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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?

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 Assumptions | Comment 2 of 8 |

I shall assume that the magic transformations apply only to the first roll, not also the second: we are told to take the "die changes as described" and the description given says "after THIS roll..." (singular).

The odds are 50/50 that an odd number (1,3,7) will appear on the top, and that an even number (2,4,6) will appear.

If an odd number is on top, the faces after the magic will be 1,9,49 for the odds squared and the unchanged evens 2,4,6.  Of these six faces, three (1,9,49) = 1 (mod 8).

If an even number is on top, the specs are ambiguous.  The "previously odd numbers" are increased by 3: presumably 1,3,7 changed to 4,6,10. ---  But "THEN all the even numbers are halved and then squared. Does this mean that the original odds (1,3,7) which have become (4,6,10) also get this second transformation? (I shall interpret THEN as meaning that the odds have already become the evens 4,6,10, and hence all six evens (2,4,6 original, 4,6,10 just changed) are then to be halved and then squared:  giving (1,2,3, 2,3,5), and then these are squared giving (1,4,9, 4,9,25). Of these six faces,four (1,9,9,25) = 1 (mod 8).

Since 7 of the 12 dice faces are = 1 (mod8), I assume this is the answer intended: odds are 7/12.  Isn't that the nice thing about invoking "magic" -- words can mean whatever you want them to mean (pace Alice).

 Posted by ed bottemiller on 2010-06-10 12:31:45

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