bobhammock
08-04-2005, 06:46 AM
Hi all,
This forum seems great.
My question is very specific and concerning an archived thread with BruceZ in the drivers seat:
http://archiveserver.twoplustwo.com/showflat.php?Cat=&Number=396447&page=&view=&sb=5&o =&fpart=1&vc=1
In Brucez's answer it follows:
" C(8,2)* 78/C(52,2) * 73/C(50,2) = 9.8%. C(8,2) is the number of ways to pick the 2 players. There are 73 pairs remaining after the first player chooses a pair"
- Now, why are there 73 pairs left after 1st player chooses a pair? I know we can't just pick from 12 ranks and pretend that the 2nd player can't pick a pair of the same rank as the first. But why 73...?
I want of course to be able to make the next term myself.
ty.
This forum seems great.
My question is very specific and concerning an archived thread with BruceZ in the drivers seat:
http://archiveserver.twoplustwo.com/showflat.php?Cat=&Number=396447&page=&view=&sb=5&o =&fpart=1&vc=1
In Brucez's answer it follows:
" C(8,2)* 78/C(52,2) * 73/C(50,2) = 9.8%. C(8,2) is the number of ways to pick the 2 players. There are 73 pairs remaining after the first player chooses a pair"
- Now, why are there 73 pairs left after 1st player chooses a pair? I know we can't just pick from 12 ranks and pretend that the 2nd player can't pick a pair of the same rank as the first. But why 73...?
I want of course to be able to make the next term myself.
ty.