If Greg rolls four fair six-sided dice, what is the probability that he rolls an equal number of 1's and 6's?
Each die has 6 faces, and the odds of a 6 or 1 showing up are 1 in 6. So, there are several possible victory conditions:
1) One of the 4 die show a 1, and none show 6.
2) Two of the die show one and one or none show 6.
3) Three of the die show one, and 2 or less show 6.
All 4 dice show 6. There are 1296 possible combinations of dice that could be rolled.
The probabilities are as follows.
4 in 81 1 of the 4 dice shows a 1, and none show 6.
1 in 81 2 of the dice show one and one or none show 6.
5 in 1296 3 of the dice show one, and the 4th die result doesn't matter, as long as it isn't 1. 1 in 1269 - All 4 dice show 6.
So, there is a 43 in 648 chance of this occurring.
Best answer I could come up with. Had to do a little googling to confirm my answer and I found this explination. More clearer than what I came up with but same answer.