Select $6$ mirrors on $6$ faces of a room (a cube). When you go to the center, how many selves you can see from one mirror?

Question

Select $6$ mirrors on $6$ faces of a room (a cube),when you go to the center, how many selves can you see in one mirror ?

Answer 1

The short answer here is that you'll get the same result as if you'd be searching for the visible points in a primitive cubical lattice, when looking from the origin (which clearly has to be a lattice point itself).

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You have to observe that a point is visible exactly when its coordinates (for comfort all being integers) would be coprime, i.e. when having $P=(x,y,z)$ and $gcd(x,y,z)=1$.

Next consider probabilities. Let be $p$ some prime number. For having $p$ dividing some integer number $x$, i.e. $p|x$, the probability is $1/p$. Thus for having all three, $p|x$, $p|y$, and $p|z$, the probability then is $1/p^3$. Thus the searched for probability happens to be

$$\prod_p \left(1-\frac1{p^3}\right)$$

If you want to calculate that thing, you'd first have to have a look onto Euler's product formula and Riemann's zeta function.

$$\zeta(s)=\sum_{n\geq 1}\frac1{n^s}=\prod_p\left(\sum_{k\geq 0}\frac1{p^{ks}}\right)=\prod_p\frac1{1-\frac1{p^s}}$$

The one but last equality is the fundamental theorem of arithmetic that every integer has a unique prime factorization. Thus multiplying out the product would produce any $1/n^s$ exactly once. - The last one then is just the geometric series formula.

Thus your searched for probability happens to be

$$\prod_p \left(1-\frac1{p^3}\right)=\frac1{\zeta(3)}\approx 0.83$$

--- rk