WHat are the odds of AA AA and KK being dealt 10 handed?
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WHat are the odds of AA AA and KK being dealt 10 handed?
Not positive on this, but I think I remember reading it is like 5000:1.
Sucks to be KK there.
with the preflop action between the other two he might be able to let his KK go, so maybe it's actaully KK lucky day
Better than being up against 1 AA, no chance for set over set.Quote:
Originally Posted by midas06
KK vs 2 with AA is doing much better than KK vs a single AA or ever KK vs AA and QQ.
Okay, let's break this down:Quote:
Originally Posted by Triptan3s
What are the odds that one of ten players will be dealt AA?
There are 4 aces in the deck out of 52 total cards, so there is a 4/52 or 1/13 chance that a player will be dealt the ace as his first card. His chance of being dealt another ace for his second card is pretty easy: 3/51 because we know he has an ace already, so there are 3 aces and 51 cards left in the deck.
Since both of these have to occur for him to have AA, we multiply the probabilities of each and get:
The chance of being dealt pocket aces is 1/13 * 3/51 = 3/663 or 1/221.
Since any of the ten players can get the first AA, we have to find the probability that one of them does. We cannot just divide by ten however, since that's not how probabilities work.
When something has multiple chances to occur, it is easier to determine the chance of it not happening and subract that from one. Since you have a 1/221 chance of getting AA, you must have a 220/221 chance of not getting it. So the chance that none of ten players will be dealt AA is the chance of one player not getting AA: (220/221) to the tenth power.
So the chance of someone at a 10 player table being dealt aces is
1-(220/221)^10 = 0.0443 or 4.43% of the time.
Now, what are the odds that one of the remaining players will get KK?
There are 50 cards left (2 aces are gone) and 4 kings remaining.
So the odds of a single player getting KK becomes:
4/50 * 3/49 = 12/2450, or about 1 in 204. Note this is slightly better than the normal odds since we know 2 non-kings have been removed. Otherwise we would get the same probability for any pocket pair as we did for AA: 1 in 221.
The odds of not getting KK are then (2450 -12) / 2450 = 2438/2450
We have 9 remaining players who could get KK, so to get the probability that one does we again subtract from one, but multiply to the ninth power:
1 - (2438/2450)^9 = 0.0432
If we wanted to get the odds of both a player getting AA and another getting KK we multiply the respective probabilities:
0.0443 * 0.0432 = 0.0019 or a 0.19% chance... or about 1 in 523.
Finally, what are the odds that one of the remaining players will get the last AA?
There are only 2 aces left out of 48 cards, so now the chance of getting AA is 2/48 * 1/47 or 1/1128
There are 8 remaining players, so our probability that one will get the other AA is:
1-(1127/1128^8) = 0.0071
Multiplying the probabilities again:
0.0443 * 0.0432 * 0.0071 = 0.000014
Last to change probabilities to odds:
subtract the probability "for" from 1 to get the probability "against":
0.999986 to 0.000014
divide both sides by the probability "for":
73,595 to 1
THanks Demi
POKERSTARS GAME #3668066632: HOLD'EM NO LIMIT ($0.50/$1.00) - 2006/01/17 - 16:25:50 (ET)Quote:
Originally Posted by midas06
Table 'Sulaphat II' Seat #4 is the button
Seat 1: molahlah ($10.15 in chips)
Seat 2: Lukieplaya ($172.60 in chips)
Seat 3: Rockymv ($117.30 in chips)
Seat 4: Junebug1979 ($97 in chips)
Seat 5: gilbert3040 ($73.20 in chips)
Seat 6: JNacci ($175.65 in chips)
Seat 7: Capstone66 ($46.70 in chips)
Seat 8: Brickeldred ($35.60 in chips)
Seat 9: Min9 ($107.70 in chips)
gilbert3040: posts small blind $0.50
JNacci: posts big blind $1
*** HOLE CARDS ***
Dealt to Lukieplaya [Kc Kd]
Capstone66: folds
Brickeldred: calls $1
Min9: folds
molahlah: folds
Lukieplaya: raises $3 to $4
Rockymv: calls $4
Junebug1979: folds
gilbert3040: folds
JNacci: folds
Brickeldred: calls $3
*** FLOP *** [7d Kh 4d]
Brickeldred: checks
Lukieplaya: bets $7
Rockymv: raises $14 to $21
Brickeldred: raises $10.60 to $31.60 and is all-in
Lukieplaya: calls $24.60
Rockymv: calls $10.60
*** TURN *** [7d Kh 4d] [3s]
JNacci said, "uh oh"
Lukieplaya: bets $40
JNacci said, "This could be ugly"
Rockymv: raises $41.70 to $81.70 and is all-in
Lukieplaya: calls $41.70
*** RIVER *** [7d Kh 4d 3s] [Td]
*** SHOW DOWN ***
Lukieplaya: shows [Kc Kd] (three of a kind, Kings)
Rockymv: mucks hand
Lukieplaya collected $163.40 from side pot
Brickeldred: mucks hand
Lukieplaya collected $105.30 from main pot
JNacci said, "BIG money"
Brickeldred leaves the table
*** SUMMARY ***
Total pot $271.70 Main pot $105.30. Side pot $163.40. | Rake $3
Board [7d Kh 4d 3s Td]
Seat 1: molahlah folded before Flop (didn't bet)
Seat 2: Lukieplaya showed [Kc Kd] and won ($268.70) with three of a kind, Kings
Seat 3: Rockymv mucked [Ah Ad]
Seat 4: Junebug1979 (button) folded before Flop (didn't bet)
Seat 5: gilbert3040 (small blind) folded before Flop
Seat 6: JNacci (big blind) folded before Flop
Seat 7: Capstone66 folded before Flop (didn't bet)
Seat 8: Brickeldred mucked [As Ac]
Seat 9: Min9 folded before Flop (didn't bet)
Thx for pwning fellow FTR memebers.
