B How to calculate this conditional probability?

[Tony Hau]

So this question arises from an online course on Bayes Theorem and Total Probability.

The question says that

1. there are three prisoners A, B and C. The King decides to release two of them and sentence the remaining one.
2. Supposed that you were A and so your chance of being released is $\frac{1}{3}$
3. Suppose you know the guard well. You can ask him about who will be one of the released prisoners, except yourself, i.e. B and C.
4. Let $G_B =$ guard says B is released, and $X_A =$ sentence A.
5. Now calculate $P(G_B \mid X_A)$.

I know the answer should be $\frac{1}{2}$, because given that $X_A$ is sentenced, it means that both B and C will be released and so the chance of the guard telling you B is released is one half.

However, when I try to apply $P(G_B \mid X_A) = \frac{ P(G_B \cap X_A)}{ P(X_A)}$, the answer I get is $\frac{3}{4}$. I get the answer $\frac{3}{4}$ because:

Event of being sentenced	Event of guard's telling
$X_A$	$G_B$
$X_A$	$G_C$
$X_B$	$G_C$
$X_C$	$G_B$

So $P(G_B \cap X_A) = \frac{1}{4}$ and $P(G_B \mid X_A) = \frac{3}{4}$. Why am I wrong?


P.S. The lecture video is here:

 

[Tony Hau]

$P(G_B \cap X_A) = \frac{1}{3}$

Can you explain why it is so? There are four events as listed in the table and so the denominator should be 4. Plus if $P(G_B \cap X_A) = \frac{1}{3}$, then the resulting probability of $P(G_B \mid X_A) = 1$

 

[PeroK]

Can you explain why it is so? There are four events as listed in the table and so the denominator should be 4. Plus if $P(G_B \cap X_A) = \frac{1}{3}$, then the resulting probability of $P(G_B \mid X_A) = 1$

Okay, I got confused. I thought two were condemed, but I see two are released. Let me look at it again.

 

[PeroK]

This is tricky. The equally likely events are $X_A, X_B, X_C$, each with a probability $1/3$. If $X_A$, then the guard tells A about B or C with equally probability, so:
$$P(X_A \cap G_B) = P(X_A \cap G_C) = 1/6$$In the other cases, the guard has no choice, so:
$$P(X_B \cap G_C) = P(X_C \cap G_B) = 1/3$$So, the four events you listed are not equally likely.

 

[PeroK]

PS you are asked to calculate $P(G_B|X_A)$, which is fairly meaningless. Note that this is not $P(X_A|G_B)$, which is the probability A is condemed, given that the guard says that B is to be released. Which is not the same as $P(X_A|\neg X_B)$, which is the probability that A is condemed given that B is released.

It's early Sunday morning. I should have had another cup of coffee before looking at this.

 

[PeroK]

PPS You can't really calculate $P(G_B|X_A)$, as it's an assumption of the problem that this is 1/2. I.e. in order to calculate anything we have to assume this. You can, however, verify Bayes' theorem in this case.

I'm starting to wake up!

 

[Dale]

There are four events as listed in the table

The probability of those four events are not equal. Listing a table and counting is only a viable strategy if all of the listed events are equiprobable.

 

[Tony Hau]

Thank you for all your responses. Let me think about it more carefully, to be honest this is my first time learning about the term "equiprobable".