Help me understand Monty Hall

Factual Questions
1

I still don’t get it, even though it’s been explained to me many times. I have yet to experience a true “AHA!” moment.

For those that don’t know the scenario: Monty Hall (Let’s Make a Deal) shows you three doors. Behind one door is a brand new spankin’ '71 Cadillac. Behind the other two doors are zonk prizes.

I choose door A. Before Monty tells me what I’ve won, he reveals what’s behind door C. It’s a zonk prize. He now asks if I want to change my mind and pick door B instead of A.

Intuitively, I think that it doesn’t matter. 50/50 chance, right? But I know I should really pick door B. I just have no idea why. Can someone explain?

2

The Master speaks.

On “Let’s Make a Deal,” you pick Door #1. Monty opens Door #2–no prize. Do you stay with Door #1 or switch to #3?

3

OK, imagine that you have an opponent. You get to choose the door and you never change your mind. The other guy is left with whichever door you didn’t choose and Monty didn’t open.

Now, since you never change your mind, it doesn’t matter to you whether or not Monty opens a door at all. So your odds of winning are 1 in 3, yes?

However, that means your opponent will then win the remaining 2 out of 3 that you don’t win.

Effectively, your opponent is always switching, and you never are. And that’s why you should switch.

Does that make sense?

4

Clone Monty and the studios, so there are three games played at the same time.

In studio #1, the contestent picks #1
In studio #2, the contestent picks #2
In studio #3, the contestnet picks #3

The big prize in behind the same curtain in each studio.

If everybody sticks with their first choice, there’s just one winner.

If everybody switches, one of them goes from winner to loser, but the other two go from loser to winner.

5

You see, Monty cheats. He knows which door is which and never accidently shows you the prize door. It is a non-chance event that in no way influances the odds of the game.

(I was hoping you were going to ask about the kinky-sex angle on Monty.)

6

The fact that Cecil got it wrong the first time, and that so many people got it argued with him when he got it right, makes me feel like less of an idiot.

We Cecil said (paraphrased) “You have a choice between door #1 and both doors #2 and #3”, the light bulb went on. Then I read this…

…and said “Well duh, of course.” OK, me get. Me big jeen-yus now.

Thanks!

7

Paul in Saudi’s point is vital: the analysis only works this way if Monty
(i) knows where the prize is;
(ii) always reveals a dud.

Assuming this, there are two possibilities:
A) You chose correctly the first time. Monty reveals a dud and the other door is also a dud.
B) You chose incorrectly the first time. Monty reveals a dud and the third door conceals the prize.

We also need to note that the probability that you chose correctly the first time is 1/3.

So, should you switch? If you stick with your first choice, then you win in scenario A and lose in scenario B.
If you switch , then you lose in scenario B and win in scenario A.

Now recall that the probability of scenario A is 1/3 and the probability of scenario B is 2/3. Thus, you should switch.

8

The thing that made me understand this better was to increase the number of doors.

For example, in one of the houses in your town I have hidden $1000. I ask you to guess which house it’s in and you randomly pick the one at 14 Myrtle Lane.

I then open the doors of every other house in the town except for the one at 10 Main Street.

I give you the opportunity to switch to the house on Main Street. Do you switch? Does it still seem like you have a 50/50 chance?

9

Hijack. Do tell!

10

But it’s a big condition, that Monty is assumed to always show you a losing door. I don’t think I’ve ever come across a statement of the puzzle where this was stated as a given. And if it’s not a given, like Cecil pointed out, you have to make some assumptions about Cecil’s motivations.

I’ve argued with several people about this point. I like to re-state the puzzle in such a way as to make it more apparent. You walk up to a street hustler offering to play the ball-under-a-cup game. He mixes them up, you make your guess, and before he reveals your guessed cup to you, he shows you one without the ball and offers to let you switch. Would you do it? It should be apparent that you would be a fool to switch, and the only difference between this puzzle and the Monty Hall puzzle is the assumed motivations of the host.

11

It is stated as a given in the OP. (…he reveals what’s behind door C. It’s a zonk prize.) Even if it were not stated as a given, it would be implicit. It’s obvious that if he shows you what’s behind the winning door, there’s no guessing left–everyone KNOWS where the prize and the contestant would naturally choose the opened winning door.

Huh? What do Cecil’s motivations have to do with it? This doesn’t make any sense to me.

I don’t see any difference between this scenario and the stated problem, and it’s not at all apparent that one would be a fool to switch. You can apply the very same analysis to this situation that was explained for the OP problem, and get the very same result. If one of the losing choices is revealed to you, 2/3 of the time you’ll go from loser to winner if you switch.

Now if said street hustler is palming the ball, you’ll always lose regardless of whether you switch or not, but that’s a different game.

12

I don’t like how anybody else explained this. The probability thing was what did it for me, but I’ll try to make it clearer than Jabba.

You’re presented with three doors, one of which has the prize. I’m going to start with some shorthand now–p[Door #n] means “the probability that the prize is behind door #n.” Now, here’s the facts:

p[Door #1] + p[Door #2] + p[Door #3] = 1 which is another way of saying that the prize is behind one of the doors.

We have no further information about the placement of the prize, except that it is behind one of the three doors and does not move based on what you pick, so the odds are the same between the three doors and we say that:

p[Door #1] = p[Door #2] = p[Door #3] = 1/3

You can change the door numbers arbitrarily, but let’s say you pick door #1. Now we can say that

p[Door #1] = 1/3

and

p[not Door #1] = p[Door #2] + p[Door #3] = 2/3

The prize remains behind one of the three doors. Monty Hall opens door #2 and there’s no prize there. This is new information.

