Yeah, if you put it that way about those 100%, I feel kind of reatarded.

I took closer look to your example. I don't have the gratest mathematical mind, but by simple logic, I think you made a little mistake, BUT correct me if I am wrong, because that is how I learn.

So.. If I understood corectly, you calculated the advantage of the A

K

(42/48)

We have 52 cards. We know 48 of them. The opponent have 6 outs (the 1 pair scenario).

**Opening the first card of the street:**
48 - 6 = 42 (the

**player2** non-outs)

42/48 = 0.875

**Opening the second card from the street:**
48 - 1 = 47 (we took one card from non-outs out)

The player still have 6 outs, thus 47 - 6 = 41

41 / 47 = 0.872

Opening third, fourth and fith cards:

40 / 46 = 0.869

39 / 45 = 0.866

38 / 44 = 0.863

Than we take the avarage precentage => 0.875 * 0.872 * 0.869 * 0.866 * 0.863 = 0.4955.. ~ 50%

And on paper this is believable, because if we take the formula from calculating pot odds => (unknown cards - outs) / outs

So lets take a look at the probability for

**player2** to improve in turn and/or river.

we have 45 unknown cards ( 52 - 2 (your cards) - 2 (known opponents cards) - 3 (the flop, where the

**player2** didn't improve) = 45 )

(45 - 6) / 6 = 6.5:1 = approx. 15.3 %

Because we open 2 cards the probability is 2 times higher, thus 15.3 * 2 = 30.6 %.

But this still feels so wrong, because it seems so unlogicall, that in that scenario, there is 50:50 chace, that

**player1** with A

K

will win

**player2**, who holds Q

J

by high card (

**player2** not improving), so than its just a coinflip? Can that be?

Where did I go wrong in this scenario? (please prove me wrong mathematically, because the learning process is what I am the most interested in).