Monty Hall Problem

5-minute read
Notes • Math

Monty Hall Problem is a classic probability puzzle based on an American show called $Let’s \ Make \ a \ Deal$. While it brings joy to the audience, it also keeps torturing math students and people wanted to have a future in this field (just kidding). In this post, I summarized the classic Monty Hall Problem and some of its variants.

picture alt

Classic Monty Hall Problem

Let’s assume you’re the participant of the show and you’re standing before 3 doors which look exactly like the others. Behind one door is a car and for the other two, goats. You don’t know what’s behind the doors. Now you have to pick one door and you’ll say which door you picked. As soon as you announce your decision, the host whose name is Monty Hall will open one of the other two doors that he knows there’s a goat behind it. Then you’ll have a chance to decide either to stick to your previous decision or to switch to the third door. Will you switch? And if you do, what’s the probability of winning the car?

Solution:

  • If you don’t switch: the event that you win is independent of Monty’s action of showing you the goat. Since your choice is fixedthe probability of winning is $\frac{1}{3}$.
  • If you switch: you’ll win only if you original choose the doowhich has a goat behind it. Since out of three doors, there’re twdoors which have goats, the probability of choosing a goat door i$\frac{2}{3}$, which is also your probability of winning.
  • By comparison, it is obvious that you should switch.
Variant 1

Suppose there’re four doors and two prizes put behind different doors which are respectively worth 100 dollars and 1000 dollars. For the other two doors, there’re nothing. Like the previous rule, you have the chance to pick one door. By the rules of the game, after your picking, the host should select one door that is not selected by you and doesn’t contain any prizes and open it. However, here’s one new rule of the game: the host never opens the fourth door unless he has to. If two of the first three doors have no prize and haven’t been picked by you, the host will randomly choose which of those two to open. After he opens the door, you’ll have the chance to decide either to stick to your previous decision or to switch.

Here’re three senarios. For each one, what’s your best strategy and your expected winnings?

  • You chose door 1 and the host opened door 2.
  • You chose door 1 and the host opened door 4.
  • You chose door 4 and the host opened door 2.

Solution:

  • For senario 1: For integers $1 \leq j \leq k \leq 4$, let $K_{j, k}$ be the eventheat the prizes are behind door $j$ and door $k$. So $P(K_{j, k}= \frac{1}{6}$. Let $D$ be the event that door 2 has been opened, then: $$ P(D | K_{1, 3}) = 1, \ P(D | K_{1, 4}) = \frac{1}{2}, \ P(D | K_{3, 4}) = 1. $$ According to the law of total probability, $$ P(D) = \frac{1}{6} \cdot 1 + \frac{1}{6} \cdot \frac{1}{2} + \frac{1}{6} \cdot 1 = \frac{5}{12} $$ So, $$ P(K_{1, 3}|D) = \frac{P(K_{1, 3} \cap D)}{P(D)} = \frac{P(D|K_{1, 3}) \cdot P(K_{1, 3})}{P(D)} = \frac{2}{5} $$ Similarly, we could calculate that, $$ P(K_{1, 4} | D) = \frac{1}{5}, \ P(K_{3, 4 } | D) = \frac{2}{5} $$ From above, we could see that there’s $\frac{4}{5}$ probabilitthat there’s a prize behind door 3, so your best strategy is tswitch to door 3. And your expected winning should be $\frac{4}{5} \cdot 550 = 440$.

  • For senario 2: since the host opens the door 4, it means there must be prize behind door 2 and door 3, so you could switch to door 2 and door 3 and your expected winning should be 550.

  • For senario 3: just like how we calculated for senario 1, we only need to change $D$ as the event that the host opens door 2, in this case, $$ P(D | K_{1, 3}) = 1, \ P(D | K_{1, 4}) = \frac{1}{2}, \ P(D | K_{3, 4}) = \frac{1}{2} $$ So, $$ P(D) = \frac{1}{6} \cdot 1 + \frac{1}{6} \cdot \frac{1}{2} + \frac{1}{6} \cdot \frac{1}{2} = \frac{1}{3} $$ And, $$ P(K_{1, 4}|D) = \frac{P(K_{1, 4} \cap D)}{P(D)} = \frac{P(D|K_{1, 4}) \cdot P(K_{1, 4})}{P(D)} = \frac{1}{4}, \ P(K_{1, 3} | D) = \frac{1}{2}, \ P(K_{3, 4} | D) = \frac{1}{4} $$ From above, we know that the probability of a prize behind door 1 or door 3 is $\frac{3}{4}$, the probability of a prize behind door 4 is $\frac{1}{2}$. So your best strategy should be switch to door 1 or door 3 and your expected winning is $\frac{3}{4} \cdot 550 = 412.5$.

Variant 2

The rule of the game is same as Variant 1, only the host changes his strategy of opening doors. Given the following senarios, what’s your best strategy and your expected winnings?

  • If you don’t choose the door with 100 dollars prize, the host opens that.
  • If you choose the door with 100 dollars prize, the host randomly chooses one of the two doors behind wiich there’s no prize.

Solution:

  • For Senario 1: if host opens the door and there’s 100 dollars prize behind it, you know that the 1000 dollars prize is behind one of the left three doors. So your probability of winning will be $\frac{1}{3}$ no matter you switch or not. Your expected winning will be $\frac{1000}{3}$ dollars.

  • For Senario 2: if the host opens the door and there’s nothing, you’ll know that you’ve chosen the door with 100 dollars prize. So for the rest two doors, each one has $\frac{1}{2}$ probability of having the 1000 dollars. You could stick to your choice and get 100 dollars, but switch to one of the doors will have expected winning of 500 dollars.