(SOLVED) Monty Hall: the number of unknowns decreases but probabilities stay the same?

Question

I recently got an explanation of the Monty Hall problem and I thought I understood it but after giving it more thought, it still looks wrong.

Instead of using goats and doors, the example used a 52-card poker deck. The cards are facing down on a table and you need to pick one card without looking at it. Let's say that the winning card is the Ace of hearts (A❤️). The probability that you got an A❤️ is 1/52.

The other 51 cards are still on the table facing down, but now the game host reveals 50 non-winning cards and asks you if you want to change the card you picked.

The explanation that I've understood as to why you should change your choice is that the card you picked (set X) has 1/52 probability of being the winning card, while the other 51 cards (set Y) have a 51/52 probability of having the winning card. And when the game host revealed 50 cards, that 51/52 probability didn't change, so basically now there is a single card with a 51/52 probability of being the A❤️.

I accepted this explanation as a non-intuitive mathematical truth. But after giving it more thought, it still doesn't make sense to me! Here my thought process:

There are 52 cards on the table and I pick one at random. Now the game host reveals just one losing card. My thought is that this is no longer a 52-card game, this is now a 51-card game ⇒ My card now has a 1/51 chance of being the A❤️, and the other (still facing down) 50 cards have a 50/51 chance of having the winning card.

The game host proceeds to reveal one more card. Now this is no longer a 51-card game but a 50-card game! ⇒ My card now has a 1/50 chance of being the A❤️, and the other (still facing down) 49 cards have a 49/50 chance of having the winning card.

[...]

The game host has already flipped 49 cards and reveals one more card. A total of 50 cards are revealed. This is no longer a 52-card game or a 3-card game, this is a 2 card game! The card I picked has a 1/2 probability of being the A❤️, and the other (still facing down) card has a 1/2 chance of being the winning card. The other card in the table does not have greater chances of being the winning card than the one I've already picked.

In my thought process, I update the the probabilities every time a new card is revealed because the number of unknowns has changed, so how on earth do the probability stay the same??? Why do they stay 1/52 vs 51/52 ???

What am I misunderstanding? Help!

Edit:

Here I'm developing an example with 4 cards so it's not too long to write down:

There are 4 cards, Ace, King, Queen, Jack (A,K,Q,J). The winning card is the Ace. Here the possible scenarios:

Scenario #1: I choose A, host shows K & Q. Second chance: choose between (hidden) A and J. If I keep, I win. If I change, I lose.

Scenario #2: I choose A, host shows K & J . Second chance: choose between A and Q. If I keep, I win. If I change, I lose.

Scenario #3: I choose A, host shows Q & J. Second chance: choose between A and K. If I keep, I win. If I change, I lose.

Scenario #4: I choose K, host shows Q & J. Second chance: choose between K and A. If I keep, I lose. If I change, I win.

Scenario #5: I choose Q, host shows K & J. Second chance: choose between Q and A. If I keep, I lose. If I change, I win.

Scenario #6: I choose J, host shows K & Q. Second chance: choose between J and A. If I keep, I lose. If I change, I win.

Summing up all possible scenarios, in 3 scenarios out of 6 I win by keeping, and in 3 scenarios out of 6 I win by changing. So it's a 50/50 probability.

I get a 1/4 vs 3/4 probability only if I bundle together scenarios #1,2,3 and treat them as a single scenario with equal probability to the other scenarios. But why? I guess this is where I'm failing to understand. Or am I doing another mistake?

Edit 2:

Already a couple of times, I've seen people mentioning that the host is flipping cards randomly. Is it so? If that's the case, then I understand better the problem (although it doesn't really make sense that the host is flipping cards randomly and yet at the same time we are guaranteed that he/she's only gonna flip the losing cards... but I guess the game just stops in case the host flips the winning card)

UPDATE:

I have finally understood the problem. The answer is in the last bit I wrote just above:

I get a 1/4 vs 3/4 probability only if I bundle together scenarios #1,2,3 and treat them as a single scenario

Scenarios #1,2,3 are actually a single scenario. And the choice of the host is a 1:1 direct reflection of the probability that I picked the right (1/4) or wrong (3/4) card.

Answer 1

Here I'm developing an example with 4 cards so it's not too long to write down:

There are 4 cards, Ace, King, Queen, Jack (A,K,Q,J). The winning card is the Ace. Here the possible scenarios:

Scenario #1: I choose A, host shows K & Q. [...]

Scenario #2: I choose A, host shows K & J . [...]

Scenario #3: I choose A, host shows Q & J. [...]

Scenario #4: I choose K, host shows Q & J. [...]

Scenario #5: I choose Q, host shows K & J. [...]

Scenario #6: I choose J, host shows K & Q. [...]

