Four-Flush in 5-Card Stud Poker

Hello Wizard, I have enjoyed your site here and the Odds site for quite some time. Especially helpful is the Carribean Stud Strategy offered for A-K hands.

My question is about 5-Card Stud that our group enjoys as 1-down, 3-up, 1 down. For a long time in 5-card stud, one of the options has been "Four-Flush beats a pair". I tend to believe thats true, but how many 4-Flushes are there (or what are the odds of getting one), presuming the 5th card can pair, but not be a hand ranked higher such as a Straight, Flush, or Str-Flush? I have not found the answer in the usual places that deal with poker, and probabilities.

Thanks much for a great pair of wizards.

98Clubs

Quote: 98Clubs

...how many 4-Flushes are there (or what are the odds of getting one), presuming the 5th card can pair, but not be a hand ranked higher such as a Straight, Flush, or Str-Flush?

I get 110,940. This is 4*combin(13,4)*39-10*4*15. So, my answer is that for your game, 4-flushes are rarer than pairs (1,063,920) and also rarer than two pairs (123,552.)

Thanks very much for that. I didn't know its rarer than 2-pairs.

98Clubs

Quote: ChesterDog

I get 110,940. This is 4*combin(13,4)*39-10*4*15. So, my answer is that for your game, 4-flushes are rarer than pairs (1,063,920) and also rarer than two pairs (123,552.)

I had the same, but I'm not sure where you're accounting for the 4;

4*combin(13,4)*39-10*15 =111,390

Quote: Kelmo

I had the same, but I'm not sure where you're accounting for the 4;

4*combin(13,4)*39-10*15 =111,390

I agreee with ChesterDog.
There are:
4 suits for your 4-flush
combin(13,4)=715 ways to select 4 of the 13 cards in that suit
13*3=39 cards that are not in the 4-flush suit.
4*combin(13,4)*39 = 111540 ways to get a 4-flush
Now subtract the ways that make a straight:
10 straights: ace-five through ten-ace
4 4-flush suits
3 other suits
5 positions for nonsuits
10*4*3*5=600
111540 - 600 =110940

Edited

Sorry for the late post, I have been looking at stud poker threads to day and found this one. We play 5-Stud with 4-Flush, Paired 4-Flush, and 4-Str-Flush allowed. SO my answer comes from a different perspective.

First there are 13C4 * 39 total 4-Flushes = 111.540
There are 4*9*46 + 4*2*47 4-Str-Flushes = 2032 these include Paired, Straights, and Flushes as a 4-Str-Flush ranks higher.

There are (715-11)*12*4 Paired 4-Flushes = 33792 (13C4 - 11 4-Straight-Flushes * 12 cards thst Pair * 4 Suits)

There are 3*10*3*4 4-Flushes in a hand ranked Straight = 360 as (X, X+2, X+3, X+ 4 || X, X+1, X+3, X+4 || X, X+1, X+2, X+4). The 5th card in these 3 types cannot make a 4-Str-Flush, a Flush, a Straight, or a pair. Else the hand is ranked higher than 4-Flush. Each of the 10 possible Straights have 9 ways to make a 4-Flush/Straight: 3 for substitution {X+1, X+2, X+3} times three for suit times four suits of four-flush.

These account for all 4-Flushes ranked greater than a 4-Flush
Subtraction accounts for the balance of 4-Flushes ranked as a 4-Flush.

111540 - 2032 - 33792 - 360 = 4-Flushes = 75356
In a game without the Paired and 4-Str-Fl options, add these back as 4-Flushes for a subtotal of 111180.

Now subtract the 4-Str-Fl that are also Straights: there are 240 of them (Ace or Jack lowcan draw 3, the rest can draw 6 cards)

Final total is 111180 - 240 = 110940

I found some clarity here . But I note there are differences in answers that I hope can be cleared-up.

N&B

**EDITED** I found an error pertaining to the 4-Straight-Flush that is also a Flush. The value of 316 should not be subtracted from 4-Flushes at all. When I adjusted this, the numbers add up to 110940. The error is that the 316 belongs only to Flushes. Double checking the Straights is correct there are a total of 600 removed 240 for 4-Straight-Flush, and 360 for 4-Flush.
The original post has been corrected.

N&B