which is higher EV for pai gow poker...
If given the choice of a Set + Pair or TwoPair + Pair, I'll take the Set + pair every time.
Quote: sodawaterwell my point is that each of the hands you get is random. i think i would rather have a random 3-pair hand rather than a random full house. my reasoning being that with 3 pair, you play the low 2-pair in the back and your front hand is very likely to be a high pair. with a random full house you can push a lot as half the pairs will be low pairs.
I would agree with your analysis. But I would guess they are quite similar, as your two pair will lose to all dealer three of a kinds, and some dealer two pairs. That would partially offset the added value of your having a higher pair in the low hand.
Your back hand, whether it be trips or two pair, is always vulnerable to something better. How often is the "something better" going to merely be trips rather than a straight or flush or merely a better two pair/trips?
If the dealer's hand is weak, it's irrelevant if you have two pair or trips in back, but important that you have a stronger front.
Having three pair means you're always playing a high pair up front, which has a high percentage of winning, making your over-all win percentage higher.
A random full house would average to a set of 8's with a pair of 8's. Impossible I know.
A random 3 pair would average to around 8's and 5's with a pair of J's.
So comparing those two hands and using the WOO power chart, the full house is slightly better.
I didn't really factor in the joker, which helps 3 pair's case more. My official guess is that it's really really close, with a slight edge to full house.
Of course, I'll take the boat because I usually throw $1-$2 on the bonus every hand.
Quote: DJTeddyBearThree pair. No question about it.
The more I think about it, the more I am convinced it is three pair. With 5 aces in the deck the top pair will disproportionally be aces. Also, the distribution will NOT be equal in the top pair. You will have far more kings up than any lower pair, far more queens up than any lower pair, and NEVER deusces or treys, and rarely 4,s 5's or 6's. I would bet that you would either have q's, k's, or aces up top well in excess of 50% of the time.
Quote: DJTeddyBearThree pair. No question about it.
This is not based on math or anything but personal experience, which is of course, limited to a very finite number of hands. An expert such as Paigowdan, mustangsally, or even the Wiz himself will have the statistically correct answer shortly...
However, I LOSE or push on 3 pair WAY more often than I win on them. It must have something to do with card distribution, but I can honestly say that if I get 3-pair in a session on average two or three times, I only win once on it. I have pushed more times than I can count, and I've had some notable losses on them, mainly a pair of Aces in the low hand that lost to a dealer full house, three of a kind in the back and aces up.
I usually half-joke with the dealer when I get three pair that I hope to get a push. He/She will say "Nah, that's a winning hand" only to pull a straight or flush or higher two pair, or a higher pair up front. They just shake their head... me, I'm glad when I do push that hand.
Give me a full house EVERY day.
Quote:I didn't really factor in the joker, which helps 3 pair's case more. My official guess is that it's really really close, with a slight edge to full house.
The joker also helps the FH hand. If the front pair is 2's and the player has the joker (or ace) and a king the right play is to play the boat in the back. I would even maybe switch it to play AQ or even replace 3's or 4's with AK and play the boat in the back.
WoO Relative Strength of these hands is 0.95 * (5-card * 2-card) - (1-five * 1-two)
8877x + TT =0.69128 - 0.01430 = 0.67698
888xy + 77 = 0.71652 - 0.01720 = 0.69932
Quote: 98ClubsYah, Full House + AK when the pair is 22, 33, or 44. Agreed there. This would cover the Dealer Straight or Flush... as usually no pair gets played. Three Pairs seem to average to 8877x + 99. Five pairs are below this as 22 33 44 55 66 and five pairs are above TT JJ QQ KK AA. If I allow for the 5th Ace... maybe 7788x + 10-10. Full House by this rationale is 888xy + 77. So it looks like 3 pairs has the higher 2-card hand.
WoO Relative Strength of these hands is 0.95 * (5-card * 2-card) - (1-five * 1-two)
8877x + TT =0.69128 - 0.01430 = 0.67698
888xy + 77 = 0.71652 - 0.01720 = 0.69932
Not a good analysis. The average pair played alone in 3 pairs will be far higher than tens, likely at least queens. Not even including the joker as the fifth ace, the odds of the first pair not being a queen, king or ace is 10/13, the second 9/12, the third 8/11. That occurs 42% of the time only. If you add the ace in the math becomes harder for me to do, but likely only a third of three pairs hands will NOT have at least a queen high pair.
