Pai Gow Question

What is the chance that both Player and Dealer have a Pai Gow Hand?

IIRC its about 1 in 6 for the Player only. Considering the rather odd removal of cards needed, is it much more difficult for the Dealer to also have one?

I don't know the answer to this. I know the calculation has been done, or can be derived, from work I've seen on here and WoO.

But I think it has to be slightly more likely that the dealer has a Paigow when you do, than when you don't. It would go to the effect of removal. You have straight blocking cards, use up flush cards, 7 pairing cards, and possibly the joker. So you make a lot of dealer hands slightly less likely (different hands each time) with what you use up.

The opposite hands make this more apparent. Let's say you get a 7 card flush in hearts. This would make the likelihood a little greater for the dealer (or other player hands) also to have at least a 5 card flush in another suit, since he's unlikely to have many hearts in his hand, but he still has to get 7 cards. He might even beat your hand because of it, depending on ranks.

I see this a lot on 2 pair hands. Several hands hold them at once a lot of the time. Then it comes down to rank and strategy, because the effect of removal of your pairs makes it slightly more likely other hands will also hold pairs. And the more of those hands there are, the more likely they are in the remaining hands.

Not sure if that's any help, as you asked for percentages, but the leveling effect of this is part of what makes the game playable, though not in a big way.

Not sure.... But given one lop lop hand, it is unlikely to hold the joker, thus other hands are more likely to hold the joker and be less likely to be a lop lop. I would guess that the joker factor is more relevant than the 'removal' factor. If someone is bored they can run a simulation of course

Yeah, I thought someone already had run this, and I did check-in at WoO. That states 1 in 6.22 for the Player (24,780,420/154,143,080). If i square that I get 1 in 38.7. I have always presumed 1 in 40, but thought it was better defined.

When the player is dealt a Pai Gow Hand, the probability of the dealer also having a Pai Gow is higher than the usual probability of a Pai Gow Hand. For instance,

With a fresh deck, the probability of a 7-card hand being a Pai-Gow is: 0.1608

But if Player: A-Q-T-8-7-5-3 (suits are 2-2-2-1)

then the probability of the dealer having a Pai Gow is 0.1666.

The exact probability of a dealer Pai Gow will depend upon the precise cards that are in the player's Pai Gow Hand, but in general, the probability of both the dealer and the player having a Pai Gow hand is very close to 0.1608 x 0.166 = 0.02669 =1/37.5