Understanding the Monty Hall Problem

Perhaps this is an even simpler explanation:

If the answer to the Monty Hall problem were 50/50, the contestant, on average, would win the car 50% of the time simply by sticking with their original guess…but one will only win a one-in-three guessing game 33.33% of the time if one sticks with one’s first choice so it can’t be 50/50, can it? It must be 33.3.3% if one sticks and, ipso facto, 66.67% if one switches, n’est-ce pas?

Palmer, it doesn’t change. You still have the same choice if Monte reveals the goat regardless if he knows or not.

The only reason he knows is that they always want to reveal a goat and give the contestants the option, because most think it’s 50/50 and will stay with their initial choice, mistakenly.

If he reveals the car the choice isn’t given and they lose anyway, if he reveals the goat, the choice is always given but the chances didn’t change.

My head just about explodes when I think of this problem. Intuitively, I say 50/50, because once the goat is revealed by Monty, you have a one in two chance in having picked the car.
I better understood the problem when I thought of a pack of cards. If I pick a card from a full deck, I have a one in fifty two chance of having picked, say, the Ace of Spades. If I am given the choice to stick with my card or swap to the remaining fifty one cards, it’s much more likely that the Ace of Spades will be in the remaining cards. That Monty turns over fifty cards of the fifty one doesn’t change the probability of the ace being in the larger pile, so swapping will yield better probability that you will get the Ace of Spades than sticking with your original choice. So it is with three cards, or three doors with a car and two goats.
I still struggle with accepting that the fifty fifty probability is wrong though, because whichever door you pick, Monty always reveals the goat. If a friend came in at that point and only saw two doors, and told me to stick with my choice, or swap, it would mean a fifty fifty chance of me having picked the car. The third door seems irrelevant, as it’s always the goat. Aaaaaaaaaaah.

Of course, #832 tommyt has thrown my head into a spin. If I am presented with three doors with one car and two goats, and one goat is already revealed, it is 50/50 whether I pick the car out of the remaining two doors. Whether I stick or swap doesn’t change the probability of 50/50, so whether the goat is revealed before or after you pick a door makes no difference to the outcome, as the goat is always revealed. That’s where I have difficulty in understanding the problem. Whether there are three doors or a million, it’s always reduced to two possibilities, regardless of which door you pick.

Hi David, great comment & scenario with the cards. It comes down to knowledge: if your friend walks in near the end of the game and sees 2 doors, they have a 50-50 chance since they don’t know anything about the choices. But you have seen the filtering process that removed bad choices from the doors you didn’t pick. Therefore, you (with this knowledge) should do better than the uninformed stranger.

I think that’s the main tension (that I had to overcome too): seeing the filtering process gives knowledge that can increase your guess beyond random chance. (Even better knowledge would be setting Monty set up the game in the beginning! Even we arrived too late.)

Where a second contestant, with no knowledge of which door the contestant first selected, enters the game after one of the two goats has been revealed by the host, there are three different probabilities:

Probability the car is behind the door the contestant first selected: 33% .

Probability the car is behind the only other closed door: 67%.

Probability the uninformed second person randomly selects the door concealing the car: 50%.

The second person can improve their chances of winning from 50% to 67% if they can discover which door was first selected by the original contestant and use that knowledge to their advantage by selecting the other door.

While it seems bizarre that the probabilities could change depending on who is choosing, the 33/67 probabilities relate to the DOORS whereas he 50% probability relates to the UNINFORMED SECOND CONTESTANT.

By analogy, imagine world No 1 ATP ranked tennis player, Novak Djokovic, is to play Bernard Tomic, the annoying Australian ranked 26 by the ATP. Who’s going to win? Let’s assume 80% chance Novak and 20% chance Tomic.

With this knowledge, you have an 80% chance of picking the winner if you play the odds, right?

Now, consider the player’s are disguised and not discernible from one another except one player is wearing green clothes and the other player is wearing blue.

