# Monty Hall

1. Feb 5, 2005

### bomba923

Why does switching give you 2/3 chance to win?

Approach #1: Let's say the winning door is door#1:

We choose door#1, he has us choose between door#1 and door #2
---------------------------------OR between door#1 and door #3.

We choose door#2, he has us choose between d#1 and d#2.

We choose door#3, he has us choose between d#1 and d#3/

By switching, we win in 2 possible ways; by not switching, we win in 2 possible ways.
________________________________________________________________

Let's assume that we have 100 doors this time. 99% of the time we will choose the wrong door, 1% we will choose correct. If we choose wrong, we must switch to win the prize. If we choose right, he can screw us up by giving us 99 ways not to switch and win! Thus, we can win in 99 ways and lose in 99 ways, so switching/notswitching doesn't really change our expectations.

Another case: again, we have 100 doors. We choose one, he opens 98 doors without a prize in them. Now we go outside and select a random bum off of the street, and force him to choose between the two remaining. What does it matter to the bum!? 50-50 chance of winning anyway!--two doors, one prize!! (to him: what are all those goats for? I just have two doors!)

It seems intuitive to assume that, if our choice has initially 1% chance of being correct, then no matter how many others open (well, as long as its between 1 and 98 inclusive), it will still have 1% chance of being correct~ But i say its not; think about it in the below example:

Suppose we have three doors; we choose one and he opens another to reveal a goat. HE GIVES US ANOTHER CHANCE--between TWO doors. We select out of TWO DOORS--NOT THREE, This time. We have 50-50 chance of it being either door.
_________________________________________________________________

Approach #2: When I think about another way, we have 1/3 chance of selecting the right door, 2/3 chance of selecting the wrong door. If we select the wrong door (2/3 chance), we will win if we switch (2/3 chance of selecting the wrong door, and thus 2/3 chance of winning in all). If I select the right door, and don't switch, i have 1/3 of winning in all (not what my previous paragraphs said!).

Now which "train of thought" or approach is correct :grumpy: !? I demand to know!! (also, plz explain why the incorrect approach is incorrect, plz?)
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Just one more thing: in approach #1, i considered Monty's actions if I chose both the incorrect and correct door. I notice that a correct initial choice will have Monty give us one of two choices, to screw us up :rofl: . Is it correct to consider all possible trials--even if one choice will have two possible courses of Monty's action!?
If not....,...well, than I can InDeeD say that switching will give us 2/3 chance towards winning, no prob. BUT WOULD THAT BE CORRECT???!!

2. Feb 5, 2005

### bomba923

Btw, what effect does Monty Hall knowing where the prize is
have on the probability?

3. Feb 5, 2005

### bomba923

Sorry for the third post in a row:

Let's say door#1 is the winning door:
(W=win, L=lose)

We choose | Switching |Notswitching
----1-----------L------------W
----2-----------W-----------L
----3-----------W-----------L

From the switching/notswitching viewpoint (the columns):
P(win by switching)=2/3
P(win by notswitching)=1/3
-------------------------------------------
But, relative to or by the viewpoint of the player, who DOESN'T KNOW the Right Door (the rows):
P(win by switching/notswitching from #1)=50%
P(win by switching/notswitching from #2)=50%
P(win by switching/notswitching from #3)=50%
because the player doesn't KNOW the Right Door

4. Feb 5, 2005

### Hurkyl

Staff Emeritus
So what are the probabilities of each?

It would be easier to analyze if you always picked door #1, and varied which door the prize is behind.

5. Feb 6, 2005

### bomba923

But then the probability of winning by switching is still greater, even if Monty doesn't reveal a goat. Let's suppose he just asks you, "Are you sure with your selection??"
then
P(door#1 is right)=1/3
P(door#1 is wrong)=2/3
---so then,
P(win by staying with 1)=1/3
P(win by switching)=P(win with 2)+P(win with 3)=2/3
-------------------------
In that case, you will have greater chance to win by switching even
if Monty does not reveal a goat! with u still have three doors left.

That doesn't make sense!! (or does it....does it matter if Monty opens up a door with a goat inside, or if he just asks you "would you like to switch?" without opening any doors)

Edit: I don't think this counts for much :surprised

Last edited: Feb 6, 2005
6. Feb 6, 2005

### bomba923

Why is my approach #1 in the original post INCORRECT??
Where did it go wrong?

Last edited: Feb 6, 2005
7. Feb 6, 2005

### bomba923

""By switching, we win in 2 possible ways; by not switching, we win in 2 possible ways.""

Are u saying the probabilities distribute as 1/6, 1/6, 1/3, 1/3 ???

Last edited: Feb 6, 2005
8. Feb 6, 2005

### cepheid

Staff Emeritus
Why don't you tell us? You're the one who wrote it! :rofl:

This is totally wrong. Say you choose door #1. By adding the probabilities as you did above, you are assuming that you can win with EITHER door 2 or door 3, which is ridiculous, because there is is prize behind only one of them, not both of them. Here is how to do it:

It's true, there is a 2/3 chance that the prize lies behind *one of the two* remaining doors that you didn't pick (2 or 3). But if you're going to switch for that reason, and *Monty hasn't shown you which of those two has a goat behind it*, then WHICH ONE should you choose to switch to? Without that extra information, it's still 50-50 that the prize is behind door 2 or door 3, so your probability of winning is (1/2)*(2/3) = 1/3, which gives you no advantage over your first choice. The whole point of the Monty Hall problem is that Monty DOES give you that extra information, which "shifts" the 2/3 probability that the prize is behind one of the remaining doors onto just one of those doors. (That's the simplest way of thinking about it)

9. Feb 6, 2005

### bomba923

Oh yeah ...i can't really have 2/3 probability of switching and winning....but i'm still wondering, what's wrong with approach#1 in my first post here?