Summing up all possible scenarios, in 3 scenarios out of 6 I win by keeping, and in 3 scenarios out of 6 I win by changing. So it's a 50/50 probability.

I get a 1/4 vs 3/4 probability only if I bundle together scenarios #1,2,3 and treat them as a single scenario with equal probability to the other scenarios. But why?

So the intuition to understand the problem is exactly in the bit that I wrote just there at the end. Scenarios #1,2,3 are actually one single scenario. If the host was picking cards randomly, then yes, they would be 3 possible scenarios with equal likelihood.

But these 3 scenarios are actually just one because the host knows the cards and I happen to have picked the winning card.

Scenarios #1,2,3 can be thought as a single scenario called "host not hiding the Ace given that I have already picked the Ace with a 1/4 probability" or equivalenty "I picked the right card with a 1/4 probability, therefore the host is hiding the Ace"

Scenarios #4,5,6 can also be thought as a single scenario called "host hiding the Ace given that I have picked the wrong card with a 3/4 probability" or equivalenty "I picked the wrong card with a 3/4 probability, therefore the host is hiding the Ace"

The choice of the host is a direct consequence of the probability that I get it wrong (3/4) or get it right (1/4).

Answer 2

"There are 52 cards on the table and I pick one at random. Now the game host reveals just one losing card. My thought is that this is no longer a 52-card game, this is now a 51-card game."

No, it is not a 51-card game. At the start, you were given a choice of 52 cards, not 51. It is always a 52-card game because your choice was one of the 52 cards.

Answer 3

The card you choose to begin with becomes immune to any further changes the moment you make that choice. So the probability of win/loss associated with that card also becomes immune to any change. It will continue to be $1/52$, no matter how many cards you remove (or turn to reveal) from the other group.

It's different with the other cards - the remaining $51$. They can undergo changes. And so the probabilities associated with them can change too.

Now, if the host removed cards randomly (without the knowledge of which card is the winning card), the probabilities associated with each card will remain the same.

For example, if the host removed $1$ card, the win-probability of each card will stay at $1/52$ (including the card that was removed). If they removed $50$ cards randomly, your probability of winning will be $1/52$ regardless of whether you made the switch or not. I hope you would agree with me here that the win-probabilities associated with each card will not suddenly become $1/2$ just because there were two cards left. Obviously, there's a very good chance the winning card was removed among the $50$ cards that were removed (randomly).

So, when the host removes cards randomly, the probability of the game ending in your win also changes with each removal. In fact, with each removal, it becomes more likely that you can't win.

This is where host knowledge of the winning card and their use of that knowledge when removing the cards becomes pivotal. Since, they will never remove the winning card, the game always retains the possibility of ending in a win. And the probability of winning is obviously distributed among the cards that are still a part of the game.

Now the part you already know. When only $2$ cards remain - the $1$ you chose initially and one from the other group - together they give a winning probability of $1$. As I stated in the beginning, the first card still has a $1/52$ probability of being the winning card. So the other one must have a $51/52$ probability.

A bit more of intuition

Also, think about it as a player. You watch the host remove cards one by one. And now there's only a single card remaining (apart from your initial choice). This card escaped being thrown out of the game fifty times while its companions - $50$ of them - couldn't. Will you not be looking at this card with some respect? Like yeah, this is very likely the card?

The card you selected at the start is also there on the table. But it doesn't carry the same awe. Why? Because the host couldn't have removed it even if they wanted to. It wasn't protected because of the host's knowledge of the winning card but by virtue of being the card you picked.

Answer 4

This and the original Monty Hall problem can be depuzzled in a very simple way:

The question to ask is: when you change the card (door) you had chosen initially what is the probability that you loose the prize.

In the classic Monty Hall problem this probability is $1/3\,.$ Therefore you chances of winning are -as we know- $2/3$ when you change.

In your version with $52$ cards and $50$ revealed the probability of loosing the prize is $1/52\,.$ So you win with a staggering probability of $51/52$ when you change.

Answer 5

Probabilities are based on information and might change if information is added or might not change.

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If our new information is exactly "the host randomly chooses a card, flips it and it turns out not to be the winning card" then you are correct in stating that this is no longer a 52-card game but is from now on a 51-card game.

However our new information is different: "the host randomly chooses a card from those cards that are non-winning cards".

If Monty would randomly choose one of the two doors and coincidently this door does not contain the car (no use of foreknowledge) then indeed from that moment on both other doors have probability $\frac12$ to be winning doors.

But the situation is different: Monty has foreknowledge and deliberately chooses a door that is non-winning. In that case the new information has no impact on the probability that behind the door chosen originally there is a car.

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Let me add a simple solution for the problem.

Someone who sticks to his original choice will win if and only if his original choice is correct, and the probability on that is evidently $\frac13$.

Someone who switches will win if and only if his original choice is incorrect, and the probability on that is evidently $\frac23$.