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?

After the first roll we'll get the numbers : 1, 2, 9 ,4 ,6 , 49 with a probability of 50%, and the numbers 4, 9, 25, 1 ,4 ,9 also with a 50% probabilty.

The first group contains 3 numbers (1, 9, 49) which are 1 mod 8, and therefore rolling the second time ( given that this is the number distribution will give a 50% result of 1 mod 8.

The second possibility of the number distribution contains three 1 mod 8 numbers (9, 25, 1, 9) which will give a 2/3 probability that given that distribution , the second roll will result in a 1 mod 8 number.

So the overall probability will be 0.5*0.5 + 0.5* 2/3 = **7/12**