Planet Hopping

Here is the H3X system, which contains 7 planets labeled A to G.

Every month, you can move to a planet which is (a) next to you and (b) connected via the H-link (represented by the grey lines). You are currently at Planet C.

If you were to randomly move to a planet each month (as long as it agrees with the rules above), which planet are you most likely to be on after 4 months?

A, C, or G A or G B, D, E, or F C All are equally likely

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1 solution

Jordan Cahn
Oct 16, 2018

Assuming that you are equally likely to move to each of a planet's neighbors (and never remain on the same planet), this situation can be described by the Markov matrix P = ( 0 0.5 0 0.5 0 0 0 0. 3 0 0. 3 0 0. 3 0 0 0 0.25 0 0.25 0.25 0.25 0 0. 3 0 0. 3 0 0 0 0. 3 0 0. 3 0. 3 0 0 0. 3 0 0 0 0. 3 0. 3 0 0 0. 3 0 0 0 0 0.5 0.5 0 ) P = \begin{pmatrix} 0 & 0.5 & 0 & 0.5 & 0 & 0 & 0 \\ 0.\overline{3} & 0 & 0.\overline{3} & 0 & 0.\overline{3} & 0 & 0 \\ 0 & 0.25 & 0 & 0.25 & 0.25 & 0.25 & 0 \\ 0.\overline{3} & 0 & 0.\overline{3} & 0 & 0 & 0 & 0.\overline{3} \\ 0 & 0.\overline{3} & 0.\overline{3} & 0 & 0 & 0.\overline{3} & 0 \\ 0 & 0 & 0.\overline{3} & 0.\overline{3} & 0 & 0 & 0.\overline{3} \\ 0 & 0 & 0 & 0 & 0.5 & 0.5 & 0 \end{pmatrix} where the rows and columns are arranged alphabetically. That is, the entry P i j P_{ij} is the probability of moving from planet i i to planet j j , with i , j = 1 i,j=1 corresponding to A, i , j = 2 i,j=2 corresponding to B, etc.

We can represent our starting state as the row vector π = ( 0 , 0 , 1 , 0 , 0 , 0 , 0 ) \pmb{\pi} = (0, 0, 1, 0, 0, 0, 0) (indicating that we start on planet C with probability 1 1 ). Then, the probabilities for each planet after four months are π P 4 ( 0.13 , 0.12 , 0.26 , 0.12 , 0.12 , 0.12 , 0.13 ) \pmb{\pi}P^4 \approx (0.13, 0.12, 0.26, 0.12, 0.12, 0.12, 0.13) Planet C is the most likely planet, with probability roughly 0.26 0.26 .

How about "It has to be C because only C is unique. All other planets have at least one other planet which is equivqlent under the terms of this problem and so could not be a unique solution."

allan brooks - 2 years, 7 months ago

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If you look at the answer options, some of them list multiple planets. It doesn't say there's a unique solution. And, in fact, that answer depends on how many months have passed (at least for small numbers of months)! If only three months had passed, the answer would be "B, D, E or F."

Jordan Cahn - 2 years, 7 months ago

For exactness, those values are 14, 13, 28, 13, 13, 13, 14 out of 108.

Jeremy Galvagni - 2 years, 7 months ago

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Thanks for that!

Jordan Cahn - 2 years, 7 months ago

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