understanding outs shouldn't this be a 4% chance to win?!?

[Fatdogs12]

Hey everyone I have an outs question. Okay here is the situation. We are playing holdem and we are on the flop.

I have AK and my opponent has J-10. The flop comes 10-9-2.
Now I have 6 outs to hit and two cards to come. That I believe gives me a 2% chance per card per out for a 24% chance of hitting my cards. The thing is my opponent has 5 outs and therefore a 20% chance of hitting his cards, wouldn't that leave me with only a 4% chance of winning?

I put it into poker stove but it said that my chances were 22% to win. Can anyone explain why?

[Josh W]

two things...

if we assume you have a 24% chance to win minus (20% x 24%), that's about 19.2% chance of winning.

Thing is, when he hits three of his outs (Jacks), you still have 4 outs win.

i hope this makes sense.

josh

[Fatdogs12]

yeah it does because I would hit my straight, the only thing though is, is there a way to do this in my head, it seems it would be very very complicated to do in one's head. Am I correct? (What I mean by that is if I have a flop and I know a persons cards and mine how would i go about figuring out who would win if we were all in)?

Thanks

[pzhon]

In Hold'em, you can get a good estimate by determining the better made hand, then count the other hand's outs. You don't count outs for the stronger hand. In a few cases, you need to consider redraws, but they usually aren't significant. One exception is a set versus a flush or straight draw, and you can just remember that situations.

[DWarrior]

Well, if you already know both hands, then you must both be all-in, and after that trying to figure out your outs is just beating a dead horse.

Other than that, I don't see a realistic way of putting a guy on JT when flop comes T92, so that's also not practical.

The reason most hand match-up calculators are done as simulations is because there is too much to consider if you were to go about figuring out the exact probability. In this case, as others have said, if a Q turns, you pick up 4 outs, and if a J turns, you go down to 4 outes. If a T turns, you're dead, if you pair your A, he goes down to 2 outs, and if you pair your K, he goes down to 6 (Q or T)

There's really no practical way to compute the exact win % for both hands.

[Greeksquared]

You simply cannot subtract probabilities in this manner. Yes, you will hit your hand 24 percent of the time and he will improve about 20 percent of the time. What you are saying is that 90 percent of the time that you hit, your opponent will hit as well. There are 30 possible turn and rivers that will both hit you...meaning some combination of of a jack or a ten AND a king or an ace. There are (45 c 2) =990 combinations of cards left to be dealt on the turn and river. This is about 3 percent. Your total hits are equal to 6*39 + (6 c 2)=15 = 249. So, of the 249 times you hit, 30 of those times will also hit your opponent. This leaves you with 219 times out of 990 that you will win. This is approximately 22 percent. This is actually fairly easy to calculate.

For situations where you need one card for the win and your opponent will need one card to beat you if you move aheard you can do the following simple calculation:

y= your outs
o = opponents outs

(y*4)-(y*o)/10 = chance at winning after flop

Another example you hold AK your opponent holds AT. Flop is AJ3. Using equation... 3*4-3*4/10 = 11

[spaminator101]

you have 6 outs twice or a 24%chance of hitting your hand
your opponent has 5 killer cards once if you do hit your hand or about a 10% chance of killing you
you will win about 22% of the time

[Greeksquared]

My counting was actually off. Here is the correct way.

pointless but whatever..

If you hit an ace it can be paired with any other non J or T.
So there are 3*34 combinations = 102
a king can be paired with any non Q, J or T so there are 3*30 = 90

Also A and K can be paired together or with another A or K 6c2=15

And finally you can win with QJ for 4*3=12 combos

SO.... 90 +102+15+12 = 219

blah