Monty Hall: What is the "unconditional probability of success" and "conditional probability of success"?

Can someone explain the highlighted paragraph from Blitzstein, Introduction to Probability (2019 2 edn), p 69? I'm not understanding the subtlety the author is trying to get at. What is the "unconditional probability of success"? How is it not conditional because we're conditioning on what door the car is behind to solve the law of total probability equation. What is the "conditional probability of success"? Are "unconditional probability of success" and "conditional probability of success" the same thing?

For the normal Monty Hall problem, it seems that the "unconditional probability of success" and "conditional probability of success" are the same; they output the same number (2/3). But they're different in op. cit. Exercise 40, p 91.

Consider the Monty Hall problem, except that Monty enjoys opening Door 2 more than he enjoys opening Door 3, and if he has a choice between opening these two doors, he opens Door 2 with probability p, where $\frac{1}{2} \leq p \leq 1$.

(a) Find the unconditional probability that the strategy of always switching succeeds (unconditional in the sense that we do not condition on which of Doors 2,3 Monty opens).

(b) Find the probability that the strategy of always switching succeeds, given that Monty opens Door 2.

(c) Find the probability that the strategy of always switching succeeds, given that Monty opens Door 3.

The answer is 2/3 for (a) because $0 \cdot 1/3 + 1 \cdot 1/3 + 1 \cdot 1/3 = 2/3$ by LOTP. Why this answer is the same as the normal Monty Hall problem, where Monty enjoys all three doors equally?

The answer for (b) is $\frac{1}{p+1}$ and for (c) it's $\frac{1}{2-p}$.

Solution 1:

The difference between unconditional and conditional is that for unconditional, we are conditioning on C, and for conditional, we are conditioning on M. Simply put, the thing being conditioned on is different.

You start by picking door 1, it seems as the problem describes.

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 1   2   3

1. The unconditional property only depends on C, the event that the car is behind door C, so we have the following, where "car" really means "get car by switching." You'll notice that this has no information about M. That is why it is the same no matter the probability that the host chooses door 2. It's not in the formula. It only depends on where the car was initially placed. Maybe this is not a satisfactory explanation, but let's move on.

$$P(car)=P(car|C_1)P(C_1)+P(car|C_2)P(C_2)+P(car|C_3)P(C_3)\\ =0*\frac 1 3+1*\frac 1 3+1*\frac 1 3=\frac 2 3$$

2. You want to find the probability that switching succeeds, given that Monty opens door 2. Let's draw a tree. Maybe take a minute or so to figure out why it is.

                                      prob
                      / car     0     0
          / M2  p     
                      \ no_car  1     p/3
/ C1  1/3 
                      / car     0     0
          \ M3  1-p   
                      \ no_car  1     (1-p)/3

                      / car     1     1/3
- C2  1/3 - M3  1    
                      \ no_car  0     0

                      / car     1     1/3
\ C3  1/3 - M2  1     
                      \ no_car  0     0

You want $P(car|M_2)=\frac{P(car, M_2)}{P(M_2)}=\frac{P(car,M_2,C_1)+P(car,M_2,C_2)+P(car,M_2,C_3)}{P(M_2,C_1)+P(M_2,C_2)+P(M_2,C_2)}=\frac{0+0+1/3}{p/3+0+1/3}=\frac{1}{p+1}$

3. Similarly, you want $P(car|M_3)=\frac{P(car, M_3)}{P(M_3)}=\frac{P(car,M_3,C_1)+P(car,M_3,C_2)+P(car,M_3,C_3)}{P(M_3,C_1)+P(M_3,C_2)+P(M_3,C_2)}=\frac{0+1/3+0}{(1-p)/3+1/3+0}=\frac{1}{2-p}$

4. Lastly, we go back to verify that the "conditional probability" of getting the car conditioning on M is the same as the "unconditional probability" of getting a car conditioning on C, which was 2/3 as determined in part 1. We need the law of total probability $P(car)=P(car|M_2)P(M_2)+P(car|M_3)P(M_3)=P(M_2)\frac{1}{p+1}+P(M_3)\frac{1}{2-p}$.

Convince yourself with the following diagram that the $P(M_2)=\frac 1 3+\frac p 3$. That is, again $P(M_2)=P(M_2,C_1)+P(M_2,C_2)+P(M_2,C_3)=P(M_2|C_1)P(C_1)+P(M_2|C_2)P(C_2)+P(M_2|C_3)P(C_3)=1*\frac 1 3+0*\frac 1 3+p*\frac 1 3=\frac 1 3+\frac 1 3 p$. Alternatively, you could have used the huge tree above. Since opening door 3 is the complement to opening door 2, $P(M_3)=1-P(M_2)=1-\left(\frac 1 3+\frac 1 3p\right)$. Now you can finally find the "conditional" probability that you win a car by switching:

$$\left(\frac 1 3+\frac 1 3p\right)\frac{1}{p+1}+\left(1-\left(\frac 1 3+\frac 1 3p\right)\right)\frac{1}{2-p}=\frac 2 3$$

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It is the same as the "unconditional" probability conditioning on C instead of M.