Because you have no knowledge of the ability or track record of each player (just like the second contestant who sees two indistinguishable doors), your chances of picking the winner are now just 50% because you are making a random selection, not an informed one.

However, if someone informs you that Novak is wearing blue, your chances of picking the winner improve from 50% to 80% if you act rationally.

Just like in Monty Hall, the unequal probabilities relate to the objects’ (doors/tennis players) chances whereas the 50% probability relates to the punter’s chances and these vary with their knowledge.

Knowledge being power, as they say.

I understand the problem perfectly. Thinking of a pack of cards, there’s a one in fifty two chance that you have picked the Ace of Spades. It’s a much higher probability that the Ace of Spades will be in the remaining fifty one cards, whether or not fifty of those cards are revealed to you before you choose to swap. Although in swapping, you are not guarantee a win, statistically, it will yield better results over repeated tries. So it is with three cards, or three doors.
It took me so long to shed the 50/50 notion though… My simply refused to see that swapping would be better!

Here is a stumper for the 50% coinflippers.

There is a game. Two teams, white and blue, play against each other. One wins, one loses; it is not possible to tie.

You pick which of the two teams will win.

Given that set of facts, it’s a 50/50 choice. And this is the essence of your argument - that in the end, you are choosing between a goat and a car. You are right or you are wrong. It doesn’t matter which door of the three you pick, because by eliminating one of the doors you didn’t pick, your second choice will be a coinflip.

However - consider this. The two teams playing are not the white team and the blue team. The game is NFL football, it’s at the end of the 2014 season, and the teams are the Patriots (12-4) and the Jets (4-12), and the game is at Gillette Stadium.

Is it still 50/50 for you? Nothing has changed - a game, one team will win and one will lose, one team is blue and the other white. But you will be lying to me if you say it’s still 50/50, so just don’t. Yes, the Jets could win. However, the weight of the new information is overwhelmingly indicative that the smart money is on the Patriots.

What does this have to do with the Monty Hall game? Nothing, really. It just has to do with your argument, which is not relevant to the Monty Hall game. Stay with me here.

In the Monty Hall game, there are three doors. One door has a car behind it. Just let that sit for a minute and forget absolutely everything else. If there are three doors and one of them has a car behind it, and one of the doors was opened at random, there would be a 1/3 chance that the open door will reveal a car. That is an absolute fact. So you, or anybody else EXCEPT for Monty Hall would have a 1/3 chance of picking a car at this point in time.

You only have this information to go on - three doors, one car. Forget that you know that one of the doors you do not pick will be eliminated, because that absolutely has no bearing on your choice right now. It’s meaningless. The whole situation is just you, three doors, and a car. So, pick a door. You have a 1/3 chance of being lucky. Are you with me so far? We can agree that with three doors and one car, 1/3 of the doors will have a car behind it. Based on the information you know, there is no difference between any single pick.

So about the Monty Hall goat elimination - he does this, and you’re down to the second choice, and you’ve got two doors, and you now know that you have one door with a goat behind it and one with a car. You also have no way of knowing whether or not you’ve chosen a goat or a car with your first pick. And your contention is that now there are 50/50 odds that either you got lucky on your first pick or you didn’t get lucky on your first pick.

But the odds are not 50/50 that you got lucky. The odds are 1/3 that you got lucky. This is because you made your decision based solely upon one car and three doors. Nothing has changed about those facts. There are three doors and one car. The one car is behind one of the three doors. It doesn’t matter that one of the doors has been eliminated as the door concealing the car. All three doors still exist. That will not change.

And here is your choice: switch doors or do nothing. I chose those words ‘do nothing’ deliberately. If you do not switch doors then you have not acted on the new information. Your decision is still based on the original choice, which was made under conditions where no one door had anything special about it. All three looked the same: one hid a car and two hid goats. That is your original choice, right? One out of three. That choice is still made with one out of three outcomes: you got lucky, you didn’t get lucky and got goat one, or you didn’t get lucky and got goat two.

