jdominik
jdominik
Joined: Mar 18, 2011
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March 18th, 2011 at 6:16:33 PM permalink
Just wondering - what pai Gow hand will win more then loose or tie? A straight or flush is a great 5 card hand but would loose the 2 card hand. I saw this in Atlantic city where a player had a straight Ace high but had 2-8 on his 2 card hand and thus a push. It seem to me that 2 small pairs win more then 50% of the time. Thoughts?
TIMSPEED
TIMSPEED
Joined: Aug 11, 2010
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March 18th, 2011 at 8:44:05 PM permalink
AAAA*KK
In other words, five Aces and a pair of Kings..unbeatable hand.
Gambling calls to me...like this ~> http://www.youtube.com/watch?v=4Nap37mNSmQ
clarkacal
clarkacal
Joined: Sep 22, 2010
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March 18th, 2011 at 10:07:18 PM permalink
AKQ*10 AA
Royal flush with a pair of aces
FinsRule
FinsRule
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March 18th, 2011 at 10:14:28 PM permalink
Quote: jdominik

Just wondering - what pai Gow hand will win more then loose or tie? A straight or flush is a great 5 card hand but would loose the 2 card hand. I saw this in Atlantic city where a player had a straight Ace high but had 2-8 on his 2 card hand and thus a push. It seem to me that 2 small pairs win more then 50% of the time. Thoughts?



Generally an average high hand is a pair of 9's. An average low hand is KQ
PGBuster
PGBuster
Joined: Jan 15, 2010
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April 14th, 2011 at 3:09:30 PM permalink
IIRC, I read somewhere that Sanford Wong's book says the magic hand is a pair of jacks with an A/8 up.
PapaChubby
PapaChubby
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April 14th, 2011 at 3:32:58 PM permalink
I'm not really sure what the OP's question is, but I stumbled across a conundrum while thinking about it.

I assert that it is impossible to devise a hand which has the same probability of winning, losing and tying. This does not seem intuitively obvious to me, but it seems to be the way the math works.

If the probability of the front hand winning is x, and the probability of the back hand winning is y, then the overall probabilities are:

win: x*y
lose: (1-x)*(1-y)
tie: (1-x)*y + (1-y)*x or 1 - x*y - (1-x)*(1-y)

For win=lose=tie=0.333, I think I determine that x and y must be imaginary. Seems weird to me.
Doc
Doc
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April 14th, 2011 at 3:38:47 PM permalink
Quote: PapaChubby

... tie: (1-x)*y + (1-y)*x or 1 - x*y - (1-x)*(1-y) ...


I think you made an algebra/multiplication error on that line. But the solution does appear to involve complex values of x and y.
Paigowdan
Paigowdan
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April 14th, 2011 at 3:58:40 PM permalink
Quote: PGBuster

IIRC, I read somewhere that Sanford Wong's book says the magic hand is a pair of jacks with an A/8 up.



PG - that's the "median" hand or average hand. The dead Center hand.
Beware of all enterprises that require new clothes - Henry David Thoreau. Like Dealers' uniforms - Dan.
Paigowdan
Paigowdan
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April 14th, 2011 at 4:01:09 PM permalink
Quote: clarkacal

AKQ*10 AA
Royal flush with a pair of aces



Correct, best hand, Royal with Aces on top.

The AAAA*/KK hand could indeed push: dealer could have three pairs 66552/KK, and win the two-side copy KK (Dealer wins copies on each side.)
Beware of all enterprises that require new clothes - Henry David Thoreau. Like Dealers' uniforms - Dan.
ChesterDog
ChesterDog
Joined: Jul 26, 2010
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April 14th, 2011 at 4:52:55 PM permalink
Quote: clarkacal

AKQ*10 AA
Royal flush with a pair of aces



This is unbeatable, but wouldn't it push against the dealer's suited AKQJ10 XY, since the dealer's copied royal flush would beat the player's royal flush?

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