Maths challenge No. 3

[Duckers]

kranky,

Monty is actually giving you 2 chances:

The contestant then chooses a door. Monty then opens a different door, revealing a goat.

He then asks the contestant whether they want to change, or whether they want to stick

You said:
The flaw in the logic is that - once you pick a door - your odds do NOT change just because Monty shows you a goat behind a door you didn't pick. The odds would change ONLY if you haven't picked yet.

Again, Monty is asking you whether you want to change or stick with the door you picked.

 

[Xede]

Assume the car is behind door #1. It's equally likely that it's behind any door, so all three cases are identical, but here we assume it's in door #1.

There's a tree shown below. The first level has three branches. Those three branches correspond to which door you pick initially. The next set of three correspond to the door you pick after he opens one.

So, if you use a switching strategy (and assuming the car's behind door #1):

Case 1: You pick door 1 to begin with. He opens either door 2 or door 3. You switch from 1 (the car) to the other closed door, a goat. You lose.

Case 2: You pick door 2 to begin with. He can't open door 1 to reveal the car, so he opens door 3 to reveal a goat. You switch to door 1. You win.

Case 3: You pick door 3 to begin with. He can't open door 1 to reveal the car, so he opens door 2 to reveal a goat. You switch to door 1. You win.

You win in 2 cases out of 3 using this strategy. 2/3 chance of winning. If you use a "never switch" strategy, your odds are 1/3, the same as your random chances at the beginning of the game, because you're ignoring whatever happens after your original pick.

..........|--[1] Car is here.
..........|
|----[1]--|--[either 2 or 3 is still closed, LOSE]
|.........|
|.........|--[either 2 or 3 opened to reveal a goat]
|
|
|.........|--[1, WIN] Car is here.
|.........|
|----[2]--|--[2]
|.........|
|.........|--[3 opened to reveal a goat]
|
|
|.........|--[1, WIN] Car is here.
|.........|
|----[3]--|--[2 opened to reveal a goat]
..........|
..........|--[3]

 

[allan120]

Viper:
I stated earlier that it didn't matter, but it seemed to matter to Demon-Xanth, so I stuck with trying to explain it using his arguments. Plus, he wanted someone to disprove his logic.

I argued the "doesn't make a difference" point for about ten minutes with an ex-gf, until she told me it in terms of a million goats and suddenly the fog was lifted.

 

[Duckers]

Xede;

You forgot Case 4:

Case 4: You pick door 1 to begin with. He opens either door 2 or door 3. You stick with door 1. You Win.

 

[Moonbeam]

The evidence for switching is persuasive.

 

[Xede]

No, Duckers, I said "So, if you use a switching strategy..."

If you're using a switching strategy, you cannot pick door 1 to begin with, NOT SWITCH, and win. That's not allowed, by definition of a switching strategy. There is no case 4.

You could use a mixed strategy where you switch sometimes and don't switch sometimes... but that just screws everything to hell. And your odds of winning would be worse than 2/3.

 

[silent tone]

the million door and lots of goats is a good demonstration. What I've successfully convinced someone of this is: The player will usually pick a goat on the first try, (2/3). So, usually monty has to remove the only other goat(2/3 of the time). So the car will also usually be behind the third door(2/3 of the time).

 

[deepcover]

The Goat is there, or The Car is there. That means 1 of 2 endings to our smelly story. 50/50 chance.

 

[whooosh]

^

 

[Argo]

It doesn't matter if you switch or not. This is coming from a guy who got A+ in Probability and Statistics course.


Here's the deal:


Initially your chances are 1/3. When monty opens one of the doors your chances immediately jump to 1/2. It doesn't matter if you switch the door or not. It'll still be 1/2. I don't believe something as simple as that caused so much argument.

 

[Argo]

Try thinking about it this way:


After monty opens one of the doors, you choose a door out of two remaining ones. Don't think of it in terms of switching but rather in terms of choosing. That way it's obvious that your chances are 1/2 no matter what.

 

[Moonbeam]

Hay, I ain't never took nutten and that's what I said, the first time.

 

[AvesPKS]

Ok...I believe everybody sees that your odds of winning the car before Monty shows you the goat are 1:3, and after, 1:2. The sticking point seems to be whether or not you should switch. The fact is, it's all chance. Your initial condition in the first tier of logic is that, behind 2 of the three doors, there are goats, and behind the remaining door, there is the car. Now, at this point, the odds are 2:3 that you will pick a goat. In the second tier of logic, your initial condition states that Monty has eliminated one of the two remaining choices as being wrong. Now, since it doesn't matter which door you initially picked, as Monty will always eliminate one choice as being wrong, your odds at this point are 1:2. No matter which one you intially picked. Because, no matter which one you picked initially, you know that, in the second tier of logic, your odds of picking the correct door are always 1:2. If you base your second tier logic decision on your ability to be lucky (or unlucky) in the first tier of logic, you are making a serious misjudgement.

