Pai Gow Poker strategy: Split aces?

AA98632

Clearly any houseway would play this hand as [98]AA632. To me, this seems like the only reasonable play as a player as well. But lately I've seen more and more players play this hand as [A8]A9632. The logic being that "98 is such a bad up hand that it's better to put an ace up." To me, this seems like a bad play. What are your thoughts?

I'd probably split it.

On the other hand, my pai gow luck has been so godawful weird lately playing with normal moves, that I got fed up and actually did [highcard]-[fullhouse] the other day, and actually won-- to a [lowcard]-[flush]. Twice.

When the pair is 44, 33, or 22 its generally correct to play Ace-King / House. I have seen quite a few Asian players put an Ace-any / House. Not a bad idea if the Dealer's hot-handed, but generally taking a winner and making it win less is not a good idea.

ALWAYS SPLIT ACES!!!!!

deleted

Quote: Deucekies

AA98632

Clearly any houseway would play this hand as [98]AA632. To me, this seems like the only reasonable play as a player as well. But lately I've seen more and more players play this hand as [A8]A9632. The logic being that "98 is such a bad up hand that it's better to put an ace up." To me, this seems like a bad play. What are your thoughts?

I've seen an awful lot of people need the roving masseuse after getting a hand like this...twisting their neck and craning to see how many aces are around them...lol. This is a hand that, if they allow any table talk, smart players will start some kind of joking (or outright) inquiry about ace placement, trying to judge how many are left for the dealer, before they decide to guard their aces or split for a push. It's also why casinos in Mississippi have (or used to have; didn't see one on my last visit) signs posted requiring English to be spoken at the table; there was a LOT of Vietnamese collusion on this point, which would go over some of the dealers' heads. Me, I keep the aces together 100% of the time to guard the main bet; other people will gamble and split them.

Keeping the aces together returns about -0.22 whereas splitting them returns about -0.32, so keep the aces together if you wish to minimize your loss.

Quote: JB

Keeping the aces together returns about -0.22 whereas splitting them returns about -0.32, so keep the aces together if you wish to minimize your loss.

If you are 100% sure that the other two aces and joker are not in the dealer's hand, what does this do to the respective EV's?

Quote: SOOPOO

If you are 100% sure that the other two aces and joker are not in the dealer's hand, what does this do to the respective EV's?

I don't know.

Quote: JB

I don't know.

Reason I ask, is when playing at a full table at my local casino there is a lot of banter, and I am often helping other players. I often have access to 35 cards. At the table minimum level they never seem to care. I often change strategy when I know all the aces are out.... Example.... Queens and fours with an ace 5 3, I'll play the A5 if I know there are no aces left, but if there are two or three left in say, 18 cards unaccounted for, I'll play 44. Same goes for full house with an A5, if there are no more aces left I LOVE playing the full house with the A5. I won one I would have pushed last time.....

Collusional knowledge is a good thing sometimes.

That bad beat... what was he thinking?!?! Quads + Any Pair 4TW. QQQQ+55 in my book. Too rare for an exception IMHO.
By the way... folks here SHOULD compare A Nevada House Way vs. E. Coast. Two Pair rules are rather different including NO SPLITS.

Never ever ever ever ever split aces. I don't even think knowing all aces are gone helps your EV.

Quote: 98Clubs

That bad beat... what was he thinking?!?! Quads + Any Pair 4TW. QQQQ+55 in my book. Too rare for an exception IMHO.

I'd have played Queens up. The way I see it, the dealer is more likely to beat 55 up than he is to beat QQ55 down.

Without knowledge of any other cards in the deck:

98 | AA -632 has an EV = -0.2465 (Wins: 0.0120, Push: 0.7302 and Lose: 0.2579)

and splitting aces yields

A8 | A9632 EV= -0.2940 (Wins: 0.1230, Push: 0.4661 and Lose: 0.4109)

If you know the dealer does not have an Ace or Joker then splitting aces has a positive EV for any AA-xxxxx hand in which the top two singletons are 98 or less. The increase in EV is about +0.0242 for the decision to split aces with the specific hand AA-98632 versus playing this hand the conventional way. The improvement in EV for splitting aces with AA-76XXX (a relatively rare hand) is about +0.053 versus not splitting the aces.

I have a computer code that solves Pai Gow Poker analytically (not Monte Carlo) and I have rigged it to provide exact calculations for certain scenarios -specifically for observations of Aces, Kings and the Joker in other players hands as a function of the number of hands that you have peeked at. If someone will explain how I can import a photo or graphics image into a post, I will post the results of my scenario code for this question of AA-98632 and splitting aces.

