In the Monty Hall problem, is it correct to think the probabilities of the two unchosen doors as shifting?

Question

I was explaining the Monty Hall problem to someone thus:

You have three doors, and you pick one, giving you a $1/3$ chance of being right. The presenter opens one of the other two doors, knowing which door knows has a goat behind it. Then:

1. you would not pick the opened door as it contains a goat, and not the expensive prize. So the probability that the opened door is a winner, is now reduced from $1/3$ to $0$.
2. the remaining closed door must now have a $2/3$ chance of winning, an increase from the original $1/3$.

The person insisted that I was incorrect in stating that the probabilities shift in 1, 2 above. I do not understand if the opposer is correct or not. In some sense, I understand what they were saying: that the outcome does not change the probability. But at the same time, if we label the doors as "door chosen, door opened by presenter, and door remaining", then the probabilities will always be, respectively, $1/3$, $0$, $2/3$, right? The car will be behind the door that is untouched 2/3 times, no matter what.

Was I wrong to suggest that as the game is played, the probabilities of the two unchosen doors shift from $1/3$ and $1/3$ to $0$ and $2/3$?

Answer 1

I think your explanation is interesting. I think the person who is complaining about your explanation is confused about the effect of the reveal upon the probabilities. That the initial chance you chose correctly is $1/3$, and the open door -- post reveal -- has 0 likelihood of being the car, I think the confusion is where did the other $1/3$ go? You assert, correctly, that it belongs to the third door which was not initially selected or revealed. But you are arguing with someone who may think that that 1/3 is now evenly shared between your original door and the third door.

I think others have already explained the correct way to solve the game above. But, I will attempt to just point out a few things you might do to improve your explanation and convince your friend.

• Approach the question from their point of view -- get them to talk about it. Do they think that there is an equal likelihood of winning the prize behind the chosen and remaining doors?
• Use a visual. Sketch out a tree diagram. The visual will help your friend to see your point of view.

Finally, I suppose I would point out that when you ask

Was I wrong to suggest that as the game is played, the probabilities of the two unchosen doors shift from 1/3 and 1/3 to 0 and 2/3?

We're all saying your result is correct. But, maybe we should also point out that there are two components to what you are saying. There is a resulting probability $2/3$ chance of winning if you change, which you correctly state. Then there is an interpretation of why this is, which you explain by "shifting probabilities." Maybe trying out a different way to explain the result will help convince your friend.

EDIT: Let's be clear. Chosen the remaining door caries with it a $2/3$ chance of winning the car.

When you say that this person is adamant about the probabilities remaining 1/3, 1/3, 1/3 at the end of the game and that you are not sure who to believe - it seems like part of your question is not just how to convince this commenter, but also how to convince yourself.

Tree diagram Explanation

Rob gave a similar/equivalent/better explanation of this above, maybe mine is slightly different, but probably not significantly. You have 1/3, 1/3, 1/3 chance of selecting each of the doors represented at the top of the diagram. You obviously can't know which door you've chosen. 2/3 of the time you'll pick one of the two incorrect doors, 1/3 of the time you'll pick the correct door. Now, the winning strategy if you picked the incorrect door is to choose the remaining door -- this is easy to see. However it's not perfect, 1/3 of the time you'll pick the correct door to begin with and changing means you lose. But the way I would explain the game is simply that since $2/3$ of the time you will select the wrong door to begin with, the best move is to choose the remaining door. We'd expect to win 2/3 of the time.

Edit 2: Equivalent Statements

In a comment you stated

Or is the probability that the remaining door has a car still 1/3? The person suggested that even though it is more advantageous to switch because the probability that you lose is 2/3 if you stay, that doesn't imply that the remaining door has 2/3 probability of winning. The probability that the car is in one of the remaining two doors is still 2/3 (obvious), and that's why switching is better. But it doesn't mean the specific door untouched has a 2/3 probability of winning.

