Tricky and Confusing Question

[AdvancedPlaya]

Not sure I am understanding probability stuff. Let us say I have Ace-King. Then on flop comes...Ace-2-9. Ok, if I want to say that I have 6 outs of making a better hand. This is because 2 other unseen Aces can come or 4 other unseen Kings can come, they add up to 6. So having 6 outs of making 2 pair or better, it seems according to probabilties and what I have understood, I will make it around 24% of the time since I have six outs. This seems waaaay to high. Almost a quarter. Am I doing something totally wrong???

[Skipbidder]

Five cards to improve your hand. The other two aces and the other three kings. There are 47 unseen cards. You probably can't put anybody on a specific card, so on the turn, you have a 5 out of 47 chance of improving. Of those times that you don't improve on the turn, you have a 5 out of 46 chance of improving on the river. If you are trying to figure out your chances of beating somebody who happened to hold two pair on that flop (in this case 9-2), then you will also improve to beat them if you get a running pair because you will have aces up and your opponent will have to play his nines. A little better than 10% to improve on the turn, plus a little better than 10% to improve on the river, plus chance of running pair, minus those times when you improve to aces and kings on the turn but he fills up on the river, gets you to a smidgeon over 25%.

[Skipbidder]

Juggle this a little bit if you have flush possibilities.

[gaming_mouse]

24% is the correct chance that an A or K comes by the river.

Btw, this question is about as basic as it gets.

[PygmyHero]

You are correct about the 24%, notwithstanding the points that have already been made in the earlier posts. Those factors will alter this number slightly, depending on the exact cards you have and that hit on the flop.

Here's an easy way to solve these problems in the future:

odds of hitting outs on turn and / or river (http://www.texasholdem-poker.com/odds_chart.php)

If your problem is with the methodology involved, well...that will take a bit longer to explain...

[gamble4pro]

First of all, you have 3 kings left, not 4, if you hold one.
If you refer to your turn odds (odds of hitting A or K by turn), the calculus should be performed as 5/47 = 10.64%.
If you refer to the event "hittting A or K by river), the "outs" are not cards anymore (5), but 2-card combinations. These counts as follows:
(AA) = C(2,2)=1
(AK) = 2*3=6
(KK) = C(3,2)=3
(Ax)= 2*42=84
(Kx)= 3*42=126
Totally, 220 favorable combinations (situations) from C(47,2)=1081 possible. So, the odds are 220/1081 = 20.35%.

[elitegimp]

Or you could look at the probability of _not_ seeing a K or A -- as mentioned above, there are 2 As and 3 Ks left in the deck of 47. So there is a 42/47 chance that neither an A nor a K comes on the turn, in which case there is a 41/46 that neither comes on the river. Since you want _both_ these events to happen, multiply the probabilities together:

P(not hitting a K and not hitting an A) = (42/47)*(41/46) = 0.7965

so

P(hitting at least one K or A) = 1 - 0.7965 = 0.2035 (20.35%) (because P(not hitting) + P(hitting) = 1)

This is the same number gamble4pro got, which is good because we calculated the same thing (using different methods)