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A circular track of length $2n$ meters has $2n$ teleporters equally spaced along the track. Each teleporter, when activated, instantaneously teleports the object inside to the antipodally opposite teleporter.

Starting at the first teleporter, a robot begins to move clockwise along the track at a rate of $1$ meter/second. As it moves along the track, the robot paints the track red. When the robot reaches a teleporter, the teleporter activates with probability $1/2$ and does not activate with probability $1/2$. The robot continues moving clockwise after being teleported.

What is the expected number of seconds until the entire track is painted red?

By heuristics, I suspect the quantity is asymptotic to $n\log_{2} (n)$, but I am having trouble proving this (or computing an exact result). Any help would be much appreciated.

Sameer Kailasa
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Note: this is a partial answer. This problem has $n$-second long cycles, and it is difficult to tell where in the cycle the robot will finish. I give you the expected number of cycles. Therefore this will be slightly larger than the true answer. Most likely I miss by less than half a cycle, since which segment is painted last is uniformly random if that segment is alone in its cycle, but if several segments are painted on the last cycle, then it is the last segment that counts.

Pair up the antipodal track segments two-by-two into $n$ pairs. Label each pair with an integer from $1$ to $n$, in the order by which the robot visits them. Now, at (or rather right after) any integer time $i$, the robot has just started painting one of the segments in pair number $i+1$ (reduced modulo $n$), and which one that is is $50$-$50$.

Let the random variable $X_i$ for $1\leq i \leq n$ be the number of times the robot passes through pair number $i$ (reduced modulo $n$) before both segments are painted, minus one. Then it is geometrically distributed with parameter $p$: $$ P(X_i = k) = \frac1{2^k},\quad P(X_i \leq k) = \frac{2^k-1}{2^k} $$ and we want the expectation of largest one of them. Following this answer we get $$ E[\max_i(X_i)] = \sum_{k \geq 1}\left(1-\left(\frac{2^k-1}{2^k}\right)^n\right) $$ (I don't know whether that sum has a closed form). Add $1$ to that, and you get the expected number of full cycles.

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Arthur
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  • The given expression simplifies to $\sum_{i=1}^{n} (-1)^{i+1} \binom{n}{i} \frac{1}{2^i - 1}$, which isn't exactly a closed form but probably as close as can be gotten. – Sameer Kailasa Mar 08 '16 at 02:06
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    @Sam To get the asymptotics, you can crudely approximate $Geom(\frac 1 2)\approx C+\exp(\log 2)$ for $C=\frac 1 2\text{ or }\log 2$ and hence $E(\max X_i)=\log_2 n+O(1)$ for a small positive $O(1)$. – A.S. Mar 08 '16 at 07:48