Recall the Prisoners problem from class. We have three prisoners (A, B, and C) on death row. The Governor
will pardon one of the prisoners, and he tells the Warden who he chose, but the warden is not allowed to
reveal who was pardoned. Lets switch things up some, Prisoner C asked the warden if he was pardoned, but
the warden only will tell him that either A or B will die. Furthermore, as with the midterm problem, it is
close election time so the Governor decided to make his decision based on the proportion of people that would
dislike his choice. Because of this, we have that A has a 1
2 probability of being pardoned, B has a 1
3 probability
of being pardoned, and C has a 1
6 probability of being pardoned. Unlike in class, C knows that the warden
likes B, and if he has a choice in who he will say will die, he will always say A will die.
De�ne events A=fA is Pardonedg, B=fB is Pardonedg, C=fC is Pardonedg, and W=fWarden says B diesg.
(a) What is the event Wc in words?
(b) Make a tree diagram showing all probabilities in the tree.
(c) [ Calculate, showing all steps, P(C/Wc).
(d) Calculate, showing all steps, P(C/W).
(e) Calculate, showing all steps, P(A/W)
I showed you how to do the last one, am leaving this more complicated version of the classic prisoner's dilemna problem for a stats teacher (I do physics)
(a) The event Wc represents the statement "The warden says B dies."
(b) Here is the tree diagram:
----A pardoned (1/2)--
/ \
Wc (1/2) ----B pardoned (1/3)------
\ / \
----C pardoned (1/6)---
(c) To calculate P(C/Wc), we use Bayes' theorem:
P(C/Wc) = P(Wc/C) * P(C) / P(Wc)
First, let's calculate P(Wc/C). We know that if C is pardoned, the warden will always say A will die, which means Wc. Therefore, P(Wc/C) = 1.
Next, P(C) is given as 1/6.
To calculate P(Wc), we need to consider all possible outcomes where the warden says B dies. These are when A is pardoned and the warden says B dies (P(Wc/A)) and when B is pardoned and the warden says B dies (P(Wc/B)).
Using the law of total probability:
P(Wc) = P(Wc/A) * P(A) + P(Wc/B) * P(B)
We know that if A is pardoned, the warden will say B dies, so P(Wc/A) = 1. Therefore, P(Wc/A) * P(A) = 1/2 * 1/2 = 1/4.
If B is pardoned, the warden will say B dies because he likes B. Hence, P(Wc/B) = 1. So, P(Wc/B) * P(B) = 1/3 * 1/3 = 1/9.
Therefore, P(Wc) = 1/4 + 1/9 = 13/36.
Now we can substitute the values into Bayes' theorem:
P(C/Wc) = P(Wc/C) * P(C) / P(Wc)
= 1 * 1/6 / (13/36)
= 4/13.
So, P(C/Wc) = 4/13.
(d) To calculate P(C/W), we also use Bayes' theorem:
P(C/W) = P(W/C) * P(C) / P(W)
First, let's calculate P(W/C). We know that if C is pardoned, the warden will always say A will die. So, P(W/A) = 1. Also, if B is pardoned, the warden will say B dies because he likes B. So, P(W/B) = 1. Hence,
P(W/C) = P(W/A) * P(A) + P(W/B) * P(B)
= 1 * 1/2 + 1 * 1/3
= 5/6.
P(C) is given as 1/6.
To calculate P(W), we need to consider all possible outcomes where the warden says B dies. These are when A is pardoned and the warden says B dies (P(W/A)) and when B is pardoned and the warden says B dies (P(W/B)).
Using the law of total probability:
P(W) = P(W/A) * P(A) + P(W/B) * P(B)
= 1 * 1/2 + 1 * 1/3
= 5/6.
Now we can substitute these values into Bayes' theorem:
P(C/W) = P(W/C) * P(C) / P(W)
= (5/6) * (1/6) / (5/6)
= 1/6.
So, P(C/W) = 1/6.
(e) To calculate P(A/W), we use Bayes' theorem again:
P(A/W) = P(W/A) * P(A) / P(W)
P(W/A) = 1, as explained earlier.
P(A) = 1/2, as given.
P(W) can be calculated as before:
P(W) = P(W/A) * P(A) + P(W/B) * P(B)
= 1 * 1/2 + 1 * 1/3
= 5/6.
Substituting these values into Bayes' theorem:
P(A/W) = 1 * 1/2 / (5/6)
= 3/5.
So, P(A/W) = 3/5.