0
$\begingroup$

The following question is taken from glassdoor.

Three doors, it takes you $1$ minute to go through the first door, it also takes you $1$ minute to go through the second door, it make your speed slow down $\frac{1}{2}$ after going through third door. Only the first door is the right way to go out. You choose each door with the equal probability. How much time will you go out in expectation?

Let $E(X)$ be the required expectation. By law of total expectation, $$E(X) = P(\text{first door})E(X\,\vert\,\text{first door}) + P(\text{second door})E(X\,\vert\,\text{second door}) + P(\text{third door})E(X\,\vert\,\text{third door}).$$ Each probability above is $\frac{1}{3}.$ We also know that $$E(X\,\vert\,\text{first door}) = 1, \quad E(X\,\vert\,\text{second door}) = 1 + E(X).$$ However, I do not know how to calculate $$E(X\,\vert\,\text{third door}).$$ Any hint is appreciated.

$\endgroup$
  • $\begingroup$ I don't understand why $E(X\mid \text{second door}=1+E(X)$ This suggests than even after you try the second door, and it doesn't work, you have a $\frac13$ probability of choosing it again. $\endgroup$ – saulspatz Feb 15 at 1:34
  • $\begingroup$ @saulspatz: the usual rule in puzzles like this is that you can't remember which door you tried the last time, so each try is random. That should be stated in the problem. If you do remember, the problem is no fun because there are just five cases to consider. $\endgroup$ – Ross Millikan Feb 15 at 1:38
  • 1
    $\begingroup$ @RossMillikan Well, it's apparently an interview question, so I didn't think it was supposed to be fun :-). $\endgroup$ – saulspatz Feb 15 at 1:41
2
$\begingroup$

Assuming each pass through the third door cuts your speed in half but you go through that door instantly you have $$E(X|\text{third door})=2E(X)$$ When you plug this into your equation you will get a contradiction, which means that the sum $E(X)=\sum_{t=0}^\infty tP(t)$ where $P(t)$ is the probability of escape after time $t$ does not converge.

$\endgroup$
  • $\begingroup$ I will indeed get a contradiction if I plug your equation above into my equation. However, I do not follow your last statement. Why would $E(X)$ diverge? $\endgroup$ – Idonknow Feb 15 at 1:42
  • $\begingroup$ Basically $P(t)$ will not decrease fast enough. It is similar to the Saint Petersburg paradox $\endgroup$ – Ross Millikan Feb 15 at 1:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.