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You have 2 people starting on opposite sides of an octahedron, each move to a random vertex each turn. you win by moving onto the vertex of the other person, what's the probability the guy who goes first wins.

How I currently attacked it

A=Probability the first person wins

P(A)=P(A|Distance of one) (as person A will always be a distance of one from B after this move)

P(A|Distance of one)= 1/4*0+1/4*(1-P(A))+1/2(1-P(A|Distance of one))

I'm not too sure about how to go about solving it from here.

user690808
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1 Answers1

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You definitely have the right idea. I will let $M_k$ be the event the the person to move wins, given a distance of $k$. Then the probability that the first person wins is $$ P(M_2) = 1-P(M_1) $$ and $$ P(M_1) = \frac14 \cdot 1+ \frac12 \cdot(1-P(M_1))+\frac 14\cdot(1-P(M_2)) $$ Using the first equation gives: $$ 1-P(M_2) = \frac14 \cdot 1+ \frac12 \cdot P(M_2)+\frac 14\cdot(1-P(M_2)) $$ which we can solve to $P(M_2)=\frac25$.

Milten
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  • What led you to structure the game based off of distances rather than the people like I did? – user690808 Nov 04 '19 at 18:40
  • Well, you already looked at the distances, right? You wrote P(A|Distance of one) for example. The main difference I made was to focus on the person making the move instead of the fixed person 1. That was mostly to avoid confusion between things like P(A|Distance of one and 1 to move) and P(A|Distance of one and 2 to move). – Milten Nov 04 '19 at 18:52
  • Your version was fine but the notation was ambiguous. I think you would get the same answer if you solved your equations. – Milten Nov 04 '19 at 18:53
  • By the way, I started out writing $P(M_2) = 1-P(M_1) = 1-(\frac14 \cdot 1 + \ldots)$. So in my head I focused on the person just like you. I just changed it in the end since it looked a bit cleaner. – Milten Nov 04 '19 at 18:56