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.

The number of combinations of rolls of the 4 die is, correctly, $6*6*6*6 = 6^4$.
The number of ways in which all 4 die roll the same value is 6 (either they all roll 1 or 2 or 3 and so on...)
The number of ways in which Exactly 3 die roll the same value is more complicated.  Consider first the number of ways in which the first 3 die roll the same value but the fourth die rolls another.  Label the die $W_1 , W_2, W_3, W_4$ for convenience.  If $W_1 , W_2, W_3$ all roll 1's for example, then the number of combinations in which $W_4$ is not 1 is just 5.
(The combinations are just (1,1,1,2), (1,1,1,3), (1,1,1,4) (1,1,1,5) and (1,1,1,6)).  Since there are 6 possible values for the first 3 die, there are 6*5= 30 ways in which $W_1 , W_2, W_3$ are all the same and $W_4$ is different.  
There are also many ways to choose 3 die out of the 4 namely, ${ 4 \choose 3} = 4$.
Putting it all together:
P(at least 3 dice are the same) = P(exactly 4 dice are the same) + P(exactly 3 dice are the same)
$$=  \frac{6}{6^4} + \frac{4*5*6}{6^4} = \frac{1+4*5}{6^3} = \frac{21}{6^3} = \frac{7}{72} = 0.09722$$
Good Question!
