odds of AA or KK vs my AK - Gambling and Probability

Two Plus Two Forums Other Topics Probability

odds of AA or KK vs my AK

Post Reply Subscribe

...

10-30-2009 , 03:25 PM

Green_29

veteran

Join Date: Apr 2009 Posts: 3,037

at a 6max table, when i am holding AK what are the chances that another player is holding AA OR KK?

working this out myself im thinking its around 0.5%, is this correct?

Quote

10-30-2009 , 03:28 PM

Green_29

veteran

Join Date: Apr 2009 Posts: 3,037

so it should happen like 1 time every 200 times i have ak, so 1 in every ~20,000 hands overall?

is this right because its happened 5 times in the last 4k hands and im pissed.

Quote

10-30-2009 , 03:49 PM

bschr04

grinder

Join Date: Jul 2008 Posts: 454

Given that you hold AK, there are C(50,2)=1225 card combos for each of your opps.

6 of these are AA or KK.

That means each opponent has a 1219/1225= .995 chance of not having AA or KK.

That meand that prob that none of the five opps hold AA or KK is

(.995)^5= 97.5%

Quote

10-30-2009 , 04:03 PM

Green_29

veteran

Join Date: Apr 2009 Posts: 3,037

Quote:

Originally Posted by bschr04

yeah thanks i got it right i just did it for 1 player not 5.

Quote

10-31-2009 , 01:25 AM

NoMeansNo

banned

Join Date: Oct 2009 Posts: 3,263

Quote:

Originally Posted by bschr04

How exactly do I calculate the highlighted parts of the equations....dropout here, thanks.

Quote

10-31-2009 , 02:18 AM

smithcommajohn

grinder

Join Date: Sep 2009 Posts: 533

Quote:

Originally Posted by Green_29

so it should happen like 1 time every 200 times i have ak, so 1 in every ~20,000 hands overall?

is this right because its happened 5 times in the last 4k hands and im pissed.

Just FYI... you should see AK in 4000 hands about 36 times... and of those 36 times you should be up against AA or KK about once... (0.877 times)

so having it happen 5 times is well beyond normal...

Quote

10-31-2009 , 11:02 AM

Green_29

veteran

Join Date: Apr 2009 Posts: 3,037

make that 6 time in 4k hands.

Quote

10-31-2009 , 11:19 AM

smcdonn2

centurion

Join Date: Jul 2009 Posts: 165

I think we should also take into account, How many times will I be up against AA or KK given he 3 bet me.

Quote

11-01-2009 , 01:45 PM

bschr04

grinder

Join Date: Jul 2008 Posts: 454

Quote:

How exactly do I calculate the highlighted parts of the equations....dropout here, thanks.

nomeansno,

C(X,Y)= X!/(Y!*(X-Y)!) where X!= X(X-1)(X-2)(X-3)...(1)

so C(50,2)= 50!/(2!*48!) = (50*49*48*...1)/(2*1)(48*47*...1)

all the factors 48 thru 1 cancel and you are left with:

50*49/2 = 1225

to calculate .995^5, just plug into google search or excel.

Quote

Post Reply Subscribe

...