NHT
Posted by NHT
Feb 16, 2016

Pai Gow Poker Question

Hello Wiz: had a question that has been occupying me lately. When playing Pai Gow Poker, how would you calculate the probability of getting 1, 2, 3 or 4 Aces in your hand? In some casinos, Joker is played as an Ace. What's the probability of getting all five Aces then. Many casinos pay big prizes for that hand. Thank you

Comments

Hittem
Hittem Feb 17, 2016

.0007318 chance you'll get five aces.



48 C 2

rdw4potus
rdw4potus Feb 17, 2016

0.000738%, maybe? So, .00000738 chance? 1:135,000, not 1:1,350...

NHT
NHT Feb 17, 2016

Thank you both for answering. What is your methodology or calculating steps. I wanted to figure out the probability for getting 1 Ace first and use the methodology to arrive to more. I understand that to get an Ace is 1 from 48. However, having 7 different hands at a game (as it is dealt in California) make any difference in the calculation?



Lastly, if you could kindly tell me what does C stand for and why the multiplier 2 is used in your posted formula: 48 C 2. Thank you

Wizardofnothing
Wizardofnothing Feb 17, 2016

Your chances of getting an ace are not 1-48 they are 5 in 53

NHT
NHT Feb 18, 2016

So is this approach correct?



1st Ace =1/53 2nd Ace =1/46 3rd Ace =1/39 4th Ace =1/32

Joker =1/53



To get 2 Aces =1/53 × 1/46

To get 3 Aces =1/53 × 1/46 × 1/39

To get 4 Aces =1/53 × 1/46 × 1/39 × 1/32

To get 4 Aces and a Joker =1/53 × 1/46 × 1/39 × 1/32 × 1/53

rdw4potus
rdw4potus Feb 19, 2016

the first ace is 5/53, the second ace is 4/52 (or (5-1)/(53-1)), the third ace is 3/51...