how to find the number of combinations that has at least 2 numbers alike in the experiment of 4 people each rolling a fair die 4 people each roll a fair die once. What's the probability that at least two people will roll the same number? I could solve the question in the following way: 
The number of the different results for the whole experiment is $6^4$. Getting the same number from at least two people is the opposite of getting no number like the other, which can be found by $6P4$. As a result, the final solution is $1-360/1296 = 13/18$.
However, I thought of a way to get the other "$1296-360=936$" different combinations, in which at least 2 numbers are similar. I tried $6(4C2)+6(4C3)+6(4C4)$ based on that getting two similar number is the same as choosing 2 numbers from the four similar numbers multiplied by the 6 different numbers etc..., but the total only summed up to $66$. I also tried permutations instead of combinations but neither did it work. So, how can I get the $936$ combinations? and where are they?
Thanks
 A: The catch is that when you do the second computation, you have to also compute the ways for the other people (i.e. those not matching anyone) to have their rolls occur. Warning: This route is messy.
Call the four people A, B, C, D.


*

*If exactly two people roll the same thing, then instead of $6(4C2)$ we need
$  (4C2) \cdot 6 \cdot 5 \cdot 4 $. The $4C2$ term designates the number of ways to select who will be in the pair; they can roll any of 6 options. Then, we consider the first person (alphabetically) who wasn't in the pair; they have 5 remaining ways to roll so as not to match the pair. The other person who wasn't in the pair then has 4 ways to roll.

*Similarly, instead of $6(4C3)$ we would need $(4 C 3) \cdot 6 \cdot 5$. The 6 again represents the number of ways for the pair to roll, and the 5 represents the number of ways for the loner to roll.

*The $6 (4C4)$ term is still correct.

*And finally, we still have to consider the possibility of having two pairs, such as A and C rolling a 6 while B and D roll a 3. The number of ways for this to happen is $\frac 1 2 (4 C 2) \cdot 6 \cdot 5$. The $4 C2$ represents the number of ways to select the "first" pair, and by exclusion the other two will be in the second pair. Here, we also have to divide by 2, since there's no real difference between the "first" and "second" pairs and we've overcounted the number of possibilities. (We could have picked A and C to be in the first group, or we could have picked B and D to be in the first group; these would each get counted separately if we don't divide by 2, but they're not actually different.) The 6 and 5 in this expression serve the obvious purpose.


Putting it all together:
$$(4 C 2) \cdot 6 \cdot 5 \cdot 4 + (4 C 3) \cdot 6 \cdot 5 + (4 C4) \cdot 6 + \frac 1 2 (4 C 2) \cdot 6 \cdot 5 = 936.$$