The new information is that there’s no prize behind door #2, so the previous statement that

p[Door #2] = 1/3 is now false because

p[Door #2] = 0

So here’s all the facts we know:

p[Door #1] + p[Door #2] + p[Door #3] = 1

p[Door #1] = 1/3

p[not Door #1] = p[Door #2] + p[Door #3] = 2/3

p[Door #2] = 0

therefore, p[Door #3] must be 2/3.

The question is, why is

p[Door #1] = 1/3

still a fact? Can’t it change to 1/2 along with door #3?

That would only be true if the game were rigged. In other words, let’s say the prize could have been behind door #2, so its odds were still 1/3 when Monty Hall chose it. Monty Hall must open a non-prize door, so if the prize were there, it must be moved. If they rigged it the fair way, they’d flip a coin and move the prize to either door #1 or door #3, so you get:

p[Door #1] = 1/3 (originally) + (1/2)(1/3) (half of the probability of the 1/3 from door #2) = 1/2

p[Door #3] = 1/3 (originally) + (1/2)(1/3) (half of the probability of the 1/3 from door #2) = 1/2

If they rigged it and, if the prize were behind door #2 they’d move it to door #1, you’d get

p[Door #1] = 1/3 (originally) + 1/3 (the probability of 1/3 from door #2) = 2/3

p[Door #3] = 1/3 (originally) + 0 (never moved from door #2 to door #3)

Likewise for the other direction.

If you assume the game isn’t rigged at all, you end up with the solution I described.

13

NeedAHobby: Well, yes, obviously if Monty can randomly choose either of the two remaining doors to move the prize to, then it’s a 50-50 chance. Otherwise you said more completely what I said. You have 1/3 chance of choosing the correct door so obviously the remaining door must be the option the other 2/3 of the time.

14

Here’s the straight dope.

If you pick a door, and then Monte opens another door (with a zonk prize), it is in your best interest to switch to the other door.

If you switch, your chance of winning is 2 out of 3.

If you do not switch, your chance of winning is 1 out of 3.
Here’s the best way to think of it:

Let’s say you pick a door. Let’s assume there is a zonk prize behind the door you picked.

With me so far?

O.K. Now let’s assume we play by the above “rules,” i.e. Monte opens another door (to show a zonk prize), and you chose to pick another door. This door must have a prize behind it – you will win. How do we know? Because 1) your initial door had a zonk prize, and 2) Monte opened another door with a zonk prize. Alas, you win.

So if you play by these rules, you will always win as long as your initial pick was a zonk prize.

And what are the chances your initial pick will be a zonk prize?

2 out of 3.

15

The important condition here isn’t whether the door Monty opens is always a loser, it’s whether he always opens a door at all. If he only opens a door some of the time, then you can’t make the same assumptions about probability. This is how the grifting game below applies.

**

Heh. True. The real question is if the grifter only occasionally turns over a cup. If so, your odds are still 1/3.

–
Justin

16

Let’s work through the possibilities:



* = prize
- = zonk

  Door   You   Monty   Do You
 1 2 3   Pick  Shows   Switch?  Win?
------------------------------------
 * - -    1      2      No      Yes
 * - -    1      2      Yes     No
 * - -    1      3      No      Yes
 * - -    1      3      Yes     No

 * - -    2      3      No      No
 * - -    2      3      Yes     Yes

 * - -    3      2      No      No
 * - -    3      2      Yes     Yes

 - * -    1      3      No      No
 - * -    1      3      Yes     Yes

 - * -    2      1      No      Yes
 - * -    2      1      Yes     No
 - * -    2      3      No      Yes
 - * -    2      3      Yes     No

 - * -    3      1      No      No
 - * -    3      1      Yes     Yes

 - - *    1      2      No      No
 - - *    1      2      Yes     Yes

 - - *    2      1      No      No
 - - *    2      1      Yes     Yes

 - - *    3      1      No      Yes
 - - *    3      1      Yes     No
 - - *    3      2      No      Yes
 - - *    3      2      Yes     No

Did I miss any?

17

Zoltar: That’s correct, but you have to be careful how you come up with odds.

Let’s take the case where the prize is behind door #1 (possibilities 1-8). Let’s assume you always switch.

If you initially pick door #1 you lose.
If you initially pick door #2 you win.
If you initially pick door #3 you win.

Odds of winning = 2 out of 3.

As mentioned by many others, if you always switch (after Monte opens a door with a zonk prize), your odds of winning are 2 out of 3.

If you do not switch, the odds are 1 out of 3.

18

It is stated as a given in the OP. (…he reveals what’s behind door C. It’s a zonk prize.) Even if it were not stated as a given, it would be implicit.
[/quote]
I guess I should have stated this more clearly. If all you know is that Monty is offering you a choice to switch this time, and you don’t know whether he is forced to allow you to switch all the times the game is played, then you don’t have enough information to solve the problem without making assumptions about the host’s (Monty’s, not Cecil’s as I mis-typed earlier) motivations.

The difference is that a street hustler will offer the opportunity to switch only if you guessed right initially, therefore you will be guaranteed to lose by switching.

19

I follow now, Curt. Thanks for the clarification.

20

My approach was like zoltar’s. If I look at his permutations, which are just like mine, there are 24 possiblities. In 8 of them, you chose the correct one in the first place. Thus, if you switch, you lose. In 16 of them, you chose the wrong one. If you switch, you win. So based on that, you’d always switch.

However, you could look at it this way, of the 24 possibilities, 12 are switches. Of those, you lose 6 and win 6. 12 are no switches, of these you lose 6 and win 6. So now it seems that switching does neither harm nor good.

This is batty, what is my mistake?