Probability of 4 players on 10 player table dealt A-K offsuit?
March 2nd, 2016 at 5:46:38 PM
permalink
What is the probability that at a 10 player hold-em table, four of the players are all dealt A-K offsuit?
I came up with ~40,000,000:1 against, but I'm not at all confident in how I did it.
Help, poker probability experts?
Also, any recommendation for books that deal in depth with the probability/combinatorics for poker?
Thanks!
I came up with ~40,000,000:1 against, but I'm not at all confident in how I did it.
Help, poker probability experts?
Also, any recommendation for books that deal in depth with the probability/combinatorics for poker?
Thanks!
March 2nd, 2016 at 6:18:16 PM
permalink
1 get 1 in 41,807,675. Can anyone confirm or deny?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 2nd, 2016 at 6:23:01 PM
permalink
I also get 1 in 41,807,675. Click on the button for how I got it.
The total number of sets of 10 hands is (52 x 51) / 2 x (50 x 49) / 2 x (48 x 47) / 2 x ... x (34 x 33) / 2.
Of these, four are AK. There are only nine ways to have four hands of AK that are all offsuit: match the Aces of Spades, Hearts, Clubs, and Diamonds with the Kings of:
H S D C
H C D S
H D S C
C S D H
C D S H
C D H S
D S H C
D C S H
D C H S
There are 10 ways to assign the hand with the Ace of Spades; for each of these, there are 9 for the Ace of Hearts hand; for each of these, 8 for the Ace of Clubs hand; for these, 7 for the Ace of Diamonds hand. There are (44 x 43) / 2 x (42 x 41) / 2 x ... x (34 x 33) / 2 ways to make the six remaining hands from the remaining 44 cards.
The probability = (9 x 10 x 9 x 8 x 7) / ((52 x 51) / 2 x (50 x 49) / 2 x (48 x 47) / 2 x (46 x 45) / 2) notice that the (44 x 43) / 2 and lower terms appear in both the total number of deals and the number with four offsuit AKs, so they cancel each other out.
This is (9 x 10 x 9 x 8 x 7 x 16) / (52 x 51 x 50 x 49 x 48 x 47 x 46 x 45) = 1 / 41,807,675.
The total number of sets of 10 hands is (52 x 51) / 2 x (50 x 49) / 2 x (48 x 47) / 2 x ... x (34 x 33) / 2.
Of these, four are AK. There are only nine ways to have four hands of AK that are all offsuit: match the Aces of Spades, Hearts, Clubs, and Diamonds with the Kings of:
H S D C
H C D S
H D S C
C S D H
C D S H
C D H S
D S H C
D C S H
D C H S
There are 10 ways to assign the hand with the Ace of Spades; for each of these, there are 9 for the Ace of Hearts hand; for each of these, 8 for the Ace of Clubs hand; for these, 7 for the Ace of Diamonds hand. There are (44 x 43) / 2 x (42 x 41) / 2 x ... x (34 x 33) / 2 ways to make the six remaining hands from the remaining 44 cards.
The probability = (9 x 10 x 9 x 8 x 7) / ((52 x 51) / 2 x (50 x 49) / 2 x (48 x 47) / 2 x (46 x 45) / 2) notice that the (44 x 43) / 2 and lower terms appear in both the total number of deals and the number with four offsuit AKs, so they cancel each other out.
This is (9 x 10 x 9 x 8 x 7 x 16) / (52 x 51 x 50 x 49 x 48 x 47 x 46 x 45) = 1 / 41,807,675.
March 2nd, 2016 at 6:31:36 PM
permalink
Does the sequence of receiving the cards make a difference? First round P1: As, P2:Ah, P3:Ac, P4:Ad vs. P1:As, P2:Kd, P3:Ah, P4:Kc. In the first instance, when all the aces are gone in the first round, you only have to worry about catching one of any three remaining good cards to start the second round; while in the second instance, there are only one "good" card left to catch.
Or does it all come out in the wash?
Or does it all come out in the wash?
Simplicity is the ultimate sophistication - Leonardo da Vinci
March 2nd, 2016 at 6:47:00 PM
permalink
Quote: AyecarumbaDoes the sequence of receiving the cards make a difference? First round P1: As, P2:Ah, P3:Ac, P4:Ad vs. P1:As, P2:Kd, P3:Ah, P4:Kc. In the first instance, when all the aces are gone in the first round, you only have to worry about catching one of any three remaining good cards to start the second round; while in the second instance, there are only one "good" card left to catch.
Or does it all come out in the wash?
The way I did it was to calculate the probability the first four players get any ace/king. Then I applied a probability of 9/24 that they are all off-suit. Then I multiplied by combin(10,4)=210 one could choose four positions out of 10.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
March 2nd, 2016 at 11:06:23 PM
permalink
Quote: WizardThe way I did it was to calculate the probability the first four players get any ace/king. Then I applied a probability of 9/24 that they are all off-suit. Then I multiplied by combin(10,4)=210 one could choose four positions out of 10.
I suspect this answers my question, but is there any difference in dealing players "packets" of consecutive cards instead of multiple rounds of single cards? If there are no differences, wouldn't it be faster, and therefore more profitable, for poker rooms to deal hands as packets instead of singles? If so, why not do it this way?
Simplicity is the ultimate sophistication - Leonardo da Vinci
March 3rd, 2016 at 1:02:39 AM
permalink
No difference.
I'd guess because players don't like that // old school. Poker isn't a carnival game. (Although with some players on this board, it might turn into one. Ahem *cough cough*. You know who you are!)
I'd guess because players don't like that // old school. Poker isn't a carnival game. (Although with some players on this board, it might turn into one. Ahem *cough cough*. You know who you are!)
