Classical, thanks for the correction as my formula does indeed evaluate as you have indicated. I never could press buttons on a calculator

.

I do believe my formula is correct though.

For player one you can choose two cards from a royal flush C(5,2) ways. There are four possible suits and hence 4*C(5,2) hole card possibilities for the first player.

EG. Say player 0ne has AQ

s, player 2 is then required to have a pair of tens,jacks or kings [

**3***C(4,2)]. Player two is not allowed a

though and hence [3*

**C(3,2)**]. There are so few possibilities here that you can even list them. Player 3's hole card combinations follow from a similar line of reasoning. This deals only with the hole cards.

Given this constraint on hole cards AND the required Showdown scenario there is only one board of five cards that will allow this.

My more complete formula:

4*C(5,2)*3*C(3,2)*2*C(3,2)

***1**/C(52,2)*C(50,2)*C(48,2)*C(46,5)

Please feel free to point out the error of my ways if I am indeed compounding earlier mistakes.