View Full Version : odds of getting either of AA, KK, AK = 2.07%?

12-18-2004, 06:33 AM
8/52 * 7/52 = 2.07%
So, every 50 hands, you'll grab a monster?

12-18-2004, 03:25 PM
unfortunately most common of them is the offsuit AK variety

but isn't it 8/52 * 7/51?

One Armed Dwarf
12-18-2004, 03:43 PM
A better way to analyze this is to use choose.

P(AA,KK,AK) = (#ways to get AA + #ways to get KK + #ways to get AK) / (# ways to be dealt two cards)
= (4ch2 + 4ch2 + 4*4) / 52ch2
= 2.11%

I think what you mean to write is 8/52*7/51. However, I don't like to do probabilities using multiplication... I'm not sure if it works very well conceptually. 8*7 counts every combination of the 8 cards twice, A1A2, A2A1. Luckily, in this case you are ok because you also count every combination twice in your denominator. Still, I think that calculating probabilities using multiplication like that might get you into trouble with more complicated calculations.

12-18-2004, 04:09 PM
In the case, the OP's method is better than yours, IMO. True, you can't generalize the method to solve many types of problems, but for this particular problem it is by far the simplest and clearest solution.

Just cause you have complex tools, it doesn't mean it's better to use them when a hammer will do the job just as well.


12-18-2004, 07:44 PM
In a far more coincidental way, I'd wondered for a time why the total number of different starting hands was exactly 169. Just multiplying the 13 cards by itself should give an incorrect answer, since that would mean AK was different than KA.

The simplicity is that for every combination that isn't a pocket pair, there are two ways for it to show up in your hand - suited or unsuited.

So the answer and method are the same, but not for the obvious reason.