Remember
this Magic Die? I've got another magic die, which is n-sided (with numbers 1 through n on its faces) and where n=0 mod 8.
After the first roll, if an odd number appears on the top face, all odd numbers on the die are squared. If an even number appears on the top face, all odd numbers are increased by 3 and then all numbers are halved and then squared.
If the given die changes as described and assuming a perfectly balanced die, what is the probability that the number appearing on the second roll of the die is 1 mod 8? How about 4 mod 8?
Not sure why nobody has tackled this yet.
Consider the numbers 4k+1, 4k+2, 4k+3, 4k+4.
When squared and converted to mod 8, they equal 1, 4, 1, and 0, respectively, no matter what k is.
Well, the the numbers on the die are equally divided among numbers of one of the above forms.
And if you add 3 to the odd numbers on the die and divide by two, the resulting numbers are still divided equally among the four forms.
(even numbers 8m+2, 8m+4, 8m+6, and 8m+8 become 4m+1, 4m+2, 4m+3 and 4m+4)
(odd numbers 8m+1, 8m+3, 8m+5, and 8m+7 become 4m+2, 4m+3, 4m+4 and 4(m+1)+1)
So, no matter whether the first roll is odd or even, the second roll has a 50% chance of being 1 mod 8, a 25% change of being 4 mod 8, and a 25% chance of being 0 mod 8.
Edited on February 28, 2012, 9:56 pm
Don't forget 0! (spoiler)
