COMBIN(44,5)/COMBIN(46,7). But again, that doesn't guarantee that they'll play them in their low hand. You would need to apply the house way to all COMBIN(44,5) = 1,086,008 possible dealer hands containing that pair to determine how many of them set the pair in their low hand.

I've edited for my correction.