Poker odds question

[MrDingleDangle]

at a 6 player table, you have pocket kings, what is the likelyhood that someone else has pocket Aces?

their answer was 44-1 and that makes no sense whatsoever to me

first of all it shouldnt matter how many other players there are, unless you know they dont have an Ace.

so can anyone come up with a way they get 44-1...i think the travel channel is on crack

 

[stan394]

The total number of players of coz matters. If there are 9 other players on the table, of course the probability of SOMEONE having pocket aces is much higher than if there are only 2 other players on the table.

 

[jaybert]

well, there is a 6% chance someone has a pocket pair. that pretty much means that there is a 30% chance that someone else has a pocket pair (6 people table minus yourself). You are looking for a specific pocket pair, which there are 13 of. so 30%/13 = 2.3%.

1/44 = 2.27%

Mine are just rough estimates, but if you do a little work thats what you get.

(I got the 6% you will get a pocket pair off of a texas hold'em website, but you can figure, (52/52) * (3/51) gives you approximately 6%. Figure it as you have a 52/52 chance of getting a first card, then after that, you have a 3/51 chance that your second car matches your first one

 

[Ketteringo]

Well, the odds of 1 person having pocket aces is .45%, x 5 people = 2.25%, or 1 in 44.

 

[HomeBrewerDude]

5*4/50*3/49

 

[stan394]

You have pocket kings. That means there are 50 cards out there. 4 out of the 50 could be aces.

Let's say there are only 1 other player.

The probability of player A has pocket aces is 4/50*3/49 ~= 1/204

Now let's say there are a total of 2 other players

The probability of player A has pocket aces is ~= 1/204
The probability of player A has exactly 1 ace is 4/50*46/49 + 46/50*4/49 ~= 1/13
The probability of player A has no aces is 46/50*45/49 ~= 1/1.18
same for B, individually

Now you can also compute the probability of these, using similar calculations
A having 2 aces AND B having 2 aces ~= 4/50*3/49*2/48*1/47 ~= 1/230300
A having 2 aces AND B having 1 ace ~= 4/50*3/49*(2/48*46/47 + 46/48*2/47) ~= 1/2500
etc for all 9 combinations

All the scenarios which satisfy "at least one other players have pocket aces":
- A has 2 aces AND B has 0 aces
- A has 2 aces AND B has 1 aces
- A has 2 aces AND B has 2 aces
- A has 0 aces AND B has 2 aces
- A has 1 aces AND B has 2 aces

Add the 5 probabilities from above, and I bet you will get a number less than 1/204.

 

[david46675]

Originally posted by: jaybert
well, there is a 6% chance someone has a pocket pair. that pretty much means that there is a 30% chance that someone else has a pocket pair (6 people table minus yourself). You are looking for a specific pocket pair, which there are 13 of. so 30%/13 = 2.3%.

1/44 = 2.27%

Mine are just rough estimates, but if you do a little work thats what you get.

(I got the 6% you will get a pocket pair off of a texas hold'em website, but you can figure, (52/52) * (3/51) gives you approximately 6%. Figure it as you have a 52/52 chance of getting a first card, then after that, you have a 3/51 chance that your second car matches your first one

correct, so take ((3/51)*5)/(13) = 2.2624%

edit: this is before any of the house cards are revealed.
strictly pocket terms.

 

[Albis]

we have stats majors here

 

[thelanx]

AA vs. KK hands are often painful for the KK guy. I just came back from being shortstacked at a SNG 4 ppl left (Top 3 get paid), stole some blinds to even up with the next shortest stack and then busted him out AA vs. JJ. Ouch.

 

[birdpup]

What does the term "pocket" mean in this context? Two or a kind, three of a kind, four of a kind, a pair of cards with no jokers? Are jokers involved? I assume only one deck of cards is used.

 

[thelanx]

Pocket pair means you have a pair as your two hole cards (the cards that are unique to you). It is powerful because your pair is hidden from everyone else. And only one deck is used with no jokers.

 

[royaldank]

"Pocket" refers to the 2 cards you own face down. Not the community cards.

 

[birdpup]

How many cards does one person hold? Five?
And two of these are face down while the other three are face up?
Does the dealer possess a hand?
If the dealer possesses a hand, is this also two face down and three face up?
If the dealer possesses a hand, is the total # of players 6 or 7?

 

[KLin]

Originally posted by: birdpup
How many cards does one person hold? Five?
And two of these are face down while the other three are face up?
Does the dealer possess a hand?
If the dealer possesses a hand, is this also two face down and three face up?
If the dealer possesses a hand, is the total # of players 6 or 7?

He's talking about Texas Hold'em.

 

[birdpup]

Thank you, I give up trying to figure this one out because I cannot figure out the rules or the assumptions.

 

[xenocyd3]

Common odds when holding unpaired hole cards:
flopping EXACTLY one pair by pairing a hole card 26.939%
flopping EXACTLY two pair by pairing a hole card AND pairing on the board 2.02%
flopping EXACTLY two pair by pairing EACH of your hole cards 2.02%
flopping EXACTLY trips by flopping two cards to one hole card 1.347%
flopping EXACTLY a full house, trips of one hole card and pairing the other 0.092%
flopping EXACTLY four of a kind, three cards to one of your hole cards 0.01%


Common odds when holding paired hole cards:
flopping EXACTLY two pair by pairing the board 16.163%
flopping EXACTLY trips by flopping a set for your pocket pair 10.775%
flopping EXACTLY a full house, a set to your hole pair and pairing the board 0.735%
flopping EXACTLY a full house, by the board tripping up 0.245%
flopping EXACTLY four of a kind, two cards to your hole pair 0.245%


Common odds when holding two unsuited cards:
flopping a four flush 2.245%


Common odds when holding two suited cards:
flopping a flush (including the slight chance of a straight flush in some cases) 0.842%
flopping a four flush 10.944%


Common odds when holding connectors from 54 to JT
flopping a straight (including the slight chance of a straight flush in some cases) 1.306%
flopping an 8 out straight draw* 10.449%


Common odds when holding one gapped connectors from 53 to QT
flopping a straight (including the slight chance of a straight flush in some cases) 0.980%
flopping an 8 out straight draw* 8.08% **


Common odds when holding two gapped connectors from 52 to KT
flopping a straight (including the slight chance of a straight flush in some cases) 0.653%
flopping an 8 out straight draw* 5.224% **


Common odds when holding three gapped connectors from A5 to AT
flopping a straight (including the slight chance of a straight flush in some cases) 0.327%
flopping an 8 out straight draw* 2.612% ***

jacked from www.flopturnriver.com

 

[chuckywang]

I get about 1/41 chance that someone has pocket aces.
Here's my logic:
There are 50 cards and 4 aces remaining for the 5 players that are not you.

There are Permute(50,10)/2^5 different hands for them. Call this A.

Out of those hands, there are:
5*6*Permute(48,8)/2^4 ways that someone will get pocket aces. Call this B. However, this is counting the number of ways for two people to get pocket aces twice.

So we subtract that off (this is known as the principle of inclusion-exclusion in mathematics):
We subtract off 10*6*Permute(46,6)/2^3. Call this C.

Therefore, the total probability is:
(B-C)/A which is about 1/40.9