10. Feb 6, 2005

### cepheid

Staff Emeritus
Hmm...first I'll formulate an explanation the way I think it ought to be done, and you tell me if it helps you out. Your approach 1 is a bit convoluted (I think you're overanalyzing here!), so I'll try and get around to critiquing it in a little while.

Cepheid Approach
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Ok. So we have three closed doors. Two have goats behind them. The third has a brand new car! Monty asks you to pick a door. Standard assumptions about Monty's behaviour: He will open a door other than the one you picked, and one that has a goat behind it (obviously). If the door you originally picked happens to have a goat behind it, then he will show you the other goat. If the door you originally picked happens to have the car behind it, then he will show you one of the goats, and there is an equal likelihood of him showing you one goat or another.

So there are no more than three scenarios:

Scenario 1: The door you pick has goat #1 behind it. Monty opens another door and shows you goat #2 and asks whether you'd like to keep your door or switch to the remaining one. YOU SWITCH, YOU WIN.

Scenario 2: The door you pick has goat #2 behind it. Monty opens another door and shows you goat #1 and asks whether you'd like to keep your door or switch to the remaining one. YOU SWITCH, YOU WIN.

Scenario 3: The door you pick has the car behind it. Monty opens another door and shows you either goat #1 or goat # 2. He asks whether you'd like to keep your door or switch to the remaining one. YOU SWITCH, YOU LOSE.

Notice that in 2 out of the 3 scenarios, switching results in a win.
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Most commonly raised objection:
"But you're still making a choice between your original door and the third one. One has a goat, one has a car. So it's 50-50, isn't it?"

NOPE. Think about it this way. IF Monty had shown you a door with a goat behind it BEFORE you had made any selections, then it would be 50-50 between the remaining two. But he shows you a door with a goat behind it AFTER you made a selection. Since there is a greater likelihood (2/3) that you originally picked a goat door, there is a 2/3 chance that the remaining door has the car behind it, just as I showed.
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11. Feb 6, 2005

### cepheid

Staff Emeritus
I hope that clears things up...

12. Feb 7, 2005

### gnome

Your approach #1 did not correctly enumerate the possible outcomes. Here, the columns headed "prize" and "picked" list all the possibilities of the original choice. In your original approach, you were in effect doubling-up lines a, e and i.

$$\begin{array}{ccccccc} &\text{prize} &\text{picked} &\text{monty} &\text{switch} &\text{stay}\\ a &1 &1 &2\text{or}3 &L &W\\ b &1 &2 &3 &W &L\\ c &1 &3 &2 &W &L\\ d &2 &1 &3 &W &L\\ e &2 &2 &1\text{or}3 &L &W\\ f &2 &3 &1 &W &L\\ g &3 &1 &2 &W &L\\ h &3 &2 &1 &W &L\\ i &3 &3 &1\text{or}2 &L &W \end{array}$$

13. Feb 8, 2005

### cepheid

Staff Emeritus
Yeah, that maybe be clearer than my commentary. BUT (just to point out) you can do the problem by assuming the prize is behind one of the doors (say, door number one, for example), and considering the outcome given the three possible selections. In other words, you need only one group of three (out of those nine) outcomes. Either abc, def, or ghi, would suffice. Makes no difference, for your method shows that switching gives you a 6/9 chance of winning, = 2/3, same answer you'd get by assuming the prize is behind one of the doors (since it doesn't matter which one you assume it is behind).

I'm not sure what you meant by your "doubling up lines" remark. In fact, in bomba's approach 1, all he did was consider your a, b, and c. Why he did not arrive at the correct answer using that method is...perplexing (as I noted in red).

Last edited: Feb 8, 2005
14. Feb 8, 2005

### chronon

This is crucial to the problem. If you chose door 1 and he opened one of doors 2 and 3 at random, and there was a goat behind it, then the probability of the prize being behind door 1 would be 50%.

The trouble is, if you have to bring in Monty's knowledge, then you should also ask what his motivation is. If you make the (not unreasonable) assumptions

1: Monty wants to keep the prize from you.
2: Monty chooses what he does in the game show.

Then when you have chosen a door, if it doesn't have the prize behind it then you would assume he would open it and say 'sorry, you've lost'. Since he does something else, you have to assume that its very likely that you have in fact chosen the correct door.

This is why people have such trouble with the Monty Hall problem, we are much better at looking out for someone trying to trick us than at calculating probabilities of made up situations.

There is another version of the problem which is more intuitive. Suppose you get to choose 2 doors at the start, so you have a 2/3 chance of getting the prize. If Monty then opens one of your doors to reveal a goat, then it doesn't change your chance of winning - you already knew that at least one of your doors had a goat behind it.