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A miner is stranded and there are three paths that he can take.

Path A loops back to itself and takes him 1 day to walk it.

Path B loops back to itself and takes him 2 days to walk it.

Path C is the exit and it takes him 3 days to walk it.

Each path has an equal probability of being chosen and once a wrong path is chosen, he gets disorientated and cannot remember which path it was and the probabilities remain the same.

What is the expected value of the amount of days he will spend before he exits the mine?

I was thinking that maybe it used the Geometric distribution but I don't think that accounts for the varying number of days. If someone could help clear this up it would be greatly appreciated!

Robert Maggio
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4 Answers4

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Let $X$ be the random number of days that he spends in the mine. Let $A$, $B$, $C$ represent the path that he chooses when he has a choice. Then $$\mathbb E[X] = \mathbb E[X \mid A]\Pr[A] + \mathbb E[X \mid B]\Pr[B] + \mathbb E[X \mid C] \Pr[C],$$ and we note that $$\mathbb E[X \mid A] = 1 + \mathbb E[X],$$ because given that he has chosen path $A$, he takes one day to return to his starting point, at which point the expected escape time is still $\mathbb E[X]$. Similarly, $$\mathbb E[X \mid B] = 2 + \mathbb E[X].$$ What is $\mathbb E[X \mid C]$? Can you finish the rest?

heropup
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  • I wonder if he chose path A and get back to the beginning, the expected value should reduce because there is no path A in our sample space, or we consider that he can choose path A again? – CuCaRot Nov 30 '20 at 13:04
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Let $d$ be the expected number of days. If you take path A or B, after taking it you still have $d$ as the expected number. So $d=\frac 13 (d+1) + \frac 13(d+2) + \frac 13 (3)$

Ross Millikan
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Call the number of days $x$. Then with probability $\frac13$, the miner chooses path A and takes $1$ day to walk, and $x$ days to go to the exit (he is once again stranded with the exact same conditions, so it will take the exact same time). Then with probability $\frac13$, the miner chooses path B and takes $2$ day to walk, and $x$ days to go to the exit. And finally, with probability $\frac13$, the miner chooses path C and reaches the exit in $3$ days.

So we get the equality $x=\frac13(x+1)+\frac13(x+2)+\frac13\cdot3$, which means x=6

Pim
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You can use your geometric random variable approach. Define a trial as taking a path, and success as choosing path $C$. Then define a random variable $T$ to be the number of trials (paths chosen). So $T$ is geometric with parameter $\frac13$; so its expected value is $3$. So on average you can expect $2$ failures before a success, and a failure takes an average of a day and a half. From there can you determine the expected number of days before our poor miner gets out?

paw88789
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