Hi, guys !
I was trying to get some poker numbers by myself and the answers that I got were a bit different when compared to the ones from this thread which seems pretty reliable.

https://forumserver.twoplustwo.com/2...d-flop-242373/

So, what are the chances for a "bad flop" when holding KK ? (ace on the flop without a set)

-chances for an ace-high flop
1 - (48/52 x 47/51 x 46/50) = 1 - 0,783 = 21.7%

In the brackets I calculated the chances that ace won't be on the flop which is 78,3% and that means that ace will be on the flop 21.7%.
This number (21.7%) should be correct as I found the same one later on the internet somewhere.

- chances to flop a set/quads when holding KK
1 - (48/50 x 47/49 x 46/48) = 1 - 0,8824 = 1 - 88.24% = 11.76%

I did the similar calculation in the brackets again. Ik that brackets are not needed but it's better visually.

And now finally !****
To have a bad flop for KK we need to:
- miss a set/quads (88.24%)
- have an ace high flop (21.7%)

88.24% x 21.7% = 19.1%****

And some numbers that were mentioned in that post were:
-20.67%
-23.43%
-22.6%
-23%


What am I doing wrong ?

This looks so nerdy, it is just a few percent difference.
But try to understand me and ty in advance !

There are lots of ways to do this. I prefer combos whenever possible. The following calculates the number of ways of getting 1, 2, or 3 aces and no king on the flop:

Pr=[C(4,1)*C(44,2)+C(4,2)*C(44,1)+C(4,3)]/C(50,3) = 20.67%

I verified this result using my flop simulation program.

I think I’m ok doing these kind of problems but I’m not that good on finding errors of others. I think your error is that you did not do the conditioning correctly given you first calculate no flop ace without regard to your KK holding.

Hi, statmanhal

I will try that now

46/50 x 45/49 x 44/48 = 0,7745

0.7745 = 77,45%
1 - 77.45% = 22.6%

88.24% x 22.6% = 20%

Hmmmm, at least I am getting closer.


Pr=[C(4,1)*C(44,2)+C(4,2)*C(44,1)+C(4,3)]/C(50,3) = 20.67%

I've seen ppl writing these equations but I don't understand them.
Can you like explain it more detailed what are you doing there exactly ?
(imagine that you are explaining it to a 10 year old, just in case )

Look at this thread in the Beginner's Forum

https://forumserver.twoplustwo.com/3...t=mathenoobics

Regarding my equation

Pr=[C(4,1)*C(44,2)+C(4,2)*C(44,1)+C(4,3)]/C(50,3) = 20.67%

Combos = number of ways of choosing r things out of n =C(n,r)= nCr = n!/[r!*(n-r)!]

C(5,2)= 5*4*3*2*1/[(2*1)*(3*2*1)]= 5*4/2 = 10

You want 1, 2 or 3 aces and no king.

You hold 2 kings leaving a deck of 50 cards. C(4,1) is the number of ways of getting 1 of 4 aces. There are 50-6 = 44 cards that are neither aces or kings. You want the other two flop cards to be from this sub-deck. The number of ways is C(44,2). If event 1 can happen in N1 ways and event 2 can happen in N2 ways, they both happen in N1 x N2 ways. That now represents the first term in my equation. Similar reasoning applies to the other two enumerator terms.

The total number of flops is the number of ways of choosing 3 cards from a 50 card deck, and that is the denominator.

Probability is then number of ways for an event to happen divided by the total number of ways.

Thx for the explanation, I understand it now.

But I really hate calculating something with formulas or methods that I don't understand and to ONLY use them because they are correct, if you know what I mean.

I rather try to do it my way regardless of the extra time or possible failure.

And that link doesn't work...

Try this

https://forumserver.twoplustwo.com/3...t=mathenoobics

I just did and it worked for me.