Bayesian updating

Question

"An object hides in one of 3 boxes. If the object is in box i, i = 1; 2; 3, and if box i is searched once, then the object is found with probability 0.7. Suppose that it is initially believed that the object is in box i with probability 0.2, 0.3 and 0.5, for i = 1; 2; 3. One always searches the box with the highest current probability; thus one would begin with box 3. If the first search fails, which box should be searched next?"

I used bayes rule: $ P(in box 3 \mid Nf)=\frac{P(Nf \mid in box 3)*P(in box 3)}{P(Nf)}$

where P(Nf)=P(Not found) and P(Nf/inbox3)=Probability that the box is not found given that the object is in box 3.

I get:

$ P(in box 3 \mid Nf)=\frac{0.3*0.5}{P(Nf)}$

I know that I have to use the law of total probability to compute P(Nf). My question is will I have to add up the probabilities of "not found" given that the object is in that specific box times the probability (0.2,0.3 or 0.5) ? This would give me:

$0.3*0.2+0.3*0.3+0.3*0.5=0.3$

$P(inbox3 \mid Nf)=\frac{0.3*0.5}{0.3}=0.5$

Well, this can inuitively not be take case. I did something wrong using the law of total probability. This cannot be the case because the conditional probability with the updated beliefs should be lower now. Can somebody show me how to do it correctly?

Answer 1

Consider $20$ trials where we test Box $3$. In $10$ of those trials, the ball will be in Box $3$. In $3$ of those $10$ trials, the search will miss the ball. In the other $10$ trials (when the ball is in one of the other two boxes), the search also won't find the ball. So there are $13$ equally likely trials where the search misses the ball, and in $3$ of those, the ball was actually in Box $3$. That means the probability the ball is in Box $3$ is now $\dfrac{3}{13}$, which means the probability it's in one of the other two boxes is $\dfrac{10}{13}$.

Our trials haven't affected the likelihood between Boxes $1$ and $2$, so the likelihood of Box $1$ is now $\dfrac{4}{13}$ and the probability of Box $2$ is $\dfrac{6}{13}$.

Answer 2

The problem may be attacked with (virtually) no Math, except an examination of the probability that the object is (still) in Box-3, despite the initial failed search.

The conditional probability that the object is in Box-3, given the initial failed search is

$$\frac{(0.5 \times 0.3)}{(0.5 \times 0.3) + (0.5)} = \frac{15}{65} < 0.3. \tag1 $$

At this point, given the nature of the asked question, it is game over. That is, clearly, no evidence has been presented that would make Box-2 less likely than Box-1.

Further, intuitively, the failed search in Box-3, must have increased the probability that the object is in Box-2. Therefore, without bothering to calculate the exact probabilities, you now know that :

• The probability of the object being in Box-2 continues to be greater than the probability of the object being in Box-1.

• The probability of the object being in Box-2 must now be greater than $(0.3).$

• Based on (1) above, the probability of the object being in Box-3 must now be less than $0.3$.

The three bullet-points above indicate that Box-2 is the percentage play for the 2nd search.