Wow, thanks Demi--very clear explanation.
Thanks for the great post demi.
I've had KK vs AA vs AA happen before, as well as set on set on set! Online poker is clearly riggd......
Quote:
Originally Posted by Lukie
Jesus.... Now calculate the odds of one person being dealt KK, two others being dealt AA, and NEITHER person with AA re-raising pre-flop. Crazy hand.
And also calculate 2 of the 3 players in the hand being FTR members.
I have a hard time wrapping my brain around Demi's method. I like counting methods better, so I'm going to throw out a second way to solve the problem (although it's more complicated than I initially thought).
Terminology:
x! = x(x-1)(x-2)...(2)(1). (a.k.a. factorial)
5! = 5 * 4 * 3 * 2 * 1 = 120
x!! = x(x-2)...(3)(1). (a.k.a double factorial, will always be odd for us)
5! = 5 * 3 * 1 = 15
(x, y) = x! / y!(x-y)! (a.k.a combination or unordered permutation)
(4, 2) = 4!/(2! * (4-2)!) = 24/(2*2) = 6
in words, choose 2 out of 4 (e.g. :heart:, :diamond:, :spade:, :club:)
:heart::diamond:, :heart::spade:, :heart::club:, :diamond::spade:, :diamond::club:, :spade::club:
for 6 pairs (note: :heart::spade: = :spade::heart:)
How many ways to deal AA, AA, KK?
How many ways to deal 10 people cards?
(52, 20) is the total number of ways to pick 20 cards out of a full deck
(20-1)!! is the number of ways to split these 20 cards into 10 hands
Multiply for the total number of hands:
(52, 20) * (20-1)!! = 8.2492 * 10^22
(this is the denominator for our final probability)
Now for the numerator:
There are 3 ways to get 2 pairs of As, (:heart::diamond:, :spade::club:), (:heart::spade:, :diamond::club:), (:heart::club:, :diamond::spade:)
There are (4, 2) = 6 ways to get 1 pair of Ks
We have used 6 cards, now we need 7 more hands:
(46, 14) pick 14 cards from the remaining 46 cards
(14-1)!! number of ways to split these cards into hands
(46, 14) * (14 - 1)!! = 3.2416 * 10^16
For the numerator we combine these three values for the total number of deals with 2 pairs of As, and 1 pair of Ks:
3 * 6 * 3.2416 * 10 ^ 16 = 5.8349 * 10^17
(we could remove the hands that contain AA, AA, KK, KK, but that is ~0.1% so I'm going to ignore it)
Total Probability:
AA, AA, KK deals divided by Total number of deals
5.8349 * 10 ^ 17 / 8.2492 * 10^22 = 7.0733 * 10 ^ -6 or
0.000 007 073
Probabilities to odds (see Demi's post):
0.999 992 927 to 0.000 007 073
141,381 to 1
Demi: this is off by a factor of 2 from yours. Either I made a mistake, or your calculations don't account for swapping the A pair hands. If that's the reason this is an example of why I like counting, I'm less likely to miss special cases.
Dang, we need Sklansky.
EV for KK vs two AA is better than against 1AA (which makes sense intuitively but still is pretty funny).Quote:
Originally Posted by finky
HA. I was about to post and mention that I had been involved in one...didn't know it was against a fellow ftr'er. This was when I was giving ring games a shot. I don't know why I didn't reraise either.
I thought you were from FTR when this happened but I wasn't 100% certain. NH :). Although, I must say, it would have been MUCH cooler if the 3 of us got it all-in preflop...Quote:
Originally Posted by Rockymv
rofl that post sounds so wrong...
nice post demi. I could have managed to get through that eventually but it's very well stated, correct, and very clear.
We should start a thread of some of the craziest things that have happened to you as a poker player...
In hold 'em, my top 3 would have to be the following:
Being dealt pocket aces 3 straight hands on the same table. (10,793,861 : 1 against.) Next hand I got dealt a suited big-slick and lost more on that hand then I won on the previous 3 combined...
Flopping a royal flush holding Ad Jd. No action :(
Hand posted above. Destacked 2 players. ;)
I think demi and swaggidy both make the computation seem harder than it really is (and demi's calculations contain inaccuracies).
The probability of one pair of players both getting AA is 4*3*2/52*51*50*49 = 1/270725. There are 45 pairs of players at a 10-seat table, so the probability of two players getting AA is 45/270275 = 9/54145.
For each of the 8 remaining players the probability of them being dealt KK is 4*3/48*47. To get the probability that at least one gets KK we multiply by 8 and substract the probability that two had KK (to compensate for double counting):
8*4*3/48*47 - 28*4*3*2/48*47*46*45 = 2063/48645.
Hence the probability of two players getting AA and at least one getting KK is 9/54145 * 2063/48645 = 2063/292653725, which is approximately 1/141858 or 7.0 * 10^-6.
I was trying to break it down step by step and came up with some conceptual errors due to miscounting.
If i were going to do the math for a test I would have done it exactly like krimson.
The "45 pais of players" he didn't explain comes from the combination (10,2) or 10*9/2 (Although many people could have figured it out themselves)