So now you have a choice. You’ve eliminated one of the doors as a door hiding the car. So, behind one door is a car, and behind the other is a goat. 50/50 right? Wrong. It’s not 50/50. If you think it’s 50/50, then you are not using all of the information at your disposal. There are not two outcomes because there are only two doors remaining which could hide the car. There are three outcomes.

Stay with me now.

Of the two remaining doors, half of them hide a car and the other half hide a goat. But there are three outcomes.

You get lucky and you choose the door with the car.

You don’t get lucky and you choose the door hiding goat #1.

You don’t get lucky and you choose the door hiding goat #2.

This is because YOU don’t know WHICH GOAT was revealed. You can’t tell one goat from the other. All you know is that there are two goats and a car. Only Monty Hall knows which goat is revealed and which one is hidden.

“How does this matter??” you shout at me.

Well, let’s see…what has actually happened here? Keep in mind that Monty Hall knows which door hides the car, and that the other two contain goats. He doesn’t even really need to know which goat is which. What he does know is whether or not HE had to make a choice when you chose your initial door.

Because he was compelled to reveal a goat before offering you your final choice, he was able to choose a course of action in only one instance - and that instance is the 1/3 chance that you got lucky with your original pick.

If you picked the door with goat #1, he had to reveal goat # 2.

If you picked the door with goat # 2, he had to reveal goat # 1.

Only if you picked the car did he have to make a choice, and that choice was which goat to reveal, and he will have to make that choice upon one out of three possible choices you made at the beginning?

How does he decide which goat to reveal? Nobody knows. Who cares. It doesn’t matter. To him, they are interchangeable. To you they are both interchangeable - you don’t care whether you are going home with goat # 1 or goat #2. But do you know who it matters to?

THE GOATS, THAT’S WHO. And you should care, too.

The goats are the KEY.

So your choice is this: do nothing and sit on your 1/3 chance, or bet on the 2/3 goat option. It’s no coinflip. Your first choice created one of three scenarios:

You got lucky and chose the car door on your first try. The other two doors held goats.

You chose goat # 1 door. The other two doors held a goat and a car.

You chose goat # 2 door. The other two doors held a goat and a car.

You made the choice that determined which of these three divisions were made - and here is where your 50/50 argument really belongs. It doesn’t matter which of the two doors you did NOT pick had a goat behind it. One of them is guaranteed to have a goat. It doesn’t matter which one of the two remaining doors hides a goat. One is no more attractive than the other, and revealing that one of the two remaining doors held that goat is not telling you something that you didn’t already know. What you don’t know is whether the remaining door hides a goat or a car…it still doesn’t matter. It doesn’t change the probability.

There was a 100% chance that you were going to be looking at a goat between your first and second choices. That does not diminish the chance that you got lucky with your fist pick, and it does not change the chance that the other two doors hide a car.

What you really want to ask Monty Hall is which goat you are looking at.

  1. “you’re confusing the aspect of “winning” with the probability of choosing between two doors”

Are you saying the solution to the MHP is 50%? If so, you have confused the choice between two doors as random. It is not; it is a deliberate selection by the contestant. A knowledgeable contestant might switch, while an instinctive contestant might stick. The MHP question refers to the chance of winning by switching. In scientific method terms, the independent variable (action) is switch or stick and the dependent variable (outcome) is win or lose.

This is SUCH a nice problem because if you look at it one way (two doors left, it could be either, so it MUST be 50-50) it seems so easy and simple, and if you look at it another way (if you just stick, you can’t improve your chances over the original 1/3) it also seems so easy and simple – and yet the answers these two perspectives give are different!

I think the best way to explain the problem with the 50-50 answer may be that when you treat the problem as a two-door problem after Monte opens a door, you are throwing away information! It is NOT a brand new game! Of course, your original explanation was excellent and provides specifics, but it can still leave the reader wondering, “But WHY?” The answer to this question is that treating the two-door situation as a brand new problem is ignoring critical information that you didn’t have at the beginning of the game.

Thanks for such a clear treatment of this classic – and still darn nice! – problem.

