I am working on problem set 2, Q4d from the MIT 6.041

  1. Oscar has lost his dog in either forest A (with a priori probability 0.4) or in forest B (with a priori probability 0.6).
    On any given day, if the dog is in A and Oscar spends a day searching for it in A, the conditional probability that he will find the dog that day is 0.25. Similarly, if the dog is in B and Oscar spends a day looking for it there, the conditional probability that he will find the dog that day is 0.15.
    The dog cannot go from one forest to the other. Oscar can search only in the daytime, and he can travel from one forest to the other only at night.

(d) If the dog is alive and not found by the Nth day of the search, it will die that evening with probability $\frac{N}{N+2}$. Oscar has decided to look in A for the first two days. What is the probability that he will find a live dog for the first time on the second day?

My conceptualisation (which is wrong): Let $F_i$ be the event the dog is found on day $i$, $S_1$ the event the dog survives the first night. We want the intersection of three events $P(F_1^c \cap S \cap F_2) = P(F_1^c)P(S)P(F_2)$ (assuming independence).

The dog could be in $A$ or in $B$. Oscar only searches $A$ and it is not stated (but I think the answer assumes) the dog will not change forests. So $P(F_1^c)$ is the probability that the dog is in $A$ and we don't find it, or the dog is in $B$. Similarly, $P(F_2)$ is the probability we find the dog on day 2, i.e. it is in $A$ and we find it.

Therefore: $$P(F_1^c) = P(A)P(F^c|A) + P(B)P(F^c|B) = 0.4 \cdot 0.75 + 0.6$$ $$P(S) = 1 - \frac{1}{3} = \frac{2}{3}$$ $$P(F) = P(A)P(F|A) + 0 = 0.4 \cdot 0.25 $$

$$ (F_1^c)P(S)P(F) = (0.4 \cdot 0.75 + 0.6)(\frac{2}{3})(0.4 \cdot 0.25)$$

However, the result is:

(d) In order for Oscar to find the dog, it must be in Forest A, not found on the first day, alive, and found on the second day. Note that this calculation requires conditional independence of not finding the dog on different days and the dog staying alive.
$$P(\text{find live dog in A day 2}) = P(\text{in A}) \times P(\text{not find in A day 1}|\text{in A})\\ \times P(\text{alive day 2}) \times P(\text{find day 2}|\text{in A})\\ = 0.4 \times 0.75 \times (1 - \frac{1}{3}) \times 0.25 = 0.05$$

One difference is they remove $P(A)$ from the second day, this makes sense as the dog will only choose what forest to go into once. However, what I really don't understand is the omission of $P(B)$ from the first term (i.e. $0.4 \cdot 0.75$ rather than $0.4 \cdot 0.75 + 0.6$). Why is it okay to disregard this information, as the question gives no indication as to whether the dog is in $A$ or $B$?

  • 1
    $\begingroup$ Joseph, please add the self-study to your post. $\endgroup$ Commented Apr 19, 2023 at 13:24

1 Answer 1


Your mistake is assuming independence of $F_1^c$ and $F_2$. Intuitively, failing to find the dog on the first day increases the probability the dog is in Forest B, and therefore increases the probability the second search will also fail. Thus they cannot be independent. By Bayes,

$$ P(A|F_1^c) = P(F_1^c|A)P(A)/P(F_1^c)\\ = 0.75\times 0.4/(0.4\times 0.75 + 0.6) = 1/3 $$

Using this new lower probability

$$ P(F_2) = P(F_2|F_1^c)P(F_1^c)\\ = P(F_2|F_1^c,A,S)P(A|F_1^c)P(S|F_1^c)P(F_1^c)\\ = P(F_2|A,S)P(A|F_1^c)P(S)P(F_1^c)\\ = 0.25\times(1/3)\times(2/3)\times 0.9 = 0.05 $$

Gives the correct result.

Their solution conditions on $A$ first, and then conditions on $F_1^c$. It may be more clear if you included the 0 term they omitted, since $P(F_2|B) = 0$

$$ P(F_2) = P(A)P(F_2|A) + P(B)P(F_2|B)\\ = P(A)P(F_1^c|A)P(F_2|A,F_1^c) + 0\\ = P(A)P(F_1^c|A)P(S|A,F_1^c)P(F_2|A,F_1^c,S)\\ = P(A)P(F_1^c|A)P(S)P(F_2|A,S) $$

  • $\begingroup$ Thanks that's a really illuminating answer $\endgroup$
    – Joseph
    Commented Apr 20, 2023 at 7:53

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