Looking for confirmation or correction.

You have KJ in hold’em. The probability of flopping exactly top pair (not two pair nor a pair on the flop) is 19.84%.

Examples: K T 6, J 5 2

Thx.

Confirmed.

You need a flop with exactly one king, no jacks, and no pair.

There are three slots on the flop. One needs a king and the other two can't be a pair or include a jack or king. The order doesn't matter.

47 non king cards in stump.

47*46*45/1/2/3=16215 flops without a king.

50*49*48/1/2/3=19600 possible flops.

must be 19600-16215=3385 flops with one king, two kings or three kings.
1 flop has three kings, 47*3 flops have two kings.
Therefore 3385-1-47*3= 3243 flops have exactly one king.

Of these, the possibilities are
KJJ 3*3=9
KJX 3*3*44=396
KXX 3*44*43/2=2838

check: 9+396+2838=3385 (checks)

KXX is what we want, and there are 2838 of these

3*44*40/2= 1 king with no pairs and no jacks = 2640
(3*11*6=198 for 1 king, one pair, and no jacks
Then 2640+198=2838 checks).

Finally 2640/19600=13.47% is my bottom line.
Thus I think 19.84% is too high.

(However, I know in my heart that you and Who's Next must be right and therefore my approach must be wrong).

Buzz

Quick Approximation: Take KJ out of deck. There are enough cards left for 50/3 = 16.67 three card flops. Only 3 of these, at most, can have a king.
3/(50/3)=9/50=18%... answer cannot be higher than 18%... and with some flops having double kings or triple kings, it's lower than 18%.

Buzz

statmanhal is looking to flop top pair with KJ, so a jack high flop also works.

And, AK4 or QJ2 would not be flopping top pair holding KJ since there is a card on the flop higher than your pair.

Here’s the general formula I used.

Let R1 be the higher rank and R2 the lower with A=14, K=13, etc. Then the probability is

Pr(top pair w/no flop pair|R1, R2) = {3*C(R1-3,2)*16 +3*C(R2-2)*16)}/C(50,3)

For the first numerator term. Assume R1=10. There are 3 tens left. Flop must have 1 of these if top pair is to be the 10. Then the other two flop cards must be from 2 lower ranks and can’t be from rank R2. That leaves 10-3= 7 ranks from which to choose 2 ranks and then 1 card from each rank. Similar thinking for the lower rank.

Ah! I see my error. (I didn't include jack high flops). Thank you.

Buzz

I got 21.6%.

PQL Code...

What is this? You looked up his HH in a database? What was the sample size of KJ hands that saw a flop?

I doubt very much that the simulation will show a nearly 1.8% absolute difference from the theoretical result (assuming the latter 19.84% is correct).

Therefore I have to think PQL's definition of floptoppair differs from what I used; perhaps it includes two pairs or flop pairs.

EDIT: Ah, I checked the site and floptoppair does include a pair on the flop so that is obviously one of the reasons for the difference.

I see the problem, I think you are excluding boards like Kc 2c 2d, since that would be 2-pair? That PQL code includes those.

To exclude those boards I used different functions, and got 19.8384% after 50,000,000 simulations.

It must be including flops which have a lower pair such as K88 or J55.

Then you'd get [3*C(40,2)+3*C(36,2)]/C(50,3) = 4230 / 19600 = 21.5816327%

The answer to statmanhal's original question is clearly:

= [C(3,1)*C(10,2)*C(4,1)*C(4,1)+C(3,1)*C(9,2)*C(4,1)* C(4,1)]/C(50,3)

= 3888 / 19600

= 19.8367347%

It was excluding 882 but including k22. I agree with 19.8%.