C(5,3)=10, (the number of ways you can select 3 of 5 dice, distinctly)
C(5,3)*6=60, (each one of the ten ways above has 6 choices. e.g.1-6)
C(5,3)*6*5*5 = 1500, (each one of the 60 choices has a remaining 5 choices each for the two remaining selections)
P(3 of a kind) = 1500/6^5.
Before, when I thought P(1 pair)=P(3 of a kind) was because I failed to factor in the remaining two numbers(thus leaving out the factor of 125 and 25 respectively). And since C(5,3)=C(5,2) thought they were equal...in case anyone was wondering what all this fuss was about.
Originally posted by talzamir Joe wrote: Can you walk me through the two pair calculation? I'm struggling to come up with it.
Actually.. I got annoyed at the mistakes I did in the first calculation, so I did it the easy way, and wrote a bit of code that went over the 7,776 combinations and checked what they are.
One pair is indeed in 3,600 combinations. In the first go I did it ...[text shortened]... ps are one of five, one of four, and one of three possibilities.
6 x 10 x 5 x 4 x 3 = 3,600.
Ok, I was asking how you calculated "two pair's" number of calculations, but don't worry about it. I did some research and found why I was twice as high, because in each one of the combinations the order of the pairs can be switched (e.g. 1,1,2,2,5 has a compliment combination where the order of the pairs are switched i.e. 2,2,1,1,5), I missed that entirely.