1.) For the ranks of 2 through King there are 220 combinations of 3-pair as 22-33-44 up to JJ-QQ-KK
a.) There are 6 ways for each pair to be arrainged by suit 6^3 ways in total = 216
b.) The 7th card cannot make a Full House + Pair hand... that eliminates 6 cards from the remaining 47 for a total of 41
c.) The subtotal of these combinations is 220*216*41 = 1,948,320
2.) Ace-high 3 pairs have 10 ways of making a pair of Aces, the other two ranks have 6 ways each 10*6*6 = 360 ways
a.) There are 66 rankings for the other two pairs from 22-33 up to QQ-KK.
b.) Rule 1.) b.) is enforced leaving 40 cards allowed as the 7th card
c.) Straights, Flushes, and Straight-Flushes are allowed as a 3-Pair hand, since generally most House Ways will make such hand play as 3-pair (the Ace-Joker is 2-card hand)
d.) The subtotal of the Ace combinations is 360*66*40 = 950,400
3.) The total of all 3-pair combinations is therefore 2,898,720
a.) Subtract the Ace-high 3-pairs (950,400)
b.) Subtract the King-high 3-pairs (55*6^3*41 = 487,080)
c.) The remainder is 1,461,240 3-pair hands ranked Queen high or less. This equals 50.41%
I think 99JJx + QQ is 50.1% The Relative Strength of 99JJx/QQ as in my previous post is 0.73739
Quote: 98ClubsI see your point... I did a better(?) analysis as follows
1.) For the ranks of 2 through King there are 220 combinations of 3-pair as 22-33-44 up to JJ-QQ-KK
a.) There are 6 ways for each pair to be arrainged by suit 6^3 ways in total = 216
b.) The 7th card cannot make a Full House + Pair hand... that eliminates 6 cards from the remaining 47 for a total of 41
c.) The subtotal of these combinations is 220*216*41 = 1,948,320
2.) Ace-high 3 pairs have 10 ways of making a pair of Aces, the other two ranks have 6 ways each 10*6*6 = 360 ways
a.) There are 66 rankings for the other two pairs from 22-33 up to QQ-KK.
b.) Rule 1.) b.) is enforced leaving 40 cards allowed as the 7th card
c.) Straights, Flushes, and Straight-Flushes are allowed as a 3-Pair hand, since generally most House Ways will make such hand play as 3-pair (the Ace-Joker is 2-card hand)
d.) The subtotal of the Ace combinations is 360*66*40 = 950,400
3.) The total of all 3-pair combinations is therefore 2,898,720
a.) Subtract the Ace-high 3-pairs (950,400)
b.) Subtract the King-high 3-pairs (55*6^3*41 = 487,080)
c.) The remainder is 1,461,240 3-pair hands ranked Queen high or less. This equals 50.41%
I think 99JJx + QQ is 50.1% The Relative Strength of 99JJx/QQ as in my previous post is 0.73739
Thanks for doing the work, 98clubs. I think then your new analysis shows 3 pair is slightly better than the full house, correct?
Quote: 98ClubsI see your point... I did a better(?) analysis as follows
1.) For the ranks of 2 through King there are 220 combinations of 3-pair as 22-33-44 up to JJ-QQ-KK
a.) There are 6 ways for each pair to be arrainged by suit 6^3 ways in total = 216
b.) The 7th card cannot make a Full House + Pair hand... that eliminates 6 cards from the remaining 47 for a total of 41
c.) The subtotal of these combinations is 220*216*41 = 1,948,320
2.) Ace-high 3 pairs have 10 ways of making a pair of Aces, the other two ranks have 6 ways each 10*6*6 = 360 ways
a.) There are 66 rankings for the other two pairs from 22-33 up to QQ-KK.
b.) Rule 1.) b.) is enforced leaving 40 cards allowed as the 7th card
c.) Straights, Flushes, and Straight-Flushes are allowed as a 3-Pair hand, since generally most House Ways will make such hand play as 3-pair (the Ace-Joker is 2-card hand)
d.) The subtotal of the Ace combinations is 360*66*40 = 950,400
3.) The total of all 3-pair combinations is therefore 2,898,720
a.) Subtract the Ace-high 3-pairs (950,400)
b.) Subtract the King-high 3-pairs (55*6^3*41 = 487,080)
c.) The remainder is 1,461,240 3-pair hands ranked Queen high or less. This equals 50.41%
I think 99JJx + QQ is 50.1% The Relative Strength of 99JJx/QQ as in my previous post is 0.73739
There is no way that 99/jj/qq is the average 3 pair hand.