This is nonsense. If the probability that you stay and lose is $2/3$, then the probability that the car is behind the only remaining door with the chance of concealing the car must be $2/3$. Your statement and your friend's statement are identical. To the person playing the game, you can't know which door the car is behind, and the car can't switch as the game is played - so it might seem logical to think that since the car can't move there must always be a 1/3 chance that the car is behind any door. Before we start the game this is exactly the situation -- there is a 1/3 chance that the car is behind any of the doors. After the game when we know the location of the car, we could I suppose say that the probabilities are $\{1, 0, 0\}$ -- two doors do not have a car, one does.

Further you said that "the probability that the car is still behind one of the remaining two doors is 2/3" -- no, the probability that the car is still behind one of the remaining two doors is 1 (unless you have some reason to doubt the construct of the game to begin with). It can't be the case that the probability of the remaining door has a car is still 1/3 and that there is a 2/3 chance of winning if you change. Since (winning by changing) and (the car is behind the remaining door) are two ways to say the same thing.

Answer 2

You might want to nail down what probabilities you're talking about. In the first case, it's "the probability that the prize is behind that door, GIVEN what you know now." In the second, it's the same thing, given what you know AFTER the revelation. The numbers don't change -- they are different numbers all along.

Consider a simpler experiment: I have two pennies -- one an ordinary one, the other a 1909 SVDB penny worth $300. I place them behind me and grab one in each hand behind my back. I don't know which hand has which penny. I ask you "what's the probability that the valuable one is in my left hand?" You say "50%".

Now I do the same thing, but this time, having grabbed the coins, I peek in my left hand and say to you that the valuable coin is there. I then ask you "What's the probability that the valuable coin is in my left hand?" Assuming you trust me, the answer is surely 100%. But as you'll observe, these are different probabilities.

The tricky thing in the Monty Hall problem is that \begin{align} Pr[\text{Door $n$ contains car} | \text{door $k$ does not contain car}] \end{align} always has the value $2/3$, but the value of that probability to you is nothing until you're informed that door $k$ does not have the car.

Answer 3

$$ P\left({\mbox{Prize is behind chosen door}}\right){}\neq{}P\left({\mbox{Prize is behind chosen door}\,|\,\mbox{Prize is not behind door with goat}}\right)\,, $$ in general.

Answer 4

I think it is a little confusing to talk of the probabilities changing. However, at the point that the presenter opens the door and reveals a goat, the probabilities of your chosen door, the open door and the other door being the one with the car are 1/3, 0 and 2/3 as you say. It is more usual to talk about conditional probabilities: what is the probability of one event given that some other event has occurred?

You can avoid talking about conditional probabilities altogether by explaining the Monty Hall problem like this: if you picked the car initially, then switching makes you lose, while if you picked a goat initially, then switching makes you win. The probability of picking the car initially is $\frac{1}{3}$ and the probability of picking a goat initially is $\frac{2}{3}$, but switching interchanges the outcomes and hence interchanges these probabilities. In a nutshell, interchanging success and failure in a situation where failure is originally more likely than success is a winning strategy.

Answer 5

First off, it is very wrong to say that probability "shifts." What really happens should be a three-step solution that, in general, works like this:

1. You start with a set of events {E1,E2,E3,E4,...EN} whose probabilities add up to 1. That is, P1+P2+P3+P4+...+PN=1.
2. Some of these events get eliminated by information you learn. The probabilities of what remains add up to less than 1. For example, eliminating E2 and E3, we get P1+P4+...+PN=X<1.
3. Since the updated probabilities need to add up to 1 again, we divide each of them by X: P1'=P1/X, P4'=P4/X, etc. This can be called "renormalization."

