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I came across the following question in my textbook and it made me think about how to calculate conditional probabilities when we consider probability as a measure of belief.

Joe is 80 percent certain that his missing key is in one of the two pockets of his hanging jacket, being 40 percent certain that it is in the left-hand pocket and 40 percent certain that it is in the right-hand pocket.

If a search of the left-hand pocket does not find the key, what is the conditional probability that it is in the other pocket?

My initial answer was "80 percent" as I took the fact that Joe was 80 percent certain that his missing key was in either one of the two pockets, he must (we can) now be 80 percent certain that it must be in the other pocket.

Whereas the answer in the book is as follows.

Let $L$ be the event that the key is in left-hand pocket, and $R$ in right-hand pocket.

$\begin{align*} \Pr(R\mid L^c) &= \frac{\Pr(R \cap L^c)}{\Pr(L^c)}\\ &= \frac{\Pr(R)}{1-\Pr(L)}\\ &= \frac{40}{60} = \frac{2}{3} \end{align*}$

(We reasonably assumed that $L$ and $R$ are mutually exclusive events.)


Now, whilst I can understand the solution in the textbook, I cannot understand why my approach was wrong. I believe that the difference is in which fact we take "primarily":

  • If we take Joe's belief that there is 80 percent chance that his missing key is in either one of the two pockets (and if we haven't given individual probabilities for each pocket), I think my approach would be sensible.
  • But if we take Joe's belief that his missing key is in either one of the two pockets with 40 percent chance each, then the textbook solution is more sensible.

Would you agree with this? If not, can you explain why the textbook solution is the only sensible way to model this problem and mine is not?

Thanks!

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My initial answer was "80 percent" as I took the fact that Joe was 80 percent certain that his missing key was in either one of the two pockets, he must (we can) now be 80 percent certain that it must be in the other pocket.

No, it was specified that the $80\%$ certainty that the key was among the two pockets was comprised of the two options, each with $40\%$ certainty, and this also means a $20\%$ certainty that the key was elsewhere.

Confirming that one of those three options was false, just leaves the odds as $40:20$ that the key was in the other pocket.   That is a probability of $2/3$.


To put it another way, you could also say that Joe was initially $60\%$ certain that the key was either in the left hand pocket or not in either pocket.   So by your reasoning, once the key was found to not be in the left pocket, it would be $40\%$ certain that it was in the right pocket (not $80\%$ as your reasoning also evaluated).

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