# Intuitively, why does $p$ vary inversely with $P(C_3 \mid D_2)$? But directly with $P(C_2 \mid D_3)$?

**I'm seeking merely intuition here** — NOT about the algebra that I know how to execute. Please see the fractions colored in red and orange at the bottom.

- $P(C_3 \mid D_2) = \color{red}{\dfrac1{
1 + p}}$ means that $p$ is inversely related with $P(C_3 \mid D_2)$.
**How can you intuit this inverse relationship?**

This is COUNTERintuitive to me! The question postulates that "you choose [...] door 1" and "Monty enjoys opening door 2 more than [...] door 3". But if $D_3$ hides the car, then Monty is forbidden from opening $D_3$ by the rules of the game, and must open $D_2$. So it doesn't matter whether Monty enjoys opening $D_2$ more than $D_3$!

N.B. $1/2 \le p \le 1 \iff 3/2 \le 1 + p \le 2 + p \iff 2/3 \ge {\color{red}{\dfrac{1}{ 1 + p}}} \ge \dfrac2{2 + p}.$

- $P(C_2 \mid D_3) = \dfrac{1}{1+(1-p)} = \color{darkorange}{\dfrac1{2-p}}.$
How can you intuit why $p ∝ P(C_2 \mid D_3)$?**Why this U-turn and turn about from part (b)? Why DIRECT relationship now???**

N.B. $1/2 \le p \le 1 \iff -1/2 \ge -p \ge -1 \iff 3/2 \ge 2 - p \ge 1 \iff 2/3 \le {\color{darkorange}{\dfrac1{2-p}}} \le 1.$

- 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 $1/2 \le p \le 1$.
To recap: there are three doors, behind one of which there is a car (which you want), and behind the other two of which there are goats (which you don't want). Initially, all possibilities are equally likely for where the car is. You choose a door, which for concreteness we assume is door 1. Monty Hall then opens a door to reveal a goat, and offers you the option of switching. Assume that Monty Hall knows which door has the car, will always open a goat door and offer the option of switching, and as above assume that if Monty Hall has a choice between opening door 2 and door 3, he chooses door 2 with probability p (with $1/2 \le p \le 1$).

(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 or 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.

## Solution:

(a) Let $C_j$ be the event that the car is hidden behind door j and let W be the event that we win using the switching strategy. Using the law of total probability, we can find the unconditional probability of winning: $P(W) = P(W \mid C_1)P(C_1) + P(W \mid C_2)P(C_2) + P(W \mid C_3)P(C_3) = 0 \cdot 1/3 + 1 \cdot 1/3 + 1 \cdot 1/3 = 2/3.$

(b) A tree method works well here (delete the paths which are no longer relevant after the conditioning, and reweight the remaining values by dividing by their sum), or we can use Bayes' rule and the law of total probability (as below).

Let $D_i$ be the event that Monty opens Door i. Note that we are looking for $P(W|D_2)$, which is the same as $P(C_3 \mid D_2)$ as we first choose Door 1 and then switch to Door 3. By Bayes' rule and the law of total probability,

$P(C_3 \mid D_2) = \dfrac{P(D_2 \mid C_3)P(C_3)}{P(D_2)} = \dfrac{P(D_2 \mid C_3)P(C_3)}{P(D_2 \mid C_1)P(C_1) + P(D_2 \mid C_2)P(C_2) + P(D_2 \mid C_3)P(C_3)}

= \dfrac{1 \cdot 1/3}{p \cdot 1/3 + 0 \cdot 1/3 + 1 \cdot 1/3} = \color{red}{\dfrac{1}{ 1 + p}}.$(c) The structure of the problem is the same as part (b) (except for the condition that $p \ge 1/2$, which was not needed above). Imagine repainting doors 2 and 3, reversing which is called which. By part (b) with $1-p$ in place of $p$, $P(C_2 \mid D_3) = \dfrac{1}{1+(1-p)} = \color{darkorange}{\dfrac{1}{2-p}}.$

Blitzstein, *Introduction to Probability* (2019 2 edn), Chapter 2, Exercise 40, p 91.

pp 15-6 in the publicly downloadable PDF of curbed solutions.

## 1 answer

$P(C_3|D_2)$ is the probability of the car being behind door 3 given that Monty opened door 2. $p$ is the probability that Monty opens door 2, under the assumption that he has a choice between it and door 3. That is to say $p$ is the probability that Monty opens door 2 under the conditions that: (i) the car is behind door 1 and (ii) the player first picks door 1.

If $p=1/2$ then this is just like the regular Monty Hall problem, and there $P(C_3|D_2)=2/3$.

If $p=1$ then you lose information about door 3. Because whether the car is behind door 3 or not, when given the chance, Monty just always shows you door 2. The game becomes completely "insensitive" to where the car is, and therefore you should expect that the probability becomes close to 1/2 (or, in this extreme case where $p=1$ then the probability becomes *exactly* 1/2). In effect, under this strategy and in this scenario, Monty is no longer giving you information about door 3.

This explains the two extreme values that $p$ could take, and this agrees with what you see in

$$ P(C_3|D_2) = \frac 1 {1+p} $$

On a more pedagogical note about intuitive explanations: Human intuition is imperfect and that's why we have algebra. So it's good to look for intuition when it's available, and so this question is fine. In this particular case, it is actually possible to get some amount of intuition.

But be ready for the fact that sometimes you just have to rely on the algebra to get all the precise details correct. Algebra is effectively a tool which allows us to export our understanding into symbols, manipulate the symbols according to valid truth-preserving rules, and then translate the answer back into real-world meaning at the end. So in that way, algebra is a tool that we can use to get a correct answer even when it's impossible for us to intuitively understand the steps needed to generate the answer (because there are too many steps or too many variables for human intuition to be able to manage them all).

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