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?
Once again, my posting had not seen Charlie's approach. I agree with the first part (if odd on top) the originally odd 1,3,7 become 1,9,49, and the evens are unchanged 2,4,6; 1,9,49 are 1 mod8, so odds in that case are 1/2.
For the second case (even on top), there are three consecutive stages (or at least so I interpreted "then"): (1) the odd numbers are each increased by 3, so 1-3-7 become 4-6-10; (2a) at this stage the six would read 4,6,10,2,4,6 -- these are all even, so all get halved, giving 2,3,5,1,2,3; (2b) these are "THEN" squared, giving 4,9,25,1,4,9 -- of which 4 are = 1 (mod8), giving a 2/3 odds for the "even on top" case.
I think my approach differed from Charlie's in the transition from (2a) to (2b). After both arriving at 2,3,5,1,2,3 from stage 2a, Charlie squared only those that were still even: which gives 4,3,5,1,4,9, of which only two (i.e. 1/3) are = 1 (mod8).
Will Merlin come from behind the curtain?
Can Merlin settle this?
