More 5Draw Probability Questions

[SwissPoker]

1. I was wondering how Mike Caro arrives at 2592 combinations to Aces-up in 5draw (no joker) when drawing 3 to 2 aces?

I tried to figure it like this:

pair combinations (3 ranks with 3 cards): 3 * C(3,2) * 45 = 3 * 3 * 45 = 405
pair combinations (9 ranks with 4 cards): 9 * C(4,2)* 45 = 9 * 6 * 45 = 2430

for a total of 2835 two pair combinations

What am I doing wrong?


2. I tried to figure how Mike Caro arrives at 1854 combinations to 3-Aces in 5draw (no joker) when drawing 3 to 2 aces?

I tried to figure it like this:

no pair combinations (3 ranks with 3 cards): 3 * C(3,2) = 3 * 3 = 9
no pair combinations (9 ranks with 4 cards): 4 * C(9,2) = 4 * 36 = 144

What am I doing wrong? And how do I arrive at the exact combinations?

Thanks for any help.

[gaming_mouse]

[ QUOTE ]
1. I was wondering how Mike Caro arrives at 2592 combinations to Aces-up in 5draw (no joker) when drawing 3 to 2 aces?

I tried to figure it like this:

pair combinations (3 ranks with 3 cards): 3 * C(3,2) * 45 = 3 * 3 * 45 = 405
pair combinations (9 ranks with 4 cards): 9 * C(4,2)* 45 = 9 * 6 * 45 = 2430

for a total of 2835 two pair combinations

What am I doing wrong?

[/ QUOTE ]

I don't follow what you are doing. But I get:

12 non-ace ranks, each with 6 (4 choose 2) possible pairs. Each of those pairs can have any of 44 possible kickers. Thus:

12*6*44 = 3186

which does not agree with either answer. Am I misunderstanding the question? I am counting the number of 3-card draws you can pull which, along with the aces you already hold, will give you 2 pair.


[ QUOTE ]
2. I tried to figure how Mike Caro arrives at 1854 combinations to 3-Aces in 5draw (no joker) when drawing 3 to 2 aces?

[/ QUOTE ]

2*( (48 choose 2) - 12*6) = 2112

[Siegmund]

Three cards, each presumably of a different rank, have hit the muck.

gaming_mouse has ignored these 3 cards entirely, and given the odds for being dealt 2 pair if your first two cards are aces. The OP took these 3 cards into account only for pairing and not for kickers.

For a rank you discarded, there are 3 possible pairs and *42* possible kickers (the kicker can't make a full house and it can't be one of the other two cards you've already mucked.)

For the other nine ranks, there are 6 possible pairs and *41* kickers.

3*3*42 + 9*6*41 = 378+2214=2592.

To draw 3 aces: there are 2 aces and 45 non-aces left in the deck...

2*45C2 = 2*990 = 1980

...but of those 45C2 odd cards, 3*3+9*6 of them would make a full house. 2* (990-63) = 2 * 927 = 1854.

[SwissPoker]

Thank you very much for your answers.

I still wonder how you calculated the kickers.

For 3 ranks *42* possible kickers: 47 cards - 2 aces - the 3 discards?
For 4 ranks *41* possible kickers: 47 cards - 2 aces - ?

Thanks for your help.

P.S. By the way, do you know of a good book or internet source, were all the math according to poker is explained. I have several books, but none of which does all the probability in depth.

[Siegmund]

Your kicker has to be one of the eleven ranks that matches neither your ace nor your drawn pair. There were 44 such cards in the deck at the start, but you mucked 3 cards already - either 41 or 42 possible blanks are left, according to whether your one of your discards was the same as your new pair.