Also, with the 5 aces - 888 / 77 is a bit low for average full house hand. I think it is okay to use 888 / 88 for the average even though that hand cannot exist. Since there are 5 aces 888 / 99 is a better estimate than 888 / 77.
Quote: FinsRuleThere is no way that 99/jj/qq is the average 3 pair hand.
Also, with the 5 aces - 888 / 77 is a bit low for average full house hand. I think it is okay to use 888 / 88 for the average even though that hand cannot exist. Since there are 5 aces 888 / 99 is a better estimate than 888 / 77.
Agree--- but even if it is 99/44 QQ I think that that still beats the full house. The three of a kind part of the full house will be disproportionately aces, though. But the number of times the pair will be deuces, or threes, is significant, and that number will be zero for the three pair hand. I did once get 22/33/44, but that is also very rare...
Is there someone skilled enough in the math to come up with the exact answer? Or is that too complicated and time consuming?
Quote: SOOPOOAgree--- but even if it is 99/44 QQ I think that that still beats the full house. The three of a kind part of the full house will be disproportionately aces, though. But the number of times the pair will be deuces, or threes, is significant, and that number will be zero for the three pair hand. I did once get 22/33/44, but that is also very rare...
Is there someone skilled enough in the math to come up with the exact answer? Or is that too complicated and time consuming?
I did 88 / 55 / JJ against 888 / 88 and the full house won.
If you make it 99 / 44 / QQ against 888 / 88 or 888 / 99 - it's going to be really close, but I still think it'll be slightly tilted to full house. But maybe 3 pair would do better when simulating every single 3 pair possibility, not just the average.- WOO is blocked on my work computer so I can't do the calcs.
My guess is the math is way too time consuming to be worth figuring out a purely academic question.
I know the Wizard has said that the Joker really makes Pai Gow Poker calculations more complicated.
if we randomly receive 3 pairs (including aces even though getting aces would be more common): 0.698353291031954
Edit to correct braino.
And can you explain what the .659 and .698 numbers mean. Thanks.
Quote: FinsRuleCan you give some background behind how you calculated those numbers?
And can you explain what the .659 and .698 numbers mean. Thanks.
Using the average WoO Relative Strength ( 0.95 * (5-card * 2-card) - (1-five * 1-two)) for each of the 156 different full houses and 286 different 3 pairs.
Note that I edited my original post due to a braino :)
Quote: FinsRuleI did 88 / 55 / JJ against 888 / 88 and the full house won.
If you make it 99 / 44 / QQ against 888 / 88 or 888 / 99 - it's going to be really close, but I still think it'll be slightly tilted to full house. But maybe 3 pair would do better when simulating every single 3 pair possibility, not just the average.- WOO is blocked on my work computer so I can't do the calcs.
My guess is the math is way too time consuming to be worth figuring out a purely academic question.
I know the Wizard has said that the Joker really makes Pai Gow Poker calculations more complicated.
The page with the math information is blocked, but the chat site is available? Workplace governance win!! :-)
I should report this to Human Resources.
Oh wait, that's me, and I can't do anything about it.
Quote: mipletUsing the average WoO Relative Strength ( 0.95 * (5-card * 2-card) - (1-five * 1-two)) for each of the 156 different full houses and 286 different 3 pairs.
Note that I edited my original post due to a braino :)
Did you take into account that each of the 286 3 pairs does NOT come up equally? That to me is the entire point of this academic discussion.... Just as the most simple example, 22,33,AA will come up far more often than 22,33,44.
Full Houses
-------------
Average low hand is 19.1% of the way between 88 and 99
Average high hand is 53.7% of the way between 888xy and 999xy
Three Pairs
-------------
Average low hand is 81.6% of the way between JJ and QQ
Average high hand is 21% of the way between 88XXy and 99XXy
The low hand has 91 possible ranks, ranging from unsuited 23 all the way up to a pair of Aces.
