Poker Probablity Question....

Keyzer

Scenario -

6 players at a table, each player dealt 2 cards each to begin with (texas holdem)...

What is the probablity or odds that a pair of jacks, a pair of queens and a pair of kings will be dealt from the intial deal to three of the players at the table....

dave_eire

go about it like this:

the probability of getting dealt any specific pair no matter the number of players is 1/221. this is worked out as follows:

the probability of getting dealt your first card as a jack (or any other specific card for that matter) is 4/52 and getting a second is 3/51. multiply these and you get 12/2652 or 1/221.

so now you need the odds of someone else getting dealt another pair (in this case queens), the odds of that are now 1/204.16 (workings: first card 4/50 chance (50 because 2 cards (the jacks are gone) and 3/49, multipled out = 12/2450 = 1/204.166)

and the last pair --> 1/48 for first card and 1/47 for second 12/2256 = 1/188

so multiplying all those out you have 1/221 * 1/204.166 * 1/188 = 1/8482688.968

which as you can see i quite small!! about 1 in 8.5 million!!

hope it helps

dave

<edit>mistake</edit>

DeVore

er... that seems wrong... My probability theory is rusty but what you seem to have worked out was the odds for EXACTLY three hands coming up EXACTLY Jacks first then QUEENS then KINGS. The odds for what the poster has requested are no where near that high.

I hated probs in college and ironically I now play a lot of poker but the odds arent that specific in the game.

I'll have a think about this when I get home.

DeV.

DeVore

ps: (anecdoteally) I was dealing in the Fitz there a while ago and dealt 99, TT , AA to three players out of seven. The flop came 9 (first guy punches the air) Ten (second hoots and hollers) and the last card on the flop was the Ace of course. Cue explosion from all the players!

DeV.

yaledo

I'll have a go, I have a habit of getting these wrong, so please nitpick.

1.
3 players A B C
The odds for A to get JJ, B to get QQ, C to get KK

4/52 x 3/51 x 4/50 x 3/49 x 4/48 x 3/47
= 1/8482720
2.
3 players A B C
For one of them to get JJ, one of them to get QQ, one of them to get KK

Basically we need (1) above to happen in one of 6 ways
(abc, acb, bac, bca, cab or cba)
so its 1/1413790

3. The original question
For 6 players A B C D E F

We need (2) above to happen to 3 out of the 6 players (20 possibilities for which 3 players out of 6 it could happen to: abc abd abe abf acd etc.)
: 1/70689

Mr. Flibble

Hows about this:

(6 * (3 / 221)) * (5 * (2 / 221) * (4 * (1 / 221))) = 1 in 14991

six players * odds of each player getting KK QQ or JJ
* remaining 5 players * odds of getting remaining two of KK QQ or JJ
* remaining 4 players * odds of getting remaining 1 of KK QQ or JJ

SeaSide

For player A is it not
12/52 x 3/51 as he can accept any of the K, Q, J

For player B it would then be
8/50 x 3/49

and for C
4/48 x 3/47

(6*((12/52)*(3/51)))*(5*((8/50)*(3/49))*(4*((4/48)*(3/47))))=11781

Mr. Flibble

Ah, my maths doesn't take into account that the K Q or J are out of the deck when the next person gets delt to.