odds question

TheCount212 · #1 10-23-2007, 11:54 AM

Attention math geeks...

I have AA in the BB. Five limpers. I raise PF, five callers. The flop comes KKx. What are the odds that someone holds a K?
Can you also give me a % if there are four limpers rather than five? Thank you...

fishyak · #2 10-23-2007, 01:03 PM

i'll get you started. 9 seats = 8 other players = 16 slots for a King to land out of 47 remaining cards. From there you have to assume something such as all Kings call PF. Because folding is not a random event, I think there is little difference between 4 and 5 limpers. Somebody who knows Excel can take this further, but it is close to 1 in 3 that a King was dealt to someone. Because of folds, your actual odds should be better than that.

Hamlet · #3 10-23-2007, 02:50 PM

I get about a 57% chance that a king was dealt to one of the eight people. Here is the calculation for the odds of no one getting dealt a king:

0.434 =(1-2/47)*(1-2/46)*(1-2/45)*(1-2/44)*(1-2/43)*(1-2/42)*(1-2/41)*(1-2/40)*(1-2/39)*(1-2/38)*(1-2/37)*(1-2/37)*(1-2/36)*(1-2/35)*(1-2/34)*(1-2/33)

Have I made an error somewhere?

threeducks · #4 10-23-2007, 02:56 PM

This web site might help.

Ricks · #5 10-23-2007, 03:02 PM

I am not a math expert but I believe this is correct.

If you assume that your five opponents play any King.

There are 47 unseen cards, consisting of 2 Kings and 45 non-Kings. The total number of ways to have 10 hole cards for the 5 opponents is c(47,10). The total number of ways to not have a King in these hole cards is c(45,10).

The probability that someone does not have a King is c(45,10)/c(47,10). 1 - c(45,10)/c(47,10) =~ 0.384 is the probability that someone does have a King.

For four: 1 - c(45,8)/c(47,8) = 1 - 215553195/314457495 =~ 0.315

Ricks · #6 10-23-2007, 03:17 PM

[ QUOTE ]
I get about a 57% chance that a king was dealt to one of the eight people. Here is the calculation for the odds of no one getting dealt a king:

0.434 =(1-2/47)*(1-2/46)*(1-2/45)*(1-2/44)*(1-2/43)*(1-2/42)*(1-2/41)*(1-2/40)*(1-2/39)*(1-2/38)*(1-2/37)*(1-2/37)*(1-2/36)*(1-2/35)*(1-2/34)*(1-2/33)

Have I made an error somewhere?

[/ QUOTE ]

A few too many.

(1-2/47)*(1-2/46)*(1-2/45)*(1-2/44)*(1-2/43)*(1-2/42)*(1-2/41)*(1-2/40)*(1-2/39)*(1-2/38)= 0.616

fishyak · #7 10-23-2007, 05:00 PM

Ricks, assumining a 9 seat table, how does the fact that all 8 opponents were given two cards that could have been Kings impact the appropriate formula? My concern is based on two facts: 1) that folding is a non-random event. The hands that folded were less likely to fold a hand with a King than the hands that didn't fold, and 2) all the 8 other players got 2 cards. Thus, 16 cards were dealt out that could have been Kings. Shouldn't the King question be applied to all who got cards, (8), whether they folded or not? If not, why not?

TheCount212 · #8 10-23-2007, 05:16 PM

If it makes it easier, I normally play 3/6, so I am okay with operating under the assumption that the holder of any K would not fold. But by all means I'm not questioning the logic of either of you gentlemen..

Ricks · #9 10-23-2007, 05:32 PM

[ QUOTE ]
[ QUOTE ]
I get about a 57% chance that a king was dealt to one of the eight people. Here is the calculation for the odds of no one getting dealt a king:

0.434 =(1-2/47)*(1-2/46)*(1-2/45)*(1-2/44)*(1-2/43)*(1-2/42)*(1-2/41)*(1-2/40)*(1-2/39)*(1-2/38)*(1-2/37)*(1-2/37)*(1-2/36)*(1-2/35)*(1-2/34)*(1-2/33)

Have I made an error somewhere?

[/ QUOTE ]

A few too many.

(1-2/47)*(1-2/46)*(1-2/45)*(1-2/44)*(1-2/43)*(1-2/42)*(1-2/41)*(1-2/40)*(1-2/39)*(1-2/38)= 0.616

[/ QUOTE ]

I didn't realize that you were doing this for eight.

Ricks · #10 10-23-2007, 06:01 PM

You are correct if we make the assumption that nobody would fold with a King in their hand. We could then say with certainty that there are only 39 unseen cards. My assumption that any of the five opponents will play any King actually has nothing to do with my calculation, now that I look at it.

1 - c(37,10)/c(39,10) = 1 - 348330136/635745396 =~ 0.452

I think that making the assumption that nobody would fold with a King is incorrect since there are many unplayable hands that include a King, even at 3/6. I understand your point though. It is certainly more likely that the hands that folded did not contain a King than hands that did contain a King. I have seen these points debated in the Probability forum and I'll see if i can dig some up.