Probability: Escaping Prisoner Question

Hint: Use the total expectation formula to condition of prisoner choice.

If $X$ is the distance traveled and $T_1,T_2,T_3$ are the events that the prisoner chooses tunnel $1,2$ and $3$ respectively, then:

$$E[X]=E[X|T_1]P(T_1)+E[X|T_2]P(T_2)+E[X|T_3]P(T_3)$$

Now $E[X|T_1]=100+E[X]$, $E[X|T_2]=40+E[X]$, and $E[X|T_3]=100$

  • in the first case after traveling 100ft the prisoner is in the same situation as in the beginning, so it takes on average E[X] distance again to get out

  • in the second case it takes only 40ft for the prisoner to be in the same situation as in the beginning

  • in the third case it takes 100ft for the prisoner to get out

and $P(T_1)=P(T_2)=P(T_3)=\frac{1}{3}$

So you can solve for $E[X]$


Let's assume he picks the tunnels with equal likelihood, i.e. $\frac{1}{3}$ each.

So, the prisoner definitely has to crawl that 100 feet to freedom at some day, but he can be crawling (for nothing) in those other tunnels as well with a likelihood of $\frac{2}{3}$ each day and for a distance of 70 feet on average on such a lost day.

OK, so let $P(i)$ be the probability of the prisoner crawling for $i$ days for nothing before reaching freedom, so $P(i) = (\frac{2}{3})^i * \frac{1}{3}$

And let $D(i)$ be the expected distance the prisoner crawls on $i$ lost days, so $D(i) = i*70$

We thus get:

$E(D) = 100 + \Sigma_{i=1}^{\infty} P(i)*D(i) = 100+ \Sigma_{i=1}^{\infty}(\frac{2}{3})^i * \frac{1}{3} * i * 70 = 100+ 70*\frac{1}{3}*\frac{2}{3}*\Sigma_{i=1}^{\infty}i*(\frac{2}{3})^{i-1}= 100+\frac{140}{9}*\frac{1}{(1-\frac{2}{3})^2} = 100+140=240$


I'll add another answer that I think is the simplest, and here is how I got to it:

I noticed that the answer (240 feet) was exactly the sum of going through (and back) both the 'dead-end' tunnels (100 feet and 40 feet respectively) and going through the 'freedom' tunnel once (100 feet) and I figured that was probably not a coincidence.

Now, using @Momo's method, I quickly realized that indeed: whatever the lengths of the tunnels are, and no matter how many tunnels there are (though still with exactly one leading to freedom), if each tunnel is picked with equal probability, then the expected distance in the end will be the sum of the distances traveled by going through each tunnel once.

OK, but even with Momo's method, there is still some algebra involved to get to this result, and I figured that there must be an even simpler argument for this result, and after some further thinking, I found it:

Given that no matter how many days it takes for the prisoner to get out, it is true that each day the prisoner picks each tunnel with equal likelihood (this is true while the prisoner has not escaped yet, as well as when the prisoner has escaped). It thus follows that the expected number of times the prisoner goes through a tunnel is the same for each tunnel as well. But since we know the prisoner that escapes goes through the freedom tunnel exactly once, the prisoner is expected to go through each other tunnel exactly once as well!

Tags:

Statistics

Probability

Probability Theory

Expectation

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