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 A Magic Die (Posted on 2010-06-10)
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?

 See The Solution Submitted by Ady TZIDON Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Can Merlin settle this? | Comment 5 of 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?

 Posted by ed bottemiller on 2010-06-10 13:21:46
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