Rolling tetrahedron

Probability Level 4

A regular tetrahedron is resting on a surface. Every second, the tetrahedron rolls over onto a different face. The direction that it rolls in is uniformly random and independent (i.e. when the tetrahedron is resting, there is a 1/3 chance that it will roll in any one of three possible directions during that second).

Let $a_n$ be the probability that after $n$ rolls (seconds), the tetrahedron will be resting on the same face that it started on.

Find:

$-\log_3(4a_{98}-1)$

Can anyone tell me if it is possible to generalize this result for other polyhedra? After some labor, I was able to derive simple explicit formulae for $a_n$ on the other Platonic solids (sans icosahedron).

The answer is 97.

2 solutions

Otto Bretscher
Jan 1, 2016

We have $a_n=\frac{1}{3}(1-a_{n-1})$ since there is a 1/3 chance that the tetrahedron will roll over to the original face if it's currently sitting on a different face. If we consider the difference from the equilibrium, $d_n=a_n-\frac{1}{4}$ , we find the recursion $d_n=\left(-\frac{1}{3}\right)^nd_0=\left(-\frac{1}{3}\right)^n\frac{3}{4}=\frac{(-1)^n}{3^{n-1}4}$ so $a_n=\frac{1}{4}+\frac{(-1)^n}{3^{n-1}4}$ . For even $n$ we have $\log_3(4a_n-1)=n-1$ , so, the answer is $\boxed{97}$ for $n=98$ .

Moderator note:

Great approach. In this case, because all faces are the same (we can go from any face to any other face), the analysis is much easier.

If we wanted to deal with the cube, then we will have to consider the opposite face as seperate.

Andreas Wendler
Jan 1, 2016

Yes Sir. Got same solution but as physicist found it by induction of course! Considering the probabilities for increasing n (3/9; 6/27; 21/81; 60/243 etc.) obtained formula: $\frac{1}{3^{n}}((-1)^{2}3^{n-1}+(-1)^{3}3^{n-2}+...+(-1)^{n}3) = \frac{3^{n-1}+(-1)^{n}}{4 \times 3^{n-1}}$ .

So finally the solution of the problem was reduced to the determination of a sum formula for an alternating row.

Rolling tetrahedron

The answer is 97.

2 solutions

Moderator note:

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