Hey guys,

today I have 3 math problems.
I hope this is the right place, to ask these questions.

#1
I know the calculation of the percents to flop a Flush or Flushdraw with two suited hole cards.

* Flush at the flop:
50 Unknown Cards / 39 Non Outs / 11 Outs =Preflop
19600 possible flops

11 Outs → 11 choose 3 = 11*10*9/3*2*1 = 165 possible one tone flops
=> 165/19600 = 0,84% to flop a flush

* Flushdraw at the flop (Hole Cards in spade)
50 Unknown Cards / 39 Non Outs / 11 Outs =Preflop
19600 possible flops

11 Outs → 11 choose 2 = 11*10/2*1 = 55 possible flops with two spade
Third card at the flop is one of the 39 non-spade cards
= 55*39 = 2145 possible flops with two spade and another tone

=> 2145/19600 = 10,94% to flop a flushdraw


!!!NOW MY QUESTION!!!
I found the percent of 6,5% generally to makes a flush with two suited hole cards.
So maybe after flop, turn or river.

Anyone an idea, how I calculate this???




#2
I know the calculation of the percents to flop one pair (one of your hole cards – no paired flop)

* AK and hit one of these:
50 Unknown Cards / 44 Non Outs / 6 Outs = Preflop
19600 possible flops

6 Outs → 44*43*42/(3*2*1) = 13244 possible flops without an ace or king

=> 13244 / 19600 = 67,57% no hit
=> 34,43% hit an ace or king


!!!NOW MY QUESTION!!!
I found the percent of 2,2% to flop two pair with my hole cards (unmatched)

Anyone an idea, how I calculate this??? Maybe with the same hand of AK.



#3
This is an outs calculation.

Following scenario:
Hero have 99
Flops: 8h9hTh

Flops a set, but villain flops a straight or flush.
So hero need a full house or quads to make the winning hand.

Villain shove all in. If hero call he will see 2 cards.

At the flop we have 7 outs
→ 3x8, 3xT, 1x9

Turn is a 2c.
We have 10 outs
→ 3x8, 3xT, 1x9, 3x2

Is it right to calculate 8,5 outs at the flop if hero is all in and will see 2 cards?
So 7 + 10 / 2 = 8,5???

Or is this the wrong way to calculate this?




Greetings,
Christian

This forum is fine to post these questions. There is also a Probability Forum as well which would also be perfectly fine.

I will answer your first question. Hopefully you'll get the idea from that.

Assumptions: If you ignore straight flushes and royal flushes (i.e., count them as "flushes"), count all flushes (even those for which you "play the board"), and do not count flushes in other suits, then the following is what you seek.

Say you have two spades. Then there are, of course, 50 other cards from which the 5 board cards emanate. Of those 50 cards, 11 will be spades and 39 will be non-spades.

So, first, note that there are C(50,5) different possible boards where C(x,y) is the choose or combination function. C(50,5) = 2,118,760.

Let's tally how many boards give you a spade flush:

5 spades on board
= C(11,5)
= 462

4 spades on board
= C(11,4) * C(39,1)
= 330 * 39
= 12,870

3 spades on board
= C(11,3) * C(39,2)
= 165 * 741
= 122,265

TOTAL
= 462 + 12,870 + 122,265
= 135,597

Probability
= 135,597 / 2,118,760
= 6.40%

Clearly if you modify any of the assumptions made at the top, you'd get a slightly different answer.

Wow, thanks for the brilliant answer!!!

So, If I have for example A K

The chance for Flush is 1,85%.

Calculation:

50 Unknown Cards / 39 Non Outs / 12 Outs = Preflop

2.118.760 different boards (50 choose 5)
50*49*48*47*46/(5*4*3*2*1) = 2.118.760

All boards with a heart:
5 heart boards
12 Outs → 12 choose 5
12*11*10*9*8/(5*4*3*2*1)
=> 792 combinations

4 heart boards
12 Outs → 12 choose 4
12*11*10*9 / (4*3*2*1)
=> 495 combinations
38 Non – Outs → 38 choose 1
38/1
=> 38 combinations

495*38
=> 18.810 combinations

heart boards total:
792+18.810
=> 19.602 combinations

heart and spade boards:
19.602 * 2
=> 39.204 combinations

RESULT:
39.204/2.118.760
==> 1,85%

Is that right???




With the two pair calculation, I still have some problems

Yes, in NLHE the prob of making either a heart or spade flush (including royal and straight flushes) when dealt one heart and one spade is 39,204 / 2,118,760 = 1.85% as you showed above.

Often times deriving these types of probabilities comes down to delineating the cases (the hard part) and then calculating the prob of each case (the easy part).

Feel free to show what you have already done for the two pair calculation if you are seeking feedback.