Acid Coder
3 min readFeb 24, 2021

Monty Hall Problem, Why 50–50 Is Impossible.

This article aims to reinforce the understanding of how the probability behind Monty Hall (or probability in general) works, it is for someone who agrees the answer is 66% but need a more persuasive explanation.

Monty Hall problem is settled, if you think you are smarter and wish to argue that the answer is other than 66%(the common wrong answer is 50–50), you should stop reading now, GTFO and google what is Dunning–Kruger effect instead.

I would not reiterate what is Monty Hall problem, I suggest you google it if you are new to the problem.

To understand Monty Hall problem, most people explain why the answer is 66%, but I think a more effective way is to explain why 50–50(host reveal a door, 2 doors left, 50–50) is impossible, to understand this, I am going to explain to you 2 critical points:

1.Why the probability changes?

2. Why the probability does not change?

And, there are few elements that lead to changes in Monty Hall problem,

  1. switching.
  2. changes in choices.

For smoother understanding, we will ignore switching for now, we going to focus on changes in choices first.

Now here is the main puzzle: what happens to the probability if the choices available changes before or after you made your choice?

To demonstrate this problem, we are going to use poker cards as exmaples, imagine there are four cards, J,K,Q,A on the table, the chance for you to flip the Ace is 25% obviously, what happens if:

A. BEFORE you pick a card:
i. I randomly remove 3 cards.
ii. I open all the non-Ace cards.
iii. I add more non-Ace cards, facing down of course.

B. AFTER you pick a card:
i. I randomly remove 3 cards, except your card.
ii. I open all the non-Ace cards, except your card.
iii. I add more non-Ace cards, facing down of course.

So what happens to the probability of getting Ace in Ai, Aii, Aiii, Bi, Bii and Biii? Keep in mind, no switching here for now.

Take your time to think before you continue to scroll down.

Ai: 25%(same)
Aii: 100%(increase)
Aiii: less than 25%(decrease)

Bi: 25%(same)
Bii: 25%(same)
Biii: 25%(same)

Did you get it right?

As you can see, the probability in all the cases of B remains the same, after you pick a card(you made your choice), whatever happens to the rest of the cards, is not your concern anymore, it WON’T change your pick, which is the reason why 50–50 is impossible.

Mean while, before you make your choice, anything that happens to the pool (choices) WILL affect your pick:

Aii: I reveal information to you, hence you know what left in the face-down card, increasing probability.
Aiii: I added more cards to the pool, reducing your chance to flip ace, decreasing probability.

now you may wonder will the action in Ai affect the probability, yes it will, but normally in a math question like this, we assume the chance of every card gets removed is the same, hence the probability remains the same in this case even though it is affected.

So what switching means? Switching is actually very simple, it is not even the main point here, it basically means you pick a new card in a new(sub) pool.

Now if you are sharp enough, by now you should notice, Monty Hall problem is made up of Bii and Aii cases, you start with Bii, if you choose to switch, you jump from Bii case to Aii(new pool)case.

Acid Coder

Typescript Zombie. Youtube Pikachu On Acid. (Unrelated to programming but by watching it you become a good developer overnight)