Four standard six-sided dice are rolled. What is the probability that the product of the numbers on the top faces of all four dice is 24? Express your answer as a common fraction.
Since the sample space of 4 dice is 6^4 =1,296, which counts all the "permutations" of the 4 dice, you should therefore count all the permutations whose product==24, since they are all included and counted in the total of 1296 permutations. Therefore, that number of permutations is:
1146 1164 1226 1234 1243 1262 1324 1342 1416 1423 1432 1461 1614 1622 1641 2126 2134 2143 2162 2216 2223 2232 2261 2314 2322 2341 2413 2431 2612 2621 3124 3142 3214 3222 3241 3412 3421 4116 4123 4132 4161 4213 4231 4312 4321 4611 6114 6122 6141 6212 6221 6411 Total = 52
Then the probability is: 52 / 1296 ==13 / 324
+118608 Using Catmg's combinations that do not appear to be in question
1, 1, 4, 6
1, 2, 2, 6
1, 2, 3, 4
2, 2, 2, 3
Assume all the dice are different in colour. (That will make it much easier)
Also assume that the dice are rolled one at a time. It is probably easier to act as if order counts.
That means they are all unique and a green 1 is different from a blue 1
The total number of combinations is 6^4=1296, just as catmg found.
Each of those 4 combinations above can be formed in 4! ways. But I do need to allow for the double counting here
1, 1, 4, 6 4!/2 = 12
1, 2, 2, 6 4!/2 =12
1, 2, 3, 4 4!=24
2, 2, 2, 3 4
total 52
So we have a prob of 52/1296 = 13/324
I am not 100% sure that this is correct ....... But it is the same as guest found so maybe it is ok.
