This is the solution to the dice problem in “First technical post!?”

Here you need the sum to be exactly within the range 1-12

So how do we proceed? The total number of possible outcomes with 2 dice is 6*6 = 36. And the total numbers in the range 1-12 is 12, since the probability should be the same, every sum can occur 36/12 = 3 times. is this fine? Now since you need a sum 1, there should be atleast 1 zero and since the sum 1 should appear 3 times there should be 3 zeroes. Now since there should be a sum 12 there should be atleast one 6 on the other die, since the sum 12 should also appear 3 times, there should be 3 sixes too. Now you have 3 zeroes and 3 sixes, so 3+3 = 6, which covers all the sides of the other cube.

If you have any other number other than 0 or 6 on the other cube, the probablity will not be the same!

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