Am I doing this right? I split the problem up into the cases of 2 same, 3 same, 4 same, but I feel like something special has to be done for 2 of the same, because what if there are 2 pairs (like two 3's and two 4's)?
This is what I have:
For 2 of the same: $5\times 5\times 6\times {4\choose 2}=900$
For 3 of the same: $5\times 6\times {4\choose 3}=120$
For 4 of the same: $6\times {4\choose 4}=6$
Combined: $900+120+6=1026$
Total possibilities: $6^4=1296$
Probability of at least 2 die the same: $\frac {1026}{1296}\approx 79.17$%
Confirmation that I'm right, or pointing out where I went wrong would be appreciated. Thanks!
Sorry if the formatting could use work, still getting the hang of it.
Best Answer
You answer for "exactly two the same" counts some cases twice - when you get two pairs ($4545$, for example.)
The case of $2$ the same the others different counts to $6\cdot 5\cdot 4\cdot\binom{4}{2}=720$.
The case of two pair is $\binom{6}{2}\cdot\binom{4}{2}=90$.
Those two values add up to $810$, and you over-counted by $90$ - that is, you counted each "two pair" result twice.
This gives a total of $720+90+120+6=936=1296-360$.
This sort of problem is much easier to do by calculating the probability of the opposite (that they are all different) and subtract that from $1$. The probability that they are all different is $\frac{6\cdot5\cdot 4\cdot 3}{6^4} = \frac{360}{1296}$.