In the classic explanation for the Monty Hall problem, we have:

• E1 = "Car behind your door"; P1=1/3
• E2 = "Car behind the next higher door, wrapping to #1 if necessary"; P2=1/3
• E3 = "Car behind the next lower door, wrapping to #3 if necessary"; P3=1/3

Say E3 is eliminated as a possibility when Monty Hall opens a door. Then:

• P1' = P1/(P1+P2) = (1/3)/[(1/3)+(1/3)] = 1/2
• P2' = P1/(P1+P2) = (1/3)/[(1/3)+(1/3)] = 1/2

Wait a minute, that's wrong. But then, so is the classic explanation for the correct set of answers, P1'=1/3 and P2'=2/3. It says that P1'=P1 and P2'=P2+P3 (that is, that probability "shifts"). Any true mathematician should be embarrassed to offer such an explanation. Yes, it gets the right answers, but for entirely wrong reasons.

And using the classic explanation is one of the main reasons the Monty Hall Problems is still controversial. It fails in two very obvious ways: it doesn't explain what is wrong with the solution I just gave; and it is less mathematically sound since it implies the improper concept of "shifting" probability. Why should a mathematical novice accept an explanation that has these obvious faults?

The correct solution, following the trivial pattern I stated above, needs four events. It is clearer if I list them backwards:

1. E4 = "Car behind the next lower door. Monty Hall opens the next higher door."; P4=1/3.
2. E3 = "Car behind the next higher door. Monty Hall opens the next lower door." P3=1/3.
3. E2 = "Car is behind your door, and Monty Hall opens the next higher door." P2=1/6.
4. E1 = "Car is behind your door, and Monty Hall opens the next lower door." P1=1/6.

Then the solution is that Monty Hall opened only one door - say it is the next higher one. This eliminates E2 and E4.

• P1' = P1/(P1+P3) = (1/3)/[(1/3)+(1/6)] = 2/3.
• P3' = P3/(P1+P3) = (1/6)/[(1/3)+(1/6)] = 1/3.

This gets the same answer as the "probability shifts" explanation, but with no shifting, and it shows what is wrong with the previous solution. The information you learn does not "eliminate" every game where there is a goat behind a certain door, it eliminates only those where that certain door was opened. Since another door can get opened even when that certain door has a goat, the two events are not the same.

Answer 6

You can make sense of probabilities "shifting". What you are doing here is assigning two probability measures; the first one $P_0$ might assign probabilities such as

• $P_0(\text{car behind door #2}) = 1/3$
• $P_0(\text{car behind door #1 and Monty opens door #3}) = 1/6$

and the second probability measure $P_1$ assigns

• $P_1(\text{car behind door #2}) = 2/3$
• $P_1(\text{car behind door #1 and Monty opens door #3}) = 0$

While it is reasonable to say that the probabilities of events change when switching from $P_0$ to $P_1$, do take care to note that $P_0(\text{car behind door #2})$ does not change.

Now, this is not normally how people use probability theory to account for new information; instead they they use conditional probabilities. In fact, your $P_1$ can be expressed in terms of $P_0$:

$$ P_1(\text{some event}) = P_0(\text{some event} \mid \text{Monty opens door #2}) $$

So the typical way to mathematically model the situation only ever uses a single way to assign probabilities to events; they work with $P_0$ throughout the entire problem. What changes with new information is which probabilities we are interested in. In particular, we expect the actual outcomes to be governed by the conditional probabilities.

Answer 7

The big question that may convince your friend:

Are you counting the possibility that the host will open a door to show you a car?

If he doesn't know what's behind the doors, this is a real possibility. (Also, are you allowed to switch doors if he shows you a car? It's not in the rules. Does it even matter?)

In probability, a "sample space" is (roughly speaking) the set of all possible outcomes that you are considering. A fair coin has 1/2 chance of showing heads and 1/2 chance of showing tails. You ignore all other possibilities (the coin lands on its edge; the coin falls into a storm drain and cannot be read; etc.).

If the host knows what's behind the doors and avoids ever showing you a car behind a door you have not chosen, then he has given you information, and the odds change accordingly.

You get the same result of change in odds (for a statistical analysis after the fact) if the host knows nothing, but you discard all occasions when he "accidentally" shows you a car.