Wikipedia has for the hold ‘em turn the probability of 3 cards in sequence but not 4 equal to 0.10544. I calculate it to be 0.09267. For the river, W has 0.19910 and I get 0.15317. Anybody care to do an independent look? I also disagree with 4 cards to a straight but not 5 on the river for which they show 0.03763.

W states that the values shown assume a “random” starting hand for the player but other ‘presumably correct’ values use C(52, x) as the total number of x board combos, meaning no consideration is given to a player’s holding.

I have confirmed Wiki's answer for the Exactly-3-cards-in-sequence for both the turn and river probabilities.

Here is what I did for the Turn calculation:

Consider A23x board.

Case 1: x is not A,2,3,4
C(4,1)*C(4,1)*C(4,1)*C(52-4*4,1) = 2304

Case 2: x is A,2,3
C(3,1)*C(4,2)*C(4,1)*C(4,1) = 288

Total = 2592

Now consider 234x thru JQKx (there are 10 of these) [for exposition, I will use 234x]

Case 1: x is not A,2,3,4,5
C(4,1)*C(4,1)*C(4,1)*C(52-5*4,1) = 2048

Case 2: x is 2,3,4
C(3,1)*C(4,2)*C(4,1)*C(4,1) = 288

Total = 2336

Consider QKAx (same as A23x)
Total = 2592.

So Grand Total = 2*2592 + 10*2336 = 28544
Prob = 28544/270725 = 0.105435405 (which is what Wiki gives)

Edit: I just ran through a program that deals out every possible board and got the same answers I derived via combinatorics above, so I "know" they are correct.


Feel free to show how you got different answers.

I have confirmed Wiki's answer for the Exactly-3-cards-in-sequence for both the turn and river probabilities.

Here is what I did for the River calculation:

Consider A23xy board.

Case 1: xy are not A,2,3,4
C(4,1)*C(4,1)*C(4,1)*C(52-4*4,2) = 40320

Case 2: x is A,2,3, y is not A,2,3,4
C(3,1)*C(4,2)*C(4,1)*C(4,1)*C(52-4*4,1) = 10368

Case 3: x is A,2,3, y is A,2,3 but x<>y
C(3,2)*C(4,2)*C(4,2)*C(4,1) = 432

Case 4: x is A,2,3, y is A,2,3 and x=y
C(3,1)*C(4,3)*C(4,1)*C(4,1) = 192

Total = 51,312

Now consider 234xy thru JQKxy (there are 10 of these) [for exposition, I will use 234xy]

Case 1: xy are not A,2,3,4,5
C(4,1)*C(4,1)*C(4,1)*C(52-5*4,2) = 31744

Case 2: x is 2,3,4, y is not A,2,3,4,5
C(3,1)*C(4,2)*C(4,1)*C(4,1)*C(52-5*4,1) = 9216

Case 3: x is 2,3,4, y is 2,3,4 but x<>y
C(3,2)*C(4,2)*C(4,2)*C(4,1) = 432

Case 4: x is 2,3,4, y is 2,3,4 and x=y
C(3,1)*C(4,3)*C(4,1)*C(4,1) = 192

Total = 41584

Consider QKAxy (same as A23xy)
Total = 51312

But note that AKQ32 would be double-counted as both A23 and AKQ.
So we have to subtract out 4^5 = 1024.

So Grand Total = 2*51312 + 10*41584 - 1024 = 517440
Prob = 517440/2598960 = 0.199095023 (which is what Wiki gives)

Edit: I just ran through a program that deals out every possible board and got the same answers I derived via combinatorics above, so I "know" they are correct.


Feel free to show how you got different answers.

Arrgh!. I forgot that given A23x, for example, x can also be A, 2, or 3, not only 5 to K. So, for the turn, my equation would be =(98*256+3*16*6*12)/COMBIN(52,4).

It's incredible how often I can mess up on these fairly mundane combinatorial problems.

Thx.

No worries.

By the way, what prob did you get for the Exactly-4-cards-to-a-Straight on the river?
I am getting a different prob than Wiki got but have not double-checked my work.

Edit: I just realized that Wiki's prob is for exactly 4 cards in sequence on the river (such as 3456K), but excludes other ways to have 4-to-a-straight (such as 3457K). Maybe this was already obvious.

Okay, here is what I got for the probability of having Exactly-4-cards-to-a-Straight on the river (not 5).

This is a little messy, but I think at this stage getting the right answer takes priority over pretty exposition.

Case I. Exactly 4-cards-in-Sequence (not 5)

Clearly A-4 and J-A are C(4,1)*C(4,1)*C(4,1)*C(4,1)*C(52-5*4,1) + C(4,1)*C(4,2)*C(4,1)*C(4,1)*C(4,1) = 9728

And 2-5 thru T-K are C(4,1)*C(4,1)*C(4,1)*C(4,1)*C(52-6*4,1) + C(4,1)*C(4,2)*C(4,1)*C(4,1)*C(4,1) = 8704

So Case I subtotal = 2*9728 + 9*8704 = 97792 (this is what Wiki gives)

Case II: 4-to-a-Straight but not 4-in-Sequence
Here it matters how the 4 cards are "distributed" such as 3457, 3467, 3567; the middle case must be treated differently

Anyway, for A-5 and T-A bookends, there are two cases of 9728 and one case of 8704 (for a total of 28160).

For 2-6 thru 9-K bookends, there are two cases of 8704 and one case of 9728 (for a total of 27136).

However, we have done a bunch of double-counting above. For example, 34678 is counted above as 3467 and 4678. It turns out that there are 9*3-1 = 26 of these boards which are being double-counted. And there are 4^5 = 1024 boards of five distinct ranks of each kind.

So Case II subtotal = 2*28160 + 8*27136 - 26*1024 = 246784

Therefore, GRAND TOTAL = 97792 + 246784 = 344,576

And Prob = 344,576 / 2,598,960 = 0.132582264

I ran this thru my deal-all-possible-boards program and got this answer, so I am pretty sure it is correct.

Oh, I'm pretty sure they meant 4 in sequence (I'm pretty sure it was previously worded that way) and I now get what wiki gets and you did too in case 1. I did not do case II.

Crapola. I thought Case II was the most "interesting".