Gordon,

Upload your picture to the internet using TinyPic. Copy the code under "IMG Code for Forums and Message Boards". Paste that into your post and reformat it like this: [ IMG = the address ] without spaces.

Many thanks, Deucekies.



This is an output of my computer model that applies to a player who is tracking kings and aces in other players hands. This particular table provides the change in EV for splitting Aces vs. not splitting Aces as a function of the number of Kings and Aces that you have seen assuming you have looked at two additional player's hands. The outputs for 1-5 additional hands are similar.

Because the jokA-98632 hand has an Ace and a joker, you are already at one ace (and zero kings) seen on this table. Thus the cases with zero aces are meaningless mathematics. Blue indicates those scenarios in which the change in EV for splitting Aces is positive.

The model has explicit algebraic formulas for the probability of about 250,000 different 7-card hands as a function of the composition of the card deck. It assumes the dealer uses the Trump Casino House Way.

I am writing a book about optimum strategy for playing pai gow poker that plows new ground by factoring in the cards that an alert player can see in the 7-card hands that nearby players hold. I have over 200 pages finished.

I'm happy to answer any questions about this (although as a new forum member in his first 30 days, the Wizard has imposed a strict limit on the number of my posts, so please keep this in mind)

Interesting findings. So with all the aces out, it's good to split aces in that case, but NOT if all the kings are also out?

Yes, although I hadn't really thought about that before. Here is what I think is going on:

When both the top ranks (aces and kings) are gone from the deck, there are only eleven ranks left in the deck. This reduces the number of No Pair hands (without a straight or flush) that the dealer can hold from 16.08% to 10.3%, making the A-High backhand less powerful when you split the pair of aces. I believe this is shifting the trade-off back in favor of NOT SPLITTING the pair of Aces.

Another consideration that I looked into: with the joker and all of the Aces and Kings 'busy' the dealer will never split his 2 Pair hands! That's because the dealer's 7-card Two Pair hand will never have higher than a Queen singleton. I expected that this would lead to a large reduction in the number of One Pair hands in the back. Thus, I thought that the winning percentage of the Pair of Aces in the backhand would be significantly reduced by the absence of Aces and Kings from the deck. This would have have created additional incentive to split the aces. Surprisingly, however, this is not true -or at least its not a big effect. This is because the absence of Aces and Kings also leads to significantly higher probabilities of One Pair hands being formed in the dealer's 7-card hand. The higher probability of one-pair 7-card hands just about offsets the fact that 2-pair 7-card hands are never split. Thus, the winning percentage of a pair of Aces in the backhand is not greatly affected by the absence of Kings and Aces.

By the way, I can't express how delighted I am to have discovered this forum and to find people like you to talk to about this. It is such a relief. I have been working on this for 4 years and none of my family and friends have an interest or can understand what I've been doing!

Quote: gordonm888

Another consideration that I looked into: with the joker and all of the Aces and Kings 'busy' the dealer will never split his 2 Pair hands! That's because the dealer's 7-card Two Pair hand will never have higher than a Queen singleton. I expected that this would lead to a large reduction in the number of One Pair hands in the back.

Think about that one again. With the aces and kings gone, the dealer will virtually ALWAYS split two pair, since they would often only keep two pair together if they could put at least a king up. And this would significantly increase the potency of a pair of aces down for the player.

Duh! Sorry. Brain fart. You are absolutely right. I must be tired.

I will slow down and try to analyze this properly. Let's remove all the aces and kings (and the joker) from the deck and check how the frequency of seven card hands changes from a fresh deck.

Frequency of 7-card Pai Gow Hands going from "Fresh Deck" to "Aces, Kings, Joker Absent"

No Pair 16.08% --> 10.3%
One Pair (No Flush, Straight) 41.7% --> 36%
Two Pair 22.3% --> 27.8%
Three Pair 1.9% --> 3.0%
Three of a Kind 5.0% --> 6.1%
Straight and flush* 10.0% --> 8.6%
Boat** 2.7% --> 4.2%
Quads and Higher 0.2% --> 0.4%

*Straights and flushes with No Pair or One Pair
** includes hands with Two Three-of-a kinds and Trips+ Two Pair

Note we have a 10% chance of getting a Queen-High No Pair or lower. Ugh!

Now let's check the frequency of times that a pair is played in the back 5-card hand.

Fresh Deck: Dealer plays one pair+ 3 singletons in the backhand 59.1% of the time
A,K, Joker missing: Dealer plays one pair+ 3 singletons in the backhand 63.8% of the time -which equals (100% of the One Pair hands plus 100% of the Two Pair hands which all split).

So, the number of One Pair hands in the backhand is not greatly different even though the Aces, Kings and joker are all absent, which is what I said (or was trying to say.)