A magic die, with the numbers 1, 2, 3, 4, 6, and 7 on its six faces, is rolled.
After this 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 the previously odd numbers are increased by 3 and then all the even 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?

(In reply to analytical solution by Daniel)

A five-sided die? {4,1,4,9,10}

On the interpretation alternative to that of ed bottemiller's, the six faces after an even roll are {4,1,6,4,9,10} and two of the faces are 1 mod 8, for probability 2/6 = 1/3.

In the case of the odd first roll, the 49 face is congruent to 1 mod 8, not 0. That increases those odds to 3/6 = 1/2.

Over all, then, 1/2*1/2 + 1/2*1/3 = 5/12, or the easier way, choosing the 5 possible out of the twelve equally likely possibilities:

1,2,9,4,6,49
4,1,6,4,9,10

where the 1, 9, 49, 1 and 9 satisfy the criterion.

 

Posted by Charlie on 2010-06-10 12:47:59