1. ## Rolling dices

Six fair dice are tossed independently. Find the probability that the number of 1’sminus the number of 2’s will be 3?

I know there will be two cases!
Case 1: 3 ones and 0 twos
Case 2: 4 ones, 1 two, and one (any 3-6)

I dont know how to calculate the probability. I keep getting the wrong answer.

2. Originally Posted by bigb
Six fair dice are tossed independently. Find the probability that the number of 1’sminus the number of 2’s will be 3?
Case 1: 3 ones and 0 twos
Case 2: 4 ones, 1 two, and one (any 3-6)
Case 1: 111XXX in any order. $X \notin \left\{ {1,2} \right\}$
$\frac{{6!}}{{\left( {3!} \right)^2 }}\left( {\frac{1}{6}} \right)^3 \left( {\frac{4}{6}} \right)^3$

Case 2: 11112X in any order:
$\frac{{6!}}{{\left( {4!} \right)}}\left( {\frac{1}{6}} \right)^5 \left( {\frac{4}{6}} \right)$

3. Originally Posted by Plato
Case 1: 111XXX in any order. $X \notin \left\{ {1,2} \right\}$
$\frac{{6!}}{{\left( {3!} \right)^2 }}\left( {\frac{1}{6}} \right)^3 \left( {\frac{4}{6}} \right)^3$

Case 2: 11112X in any order:
$\frac{{6!}}{{\left( {4!} \right)}}\left( {\frac{1}{6}} \right)^5 \left( {\frac{4}{6}} \right)$
I was being really really dumb...its should be 4/6 not 5/6, which is what i kept saying.You dont want 1 or 2 to come up again. Thanks for the help man!