After taking your 2 hole cards out of the deck the total number of flops from the 50 unseen cards is 50 C 3 (or "50 choose 3") which is equal to (50*49*48)/(3*2*1)
There are 2 cards of the remaining 50 in the deck which will make you a set. The number of flops containing exactly one of those 2 cards is (2 C 1) * (48 C 2) = (3/1) * ( (48*47)/(2*1) )
The probability of flopping a set is the ratio of these, which can be simplified to: (47*3*2)/(50*49) = 0.115 = 11.5% Converting percentage to odds against gives about 88.5 to 11.5 which reduces to 7.7 to 1
why divided by (3*2*1)?
huh@#$? why (3/1)?
I'm lost on the dividing parts.
Comments
(50*49*48)/(3*2*1)
There are 2 cards of the remaining 50 in the deck which will make you a set. The number of flops containing exactly one of those 2 cards is
(2 C 1) * (48 C 2)
= (2/1) * ( (48*47)/(2*1) )
The probability of flopping a set is the ratio of these, which can be simplified to:
(47*3*2)/(50*49)
= 0.115 = 11.5%
Converting percentage to odds against gives about
88.5 to 11.5
which reduces to
7.7 to 1
ScottyZ
why divided by (3*2*1)? I figured the 50*49*2 <-- cause there are two of your missing cards in the deck, right?
:banghead: huh@#$? why (3/1)? I'm lost on the dividing parts. I took this s#it in high school......
This represents the 6 different ways of ordering 3 distinct objects (i.e. the 3 flop cards).
Sorry, I botched that one. It should have been
2 C 1 = 2/1
This is just using the formula for combinations:
n C r = n!/( (n-r)!*r! )
Proving that this really is the correct formula for "r objects taken from n without replacement disregarding order" takes a lot of work.
Check out
http://mathforum.org/dr.math/faq/faq.comb.perm.html
or
http://mathforum.org/dr.math/faq/faq.mcdonalds.html
for some more information on combinations (i.e. the "50 choose 3" stuff).
ScottyZ