@Mike,
As I said analyse the outcome of 300 games: Monty will reveal the car 100 times (and you lose straightaway). In the 200 games where Monty reveals a goat you’ve already picked the car 100 times - there is no advantage in switching.

Palmer, I’m saying that after you pick a door and a goat is revealed by the host, regardless if he knew or not, you will always have a 2/3 probability if you switch. Nothing in that scenario changes the decision by what he knows.

You pick door 1, he reveals a goat behind door 2 without knowing it was there, do you switch? Yes
You pick door 1, he reveals a goat behind door 2 because he knew it was there, do you switch, Yes

@Mike
" a goat is revealed by the host, regardless if he knew or not, you will always have a 2/3 probability if you switch"
And that is where you are, quite simply, wrong. I’ve explained how that isn’t the case using a sample of 300 games. In the 200 games where Monty reveals a goat you are no better off switching doors. If it was the case, you’d only win 67 games (out of 300) by staying, which is clearly incorrect

“after you pick a door and a goat is revealed by the host, regardless if he knew or not, you will always have a 2/3 probability if you switch.”

Read it again. I’m only referring to the games where the goat is revealed and the decision there after. Knowledge of the goat doesn’t matter if the choice is only given when the goat is shown. That decision will result in a win 2/3 of the time. You’ve even shown very good examples of this in previous posts.

Let me present you with this, you pick door 1 and Monty asks someone in the crowd to open a door, they open door 2 to reveal a goat. Do you switch?

“It’s imperative that Monty does actually show a goat for your chances to increase to 2/3”

I agree, Palmer?

@Mike,
In order for switching to be a 2/3 chance it has to be guaranteed that Monty will reveal a goat from the 2 doors you didn’t pick. If there is a CHANCE that he will reveal the car instead, then in those games where he does (accidently) reveal a goat there is no advantage in switching.
In the standard Monty it is 100% certain Monty reveals a goat, in this version there is only a 66.7% chance Monty reveals a goat - that changes the conditional probability (i.e after Monty reveals a goat) that the door you picked originally contains the car. You will still win only 1/3 of ALL games by staying, but since Monty reveals the car in 1/3 of ALL games (and you lose) you only win 1/3 of ALL games by switching.

Put another game, the 1/3 of games you lose when Monty reveals the car, are games you’d normally win by switching. When he reveals the car you don’t get the opportunity to switch.

"” a goat is revealed by the host, regardless if he knew or not, you will always have a 2/3 probability if you switch”
And that is where you are, quite simply, wrong.

Exactly, so how am I wrong here, I’ve always said that the goat had to be revealed, knowledge of the goat doesn’t matter, only that it is revealed.

“Put another way” I meant to say

When Monty Opens a Door Randomly aka Monty Hall II

The revelation at random of a goat is purely coincidental rather than meaningful. It does not mean, for instance, that there is a greater chance that the car is behind one door rather than any other of the two remaining. There is a 50% chance the car is behind each of the two remaining doors and a 100% chance the car is behind either of those two remaining doors.

From the outset, the chances of a goat being shown at random was 67% and the chances of a car 33%.

If a car is revealed the contestant © immediately loses.

If a goat is revealed upfront, C should go on to win 33 games out of the 67 games when the car was not revealed upfront. The host will win 33 times upon the first random reveal and 33 times upon the contestant showing he has stuck with a goat.

If the contestant switches the 67 times a goat is revealed upfront, C will win 50% of those times out of 67, or 33 times.

Therefore, 50% of the time, switching upon seeing the goat will result in a loss, not a win, a result that is 50/67 as likely to yield a car as the likelihood that Mike postulates is the case.

If that is not clear, imagine laying down 100 sets of 3 playing cars, Joker / Joker / Ace.

Reveal one card from each row. Aces are revealed in 33 rows. You cannot win the games in those rows. Eliminate those rows. 67 rows with two face-down cards and one upturned Joker in each remain. Neither of the two down facing cards is more likely than any other to be the Ace, whether you made a guess before or after the Ace was revealed.

That’s Monty Hall II.