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?
for the first roll, we have to consider two possibilities: either an odd or even number.
Odd: probability 3/6=1/2
after this roll the die becomes {1,2,9,4,6,49}
mod 8 these are {1,2,1,4,6,0} thus after an odd number on the first roll the odds of getting 1 mod 8 on the second one is 2/6=1/3 thus the overall odds of this outcome is 1/2*1/3 = 1/6
Even: probability 3/6=1/2
after this roll the die becomes {4,1,4,9,10}
mod 8 these are {4,1,4,1,2} again the odds of getting 1 mod 8 are 1/3 and thus the overall odds are 1/2*1/3 = 1/6
thus the total odds of getting 1 mod 8 on the second roll is
1/6 + 1/6 = 2/6 = 1/3
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Posted by Daniel
on 2010-06-10 12:03:08 |