When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than

Hi Mellie,

It is very late and I am not going to get a final answer tonight but I will show you where my thoughts are taking me.

I'm going to let 1 and 6 be the biased sides so 2,3,4 and 5 all have a chance of 1/6 of being rolled.

so the chance of throwing a 1,2,3 or 4 is      $${\frac{{\mathtt{4}}}{{\mathtt{6}}}}$$   =  $${\frac{{\mathtt{2}}}{{\mathtt{3}}}}$$

the chance of throwing a 1 is    $${\frac{{\mathtt{m}}}{{\mathtt{n}}}}$$        and the chance of throwing a 6 is      $${\frac{{\mathtt{1}}}{{\mathtt{3}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{m}}}{{\mathtt{n}}}}$$

now to roll a 7 these are the possibilities

$$\\P(1,6) =\frac{m}{n}\times\frac{n-3m}{3n}=\frac{m(n-3m)}{3n^2}\\\\ P(2,5) = \frac{1}{6}\times\frac{1}{6}=\frac{1}{36}\\\\ P(3,4) = \frac{1}{6}\times\frac{1}{6}=\frac{1}{36}\\\\ P(4,3) = \frac{1}{6}\times\frac{1}{6}=\frac{1}{36}\\\\ P(5,2) = \frac{1}{6}\times\frac{1}{6}=\frac{1}{36}\\\\ P(6,1) = \frac{n-3m}{3n\times}\frac{m}{n}=\frac{m(n-3m)}{3n^2}\\\\ NOW\\\\$$

$$\\\frac{4}{36}+\frac{2m(n-3m)}{3n^2}=\frac{47}{288}\\\\ \frac{32}{288}+\frac{2m(n-3m)}{3n^2}=\frac{47}{288}\\\\ \frac{2m(n-3m)}{3n^2}=\frac{15}{288}\\\\ \frac{m(n-3m)}{n^2}=\frac{15*3}{288*2}\\\\ \frac{m(n-3m)}{n^2}=\frac{5}{64}\\\\ $Now if we are lucky n will equal 8$\\\\ m(8-3m)=5\\\\ 8m-3m^2=5\\\\ $Just by sight I can see that m=1 will work.$\\\\ $So one possibility, maybe the only possibility is $ n=8 \;and\; m=1\\\\ m+n=9$$

There is probably a much quicker way of doing this.  Plus I have not shown that this is the ONLY solutions. 

What is $F$  ?

The sides are numbered 1,2,3,4,5, and 6   So what is F  ?