# Three dice having sides labelled $e,i,π,1,0,\sqrt2$ are rolled. Find the probability of getting the product of the three results a real number.

There are three fair six-sided dice with sides $0,1,e,\pi,i,\sqrt2$. If these dice are rolled, the probability that the product of all the numbers is real can be expressed as $\frac ab$ where $a$ and $b$ are positive, co-prime integers. What is $a+b$?

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 $\frac{99}{216}$).

• Seems like an AIME problem ;) – 6005 Oct 1 '16 at 19:56
• Which of the initial six numbers is real? When is the product of two (three) numbers real, depending on the realness of the initial numbers? – TMM Oct 2 '16 at 3:02

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 $167+216=383$.
• @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