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Three fair 6-sided dice each have their sides labeled $0\,,\,1\,,\,e\,,\,\pi\,,\,i\,,\,\sqrt 2\,$. If these dice are rolled, the probability that the product of all the numbers is real can be expressed as a/b, where a and b are positive, co-prime integers. What is the value of a+b?

When I tried I got the total possibilities to be 216 and the rest I got wrong. Can you help me find the answer to the problem? (I think it is above 99/216).

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1 Answer 1

up vote 4 down vote accepted

For the product to be real, either at least one $0$ has to be rolled ($6^3-5^3=216-125=91$ possibilities), or there must be an even number of occurrences of $\mathrm i$, which makes $4^3=64$ possibilities for zero occurrences and $3\cdot4^1=12$ possibilities for two occurrences, for a total of $91+64+12=167$ possibilities. The fraction $167/216$ is already reduced, so the sum is $167+216=383$.

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Thanks that is the answer –  chndn Dec 4 '12 at 11:32
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@joriki, won't there be a case of even number of $\sqrt 2?$ –  lab bhattacharjee Dec 4 '12 at 11:40
    
@lab: I included both even and odd numbers of $\sqrt2$. Do you mean a restriction to even numbers of $\sqrt2$? That's not necessary because the product is only required to be real, not rational. –  joriki Dec 4 '12 at 12:12

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