View Full Version : AK probability
livinitup0
02-27-2005, 10:57 PM
Im trying to figure out the odds of getting AK, getting 1 ace is 1 out of 13 hands...after that getting a K is 4 out of 51 or 1 in 12.75
added together thats 25.75
so does that mean I should get dealt AK once in 25.75 deals???
olavfo
02-27-2005, 11:14 PM
The math goes like this:
<ul type="square">
Possible 2-card combinations: 52!/(2!(52-2)!) = 1326
Possible AK-combinations: 4*4 = 16
Probability of being dealt AK: 16/1326 = 0.012 = 1.2%
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If you don't know the factorial function (!) you can use a simpler expression for the total number of 2-card combinations: 52*51/2.
You got 52 choices for card 1 and 51 choices for card 2. The order in which you get the cards does not matter, so you need to divide by 2 to avoid double counting.
The number of AK combinations is just the number of aces times the number of kings (4*4).
olavfo
livinitup0
02-27-2005, 11:17 PM
Thanks for correcting me...Pretty new at this and should have taken the time to figure it out, instead i went to some bogus holdem site and got that info off there...last time i take the easy way out...thanks
livinitup0
02-27-2005, 11:23 PM
so does that mean its dealt once out of about 82 hands on average? 1326 divided by 16?
olavfo
02-27-2005, 11:27 PM
Almost correct. The odds are
1/probability = 1326/16 = 83 (rounded)
so you will get AK 1 out of 83 times. The odds can also be written as 82 : 1.
olavfo
livinitup0
02-27-2005, 11:33 PM
now if I wanted to do the same for AKs its the same execpt multiply 83 by 4 (since only 4 combination) right? so say 331 : 1
olavfo
02-28-2005, 01:03 AM
Correct, since you have a 4/1326 chance for those.
olavfo
dabluebery
02-28-2005, 10:44 AM
For those of you with aversions to "combination" math, a simple calculation like that of a pocket pair, or of two suited / unsuited cards can be done like this;
Probability of AK;
2/13 * 4/51 = (8/663)
There's a 2 out of 13 chance of grabbing an A or K with your first card. Once that happens, there are 4 other cards (A or K) of the 51 remaining.
AKs:
(8/52 * 1/51) = 2/663 ... Just one card left to satisfy "AKs" once you draw an ace or a king.
Best bet is to just memorize the number of combinations, and then all you'll have to do is figure out the numerator of whatever calculation you need. (16 for AK, 4 for AKs, etc.)
Rob
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