The high hand has fewer than 7,463 possible ranks (the standard 7,462 plus one more for Five Aces). Some 5-card hands are not possible, such as 23457 unsuited, but let's ignore these for now and go along with the 7,463.
With Three Pairs, the average low hand is 3.625 ranks higher than the Full House's average low hand, an improvement of 3.625/91 = 4% of the range of possible low hands.
With Full Houses, the high hand's average rank is 858 ranks higher than the Three Pair's average high hand, an improvement of 858/7463 = 11.5% of the range of possible high hands. This is before eliminating the impossible high hands, which if we do, the 7463 would become smaller and therefore the 11.5% would become larger.
Since the Full House boosts the average high hand by a greater percentage than the Three Pair boosts the average low hand, I think the Full House is the better starting hand.
QQ + (99 wirh 88 or less) averages is 0.7161
Miplet's random three-pairs is between these two.
EDITED here for other responses...
My calculation does take that into account, thats why there are significantly more AA high three pairs (950,400) than any other X-high three pair hands. Note that with only the KK high three pairs added in, that nearly HALF of all possibilities are accounted. This would leave the other half ranked as QQ high three pair or less. The Confusing (and understandably so) Issue is "The Relative Strength" of the hand. This proceeds from a different point of view entirely. Is JJ/TT/99 a better hand by Relative Strength than QQ/33/22 ? The answer is yes even though QQ high out-ranks JJ high. The other two pairs contribute to the overall Relative Strength.
Quote: JBLooking only at 7-card hands that contain exactly 3 pair or exactly a full house (hands with trips+two pair, four of a kind, and five of a kind were skipped), here are the results:
Full Houses
-------------
Average low hand is 19.1% of the way between 88 and 99
Average high hand is 53.7% of the way between 888xy and 999xy
Three Pairs
-------------
Average low hand is 81.6% of the way between JJ and QQ
Average high hand is 21% of the way between 88XXy and 99XXy
The low hand has 91 possible ranks, ranging from unsuited 23 all the way up to a pair of Aces.
The high hand has fewer than 7,463 possible ranks (the standard 7,462 plus one more for Five Aces). Some 5-card hands are not possible, such as 23457 unsuited, but let's ignore these for now and go along with the 7,463.
With Three Pairs, the average low hand is 3.625 ranks higher than the Full House's average low hand, an improvement of 3.625/91 = 4% of the range of possible low hands.
With Full Houses, the high hand's average rank is 858 ranks higher than the Three Pair's average high hand, an improvement of 858/7463 = 11.5% of the range of possible high hands. This is before eliminating the impossible high hands, which if we do, the 7463 would become smaller and therefore the 11.5% would become larger.
Since the Full House boosts the average high hand by a greater percentage than the Three Pair boosts the average low hand, I think the Full House is the better starting hand.
Thanks, JB. But your analysis implies that each of the hands comes up proportionally. We all know that that is not true. Simplest example... AK is very common, A2 is very rare. I would guess the added EV from changing from A2 to A3 is almost zero, but the added EV from changing from AK to 22 (as player) is HUGE. Am I correct?
Quote: SOOPOOThanks, JB. But your analysis implies that each of the hands comes up proportionally. We all know that that is not true. Simplest example... AK is very common, A2 is very rare. I would guess the added EV from changing from A2 to A3 is almost zero, but the added EV from changing from AK to 22 (as player) is HUGE. Am I correct?
There is some truth in that, and my answer is that I don't know. Pai Gow Poker analysis is difficult unless you pretend that the dealer and player hands are dealt from separate decks; when you do that, analysis is (relatively) quick and easy, albeit slightly inaccurate.
The ideal way to answer the original question would be to:
A) Determine what strategy the dealer (or opponent) is using
B) Who is banking
C) What does a winning bet pay (is commission prepaid, do they only charge $1 when betting $25, is there no commission, etc.)
D) Analyze all full house hands against all possible opponent hands using the above knowledge
E) Analyze all three pair hands against all possible opponent hands using the above knowledge
F) Compare the results
That could take days to do a thorough analysis, or a couple of hours if cheating (pretending that the two hands are dealt from separate decks). I don't consider my previous answer final unless something like the above has been done (which I'm sorry to say I will not be doing), but I will cautiously stand by it in the meantime.
