The odds of both holecards being suited?

[MortalWombatDotCom]

i was looking over my post and realized that the quotient could actually be simplified alot. in fact, gaming mouse will probably explain why jumping to the simplified version is the easy way [img]/images/graemlins/wink.gif[/img]

anyway, its the same sum, but
39! * 11! / 50! is the same as 1/(50 choose 11)
and
*(50 - 2k)! / (39 - 2k + i)! * (11 - i)!
is the same as
((50 - 2k) choose (11 - i)

so you get

2^i * (k choose i) * ((50 - 2k) choose (11 - i)) / (50 choose 11)

programming this into my programmable calculator, i can then come up with #opp: %chance as follows

1: 4.5%
2: 8.8%
3: 13%
4: 17.1 %
5: 21%
6: 24.8%
7: 28.5%
8: 32.1%
9: 35.5%

since i did the numbers for 1 and 2 opps by hand the long way before coming up with the summation, and got the same answer both ways, i'm pretty confident in this formula.

hope this answered your question.