Face Up Pai Gow Poker @ NYNY
The Ace High push rule is gone. The new rule is something like all hands automatically losing if the dealer has a pair of jacks in the 5 card hand (or maybe it was a pair of jacks anywhere in the 7 cards).
Unfortunately I did not have time to head up strip to Circus Circus to investigate.
That would happen 9.4% of the time. Your win rate with dealer Ace high 5 card hand would have been approximately 87.4% so this was a net hit of 0.5*0.094*0.874 = 0.041 on your EV. The factor of 0.5 is meant to account for the fact that this is only one of your two hands that determines the outcome of your wager.
2. In the newly reported version of Faceup Paigow poker I calculate that dealer will have two jacks and five singletons in their hand with a frequency of about 0.052. Normally, you would beat a dealer with exactly 2 jacks and five singletons.in their hand about 50.2% of the time.
I think the hit on player EV with the "automatic loss with two jacks in 5 card hand rule" is approximately ΔEV = 0.5*2*0.052*0.502 = 0.026. The factor of two in that equation is to account for a winning hand being converted into a losing hand, rather than into a push.
So, if I've done this correctly, then this "Dealer two jacks =automatic winner" version of FaceUp PaiGow may have a Player EV that is higher by 0.015 than the original version with the "A High is an automatic Push" rule.
Qualifier: This was an approximate calculation to quickly come up with a ballpark estimate, not a highly rigorous calculation. Also, it is subject to the very large probability that I made an error of some kind in the analysis, so review by other math experts would be appreciated.
Quote: gordonm8881. In the old Face Up Paigow Poker you would push when dealer had an Ace high in his 5 card hand.
That would happen 9.4% of the time. Your win rate with dealer Ace high 5 card hand would have been approximately 87.4% so this was a net hit of 0.5*0.094*0.874 = 0.041 on your EV. The factor of 0.5 is meant to account for the fact that this is only one of your two hands that determines the outcome of your wager.
2. In the newly reported version of Faceup Paigow poker I calculate that dealer will have two jacks and five singletons in their hand with a frequency of about 0.052. Normally, you would beat a dealer with exactly 2 jacks and five singletons.in their hand about 50.2% of the time.
I think the hit on player EV with the "automatic loss with two jacks in 5 card hand rule" is approximately ΔEV = 0.5*2*0.052*0.502 = 0.026. The factor of two in that equation is to account for a winning hand being converted into a losing hand, rather than into a push.
So, if I've done this correctly, then this "Dealer two jacks =automatic winner" version of FaceUp PaiGow may have a Player EV that is higher by 0.015 than the original version with the "A High is an automatic Push" rule.
Qualifier: This was an approximate calculation to quickly come up with a ballpark estimate, not a highly rigorous calculation. Also, it is subject to the very large probability that I made an error of some kind in the analysis, so review by other math experts would be appreciated.
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Also unclear that the reported rule is correct. And your analysis I think ignores two pair hands of say JJ and 22 where the 22 would go on top.
Quote: unJonQuote: gordonm8881. In the old Face Up Pai Gow Poker you would push when dealer had an Ace high in his 5 card hand.
That would happen 9.4% of the time. Your win rate with dealer Ace high 5 card hand would have been approximately 87.4% so this was a net hit of 0.5*0.094*0.874 = 0.041 on your EV. The factor of 0.5 is meant to account for the fact that this is only one of your two hands that determines the outcome of your wager.
2. In the newly reported version of Faceup Paigow poker I calculate that dealer will have two jacks and five singletons in their hand with a frequency of about 0.052. Normally, you would beat a dealer with exactly 2 jacks and five singletons.in their hand about 50.2% of the time.
I think the hit on player EV with the "automatic loss with two jacks in 5 card hand rule" is approximately ΔEV = 0.5*2*0.052*0.502 = 0.026. The factor of two in that equation is to account for a winning hand being converted into a losing hand, rather than into a push.
So, if I've done this correctly, then this "Dealer two jacks =automatic winner" version of FaceUp PaiGow may have a Player EV that is higher by 0.015 than the original version with the "A High is an automatic Push" rule.
Qualifier: This was an approximate calculation to quickly come up with a ballpark estimate, not a highly rigorous calculation. Also, it is subject to the very large probability that I made an error of some kind in the analysis, so review by other math experts would be appreciated.
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Also unclear that the reported rule is correct. And your analysis I think ignores two pair hands of say JJ and 22 where the 22 would go on top.
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You are correct, but we're both uncertain what the rule is. And I just checked and the House Edge on conventional FaceUp Pai Giow Poker is 1.8%, so my projected improvement in player EV of 1.5% (that I reported in the prior post) does seem incredibly unlikely.
However, after a closer look, I now estimate that the "dealer Ace-High Pushes" rule only increased the House Edge by about 0.035 or 3.5%.* So, if the rule "Player loses when dealer has JJ-xyz in the bottom hand" is indeed worth about 2.6%, this would imply that the House Edge of the new variant would be about 0.009 or 0.9%. That still seems too good to be true.
*ΔEV = 0.05 * 0.052 * (0.874-0.126) = 3.5%
1. Regarding the two pair hands, JJ-XX with XX = 22-TT, my previous post included those kind of hands when I estimated that a pair of jacks with 3 singletons would be in the bottom hand with a frequency of 5.2%. That value of 5.2% came from Table 14 of Stanford Wong's "optimal Strategy for Pai Gow Poker" and is Wong's calculated frequency of a pair of jacks (with three singletons) being arranged in the bottom hand. I was indeed guilty of mis-stating what that 5.2% meant in my first post.
2. If we're concerned about occasions when dealer has trip jacks or quad jacks in the bottom hand, well the dealer would be winning those (relatively rare) bottom hands with a high probability in any case so the Automatic Dealer Win rule probably would have little effect.
3. If the rule is interpreted as any dealer 7-card hand with at least 2 jacks (including J-Joker with 5 singletons that are 10 or lower) then such a hand would occur with a frequency in the ballpark of roughly 14%, However, this would make it unclear how to arrange the top 2-card dealer hand to determine the top hand winner - so I still suspect that the requirement in the rules is that the dealer's bottom hand must be arranged naturally as a pair of jacks.
Obviously we need a clarification of what this new rule actually means, and a more rigorous mathematical analysis of the House Edge than what i have done.
https://www.indiangaming.com/ags-dealer-open-pai-gow/
“Pai Gow wager loses when dealer has two or more jacks”
Quote: ams288Okay, apparently it’s called “Dealer Open Pai Gow.”
https://www.indiangaming.com/ags-dealer-open-pai-gow/
“Pai Gow wager loses when dealer has two or more jacks”
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Ouch. So two Jacks and any other five cards, it sounds like. Pair of Jacks, trip Jacks, quad Jacks, two pair with Jacks, all of the above.
