moelipp
moelipp
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August 14th, 2015 at 7:22:23 PM permalink
In Pai Gow poker, my wife had a royal flush with a pair of aces, I assume the best hand you could ever have while not the highest bonus (7-card SF). So she collected her 150-1 odds which is probably nowhere near the true odds but that's OK and she won the hand. Later on when she was telling newcomers to the table, they alerted us that if she had a suited King-Queen instead of the aces it would have paid 2,000-1. I immediately thought that that hand can easily become a push if the dealer had two pair, the aces is a better hand, why not warrant a huge bonus? Can you tell me which hand is harder to get, the so-called royal match or the royal with aces up front? Thank you very much!
Mission146
Mission146
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August 14th, 2015 at 10:29:29 PM permalink
Since the question is limited, the easiest way to figure out the answer is to assume the Royal which leaves you 47 cards of which you get two. There are Three Aces and Three Kings and Three Queens in there.

(3/47 * 2/46) =.0027752

Or:

(6/47 * 1/46) = .0027752

Therefore, the two possibilities are equally likely.

You'll notice 3*2=6 and 6*1=6 and the denominators are the same, of course.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
PlayYourCardsRight
PlayYourCardsRight
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August 15th, 2015 at 12:39:12 AM permalink
I am not a math guy, but I disagree that the two are equally possible.

Assuming the royal flush did not include the Joker, this is what I have:

For the royal match, there are three kings and tgree queebs remaining and each can a pair up only with one specific card. Therefore, the odds would be 6/48 * 1*47 or .27 percent.

For the Aces, there are four aces remaining, since the Joker also counts as an ace. Therefore, the odds would be 4/48 * 3/47 or .53 percent.

If the Joker is included as part of the royal and replaces the ace, the odds remain the same for the aces, since four aces remain available.

If the Joker is included, and replaces any other part of the Royal, the odds for aces match the odds for the Royal match, since only 3 aces remain (3/48 * 2/47 or .27 percent). The odds for the Royal match remain the same since you have to have the king and queen of a different suit so there are still only three possible royal matches available.
beachbumbabs
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beachbumbabs
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August 15th, 2015 at 12:28:04 PM permalink
Quote: Mission146

Since the question is limited, the easiest way to figure out the answer is to assume the Royal which leaves you 47 cards of which you get two. There are Three Aces and Three Kings and Three Queens in there.

(3/47 * 2/46) =.0027752

Or:

(6/47 * 1/46) = .0027752

Therefore, the two possibilities are equally likely.

You'll notice 3*2=6 and 6*1=6 and the denominators are the same, of course.



I don't think they're equal probability because of the Joker, which can serve as one of the Aces in the low hand, but can't as either the K or Q to complete the Royal Match.

Also, if it's a natural RF, there are 4 cards (including the Joker) of which any 2 will do, but with the RM, whichever K you get, the Q has to be that suit as well, so there's only 1 match for each of those.

So, it's (4/48*3/47) for a natural RF (*4 suits), plus (4/48*3/47) if the wild RF is using the Joker as an Ace (*3 suits), plus (3/48*2/47) if the RF is using the Joker as anything other than the Ace (*4 subst cards*4 suits) . (Working with 53 card deck incl. bug joker). (Not sure if parts 1 and 2 aren't an incorrect duplication/overlap of the calculation; 1st one could be natural the whole way but does allow for the Joker to be an ace, 2nd one accounts for a hand that doesn't include the natural RF ace). Would appreciate a math guy eval of this.



OTOH, for RF+RM, once all permutations of the RF both natural and wild (*6), that has to be multiplied by (3/48*1/47) since the suit has to match, and I think all that *4 Suits.

I might be making this harder than it needs to be, but I don't think these numbers will be the same, and I think the RF+RM will be rarer.

So, perhaps wasting my time on bad methodology, I ran these numbers.

Aces:
Natural RF =(4/48*3/47)*4 =0.021276596
RF w/J as A =(4/48*3/47)*3 =0.015957447
RF w/J as non-A=(3/48*2/47)*4*4 =0.042553191

Sum = 0.079787234

RF+RM:
Nat or wild RF=(3/48*1/47)*6*4 = 0.031914894
If the House lost every hand, they wouldn't deal the game.
PeeMcGee
PeeMcGee
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August 15th, 2015 at 6:35:18 PM permalink
The number of combinations for the following…

Natural Royal + AA = 4*c(4,2) = 24
Wild Royal w/joker = Ace + AA = 4*c(4,2) = 24
Wild Royal w/joker != Ace + AA = 4*4*c(3,2) = 48

Total number of combinations of Royal and Aces = 24+24+48 = 96


Natural Royal + KQs = 4*3 = 12
Wild Royal + KQs = 5*4*3 = 60

Total number of combinations of Royal Match = 12+60 = 72


Therefore, the Royal Match is the more difficult hand to get.


Disclaimer: I’m about 73% confident in these numbers.
Paigowdan
Paigowdan
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August 16th, 2015 at 12:48:11 AM permalink
This info (most of it) is at the wizard of Odds Pai Gow poker side bet math area.

Mike Shackleford has Royal + Royal match at 72 out of 154,143,080, and has natural Royal + natural aces as 12 out of 154,143,080.
Beware of all enterprises that require new clothes - Henry David Thoreau. Like Dealers' uniforms - Dan.
Mission146
Mission146
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August 17th, 2015 at 7:23:36 PM permalink
I forgot about the Joker, sorry about that.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
moelipp
moelipp
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August 21st, 2015 at 7:40:13 AM permalink
OK, thank you. She had the joker. I thought that since the royal with KQ might not even win the hand, it would be less valuable than the highest possible Pai Gow hand, royal with aces up front.

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