My last post, on the 17 x 17 challenge, ended with a solution to a problem which I thought sounded incredibly naïve* but which turned out to be correct. It got me thinking about a problem that I was posed when I was interviewing to work at investment banks. In fact I was asked the same question in interviews for two different banks (potential applicants take note).

Tunnels

Sketch of the problem as presented to me

It goes like this: You find yourself in a dark cave, which has three tunnels leading from it (see the crude sketch on the right). One of the tunnels leads outside, and it takes 3 hours to traverse it. The other two tunnels connect to each other, returning you to the cave, and they take 1 hour to walk down. If you choose tunnels at random each time you’re faced with a choice, how long will it take you on average to get outside?

I’m about to tell you the answer, so maybe you want to stop and think about it for a bit. The point I’m making is that even though the calculation you have to do is a bit involved (although simple to anyone who’s ever studied Markov chains) there’s a cheeky shortcut: you can get the answer just by assuming that you take each of the tunnels once, and add up the times to traverse them, getting an answer of 1+1+3=5 hours.

Is this just a coincidence? To find out, let’s generalise the problem. Now we find ourselves in a cave with n tunnels, of which m of them lead to an exit and n-m dump you back in the cave. Each of the tunnels takes time \tau_i to traverse, for i=1,…,n, and is chosen with probability p_i. We’ll adopt the convention that the first m tunnels are the ones that lead to an exit, and the last n-m are the return tunnels (notice that our generalisation allows the possibility that a return tunnel might take a different amount of time to traverse if we go in the opposite direction).

To work out what the average escape time is, we notice that if we pick an escape tunnel then we’re free in the time it takes to traverse the tunnel. If we pick a return tunnel, then we’ve increased our expected time to leave by the time that it took to traverse the tunnel. If the time to leave is T, then we write the expected leaving time as E[T], and

E[T] = \sum_{i=1}^m p_i\tau_i + \sum_{j=m+1}^n p_j(\tau_j + E[T]).

After a couple of lines of algebra we see that the expected time to leave is given by

E[T] = (\sum_{i=1}^n p_i\tau_i) / (1 - \sum_{j=m+1}^n p_j)

or, written in everyday language:

Avg time to leave = (Avg tunnel length) ÷ (Chance of picking an exit tunnel)

That’s it! The average time to escape only depends on the average time it takes to traverse a randomly picked tunnel, and the probability of picking a tunnel which leads to an exit. It’s completely insensitive to the specifics of the problem, namely how long the individual tunnels are, and what the precise probabilities of picking each tunnel are. In the special case where each of the tunnels is equally likely, all you have to do is add up the time it takes to traverse each of the tunnels, and divide by the number of tunnels that lead to an exit.

This appeals to me because a problem which has a lot of fine detail turns out to be insensitive to the fine detail. I’d be interested to hear about other examples of this phenomenon.

* Is it pretentious to write ‘naïve’ instead of ‘naive’? Answers, opinions and furious rants on a postcard, please.

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