Poker question for the intelligent denizens of ET

A

alfonso

Gentlemen (and Ladies, if there are any here), I have a question regarding poker hands for you.

Any help with solving this puzzle will be most sincerely appreciated.

From a deck of 52 cards, the total possible number of hands

= 52!/[(52-5)!.5!] *

or, in other words, C (52, 5) **

That's easy enough.

The total number of hands that include at least one Ace, by my reasoning, would be

Total Possible Hands - Number of Hands containing No Aces,

C (52, 5) - C (48, 5)


My problem is working out how many hands there are that contain
at least one Ace and Exactly one Heart?
(Or, at least one Ace and Exactly 2 hearts, or at least one Ace and exactly 3 diamonds -- the questin is the same)

The total hands with exactly one Heart would be

C (13,1) x C (39, 4) right? (There are 13 Hearts for the first card, and 39 cards for the other four cards)

How do I combine this with the cards containing at least one Ace?

(I get stumped over the situation where the Heart is the Ace of Hearts.)

I know many of you traders are poker buffs, so could you please give me a hand? (:d bad pun!)



* n! means (n) x (n-1) x (n-2) ... x 1
so, 5! = 5 x 4 x 3 x 2 x 1

** C (n, r) means n!/[(n-1)! x r!]

Good Trading!
(and yes, I'm holding overnights! [gulp])
 
Once you have the number of hands for your desired outcomes...

By dividing by the total number of hands, C (52,5), you have the probability of getting at least one Ace and the probability of getting exactly one Heart. Let's call these probabilities A and H. They will be 0 < A, H < 1.

A and H are independent if we define the case of getting the Ace of Hearts as a hand that has both an Ace and a Heart. Once they are independent, the joint probability is just A x H.

If, when you get the Ace of Hearts, you want to require there be <i>another</i> Heart, then the probabilities are dependent and the joint probability is significantly more complicated. Possibility enough so, that numerical simulation would be the way to find the answer.

Let me know if you find any errors in my reasoning.
 
You must log in or register to reply here.
Back
Top