Probablity of three-of-a-kind or better in a roll of four dice So. I took a math competition and one of the questions seemed simple enough.
"Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value"
Hm. Easy.
Tried solving it.
I couldn't get it.
How can I solve this?
I know you can probably use Combination, and if so, can you step by step explain how (never learned combinations yet).
Or if it's possible without combination, can you explain how to solve it logically? thanks.

Imagine the four dice are rolled one after another.  There are $6\times6\times6\times6=1296$ different possible outcomes.  Of these, $6$ have all four dice showing the same value and $4\times6\times5=120$ have three dice showing the same value and the other die showing a different value.  So the probability that at least three of the four dice show the same value is
$${6+120\over1296}={126\over1296}={7\over72}$$
The $4\times6\times5$ can be understood as follows:  If exactly three dice have the same value, then the odd die can be any one of the $4$ dice, it can have any of $6$ values, leaving any of $5$ values for the three equal dice.
