Is it 2.79%?

Thanks

Could you show how you got the 2.79% figure?

These types of problems are straightforward if you logically determine which hands you are trying to tally and then apply the "Choose" method relentlessly. You probably know that C(X,Y) is the number of ways to choose Y items out of X items. If you don't know the formula for C(X,Y) it can be found in many places on the internet. Also, the evaluation of C(X,Y) is given by many calculators and computer programs.

This is how I would approach it. Fill in the blanks in the following: To get exactly four out of five cards to be the same suit the suit can be A chosen out of B suits, once the suit is chosen I need to choose C out of D cards of that suit, and the suit of the fifth card can be E chosen out of the remaining F suits, and I need to choose G out of H cards of that suit. Since these choices are independent, you simply multiply the various choice tallies together to get the total number of ways to do them all. Of course, the total number of five-card hands is J chosen out of K cards. Then we divide the desired tally by the total.

I suggest trying to find the answer yourself using the above as a guide. But if you are really only looking for the answer, feel free to say that.

I just tried to extrapolate it from Holden. I know it is 23.53% to get dealt two suited cards and then it is 11% to hit a flop with two of your suit so I just multiplied these amounts to get 2.59% (not 2.79%)

I forgot how to do the Choose "stuff". I should relearn all of that and plan to. But wouldn't mind just getting a yay or nay as of right now.

I'm written some articles on Archie for 2+2 Magazine and just need some facts on how often you are dealt certain hands to fill out a suggested opening hand distribution

Your solution is for hold'em whereas the thread title indicates five card draw. They are not the same. The hold'em case specifically limits how the flush cards fall.

So nay to your answer.

The correct answer using combinatorics is

What about one in your hand three on the board?

Thanks man. Yeah I know they were different games, I just thought I could figure it out that way.

Thanks, that was probably the difference

Answer already given above but just to complete my "word problem" given in post 2 ...

The number of five-card hands with exactly four of one suit is:

C(4,1)*C(13,4)*C(3,1)*C(13,1) = 111,540

The total number of five-card hands is:

C(52,5) = 2,598,960

So percent of five-card hands that have exactly four of one suit is:

111,540 / 2,598,960 = 4.29%

Thanks gents!

So given that someone has a four flush the probability that they do not have an ace high flush draw = (12/13)*(11/12)*(10/11)*(9/10) = 69%

Right?

Then no ace or king equation would be the same except the numerators would all be one less. Is that right?

Yes