AK probability

[livinitup0]

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]

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%[/list]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]

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]

so does that mean its dealt once out of about 82 hands on average? 1326 divided by 16?

[olavfo]

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]

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]

Correct, since you have a 4/1326 chance for those.

olavfo

[dabluebery]

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