 

[kranky]

Argo, I'm glad you got an A+ in Prob. and Statistics, but you are wrong on this one. Believe me, if I wasn't sure, I wouldn't be trying so hard to show you guys the answer!

Argo said: <<Initially your chances are 1/3. When monty opens one of the doors your chances immediately jump to 1/2. It doesn't matter if you switch the door or not. It'll still be 1/2. >>

Since you agree that the initial chances are 1/3, let's stick with that approach. Of the two doors NOT chosen, we know for sure at least one of them hides a goat, right? We already know there's only one car.

So you are saying that your odds are 1/3 when you don't get to see the goat that you already know is there, but 1/2 if you see it. That is illogical. Your odds DO NOT change just because you saw it.

If you pick first you have a 1/3 chance. If I take the other TWO doors, I have a 2/3 chance. Now if I show you one of the two doors I have has a goat behind it, that doesn't change my 2/3 odds, because we all knew there was a goat behind one of them. As I mentioned above, switching gives you the equivalent of taking TWO doors instead of one. That's why you are better off switching.

Argo said: << Don't think of it in terms of switching but rather in terms of choosing. >>

Perhaps that's the source of the confustion - that's NOT the same problem as the one being discussed. I agree that you have a 1/2 chance if you FIRST choose AFTER you are shown one of the goats.

 

[callspread]

Ha ha ha - it always amuses me when I see people trying to argue that switching is better. Think of it this way - once you make your original choice, one of the goats is revealed, and that door is removed from the stage (I know that this has not been said to happen, but it helps the visualisation). You are now left with 2 doors, each of which has a 50:50 chance of having the car behind it.
There is no way that the chance can be anything other than 50:50 with only 2 doors involved, so switching cannot make any difference to the chance of choosing the car.
The chance of the original choice being correct (1 in 3) is no longer relevant once one of the doors has been eliminated. As soon as the goat is revealed, the number of doors is effectively decreased, so the chance goes from 1:3 to 1:2. The doors which have not been eliminated do not retain their original 1:3 chance since there are no longer 3 doors in the equation.

 

[Josh]

Viper is smart. Listen to him!

 

[callspread]

Viper said:
'Many people here are overlooking the fact that revealing the contents of a door does not change the original probability that you chose the correct door. If you were to re-distribute the doors' contents randomly &amp; then chose a door again, you're correct... The odds would be 50/50. But because the contents have remained constant the whole time, your odds are greatly increased by switching.'

This logic implies that if all the doors are revealed expect yours, leaving only one door, the odds for your door will still be 999999 to 1. This is obviously incorrect, and should help you to understand the problem. The confusion arises from you realising that your original odds never change, but not realising that a new set of odds in applied every time a door is revealed.

 

[hatboy]

I know that it's really hard to see that switching is better (I didn't see it for a long time myself). Here's an interesting way to think about it. Imagine that you pick a door, and then you are shown the contents of the other two doors, but you are never given the chance to switch. By the arguments some people have made, after you were shown one door, your chances would be 1 in 2, even though you didn't switch. Then, after you are shown the second door you didn't pick, by this same logic, your chances of winning would have to be 100%, since there's only one door left. Obviously, this is wrong. The thing you're all missing is that if you don't switch, being shown a goat MAKES NO DIFFERENCE. The odds don't change after you are shown something. If you just pick 1 door out of 3, no matter what happens afterword, your odds of picking the right door have got to be 1 in 3. The thing that's maybe even harder to see is that if you switch, the odds of winning are 2 in 3. Why is this? It's because basically, you're given the chance to have the contents of two doors. I hope that you can now see that the odds when you don't switch are 1 in 3. But if you do switch, it's analagous to picking both the door you were shown and the other remaining door. Another way of thinking about it: since all probabilities must add up to 1, and the probability that you're right without switching is 1 in 3, it just has to be 2 in 3.

For those of you who want to call it &quot;choosing&quot; rather than &quot;switching&quot;: I'll grant you that if you randomly choose one remaining door after being shown a goat, you're odds will be 1 in 2. But the reason for this is that half the time, you would switch, making your odds 2 in 3, and the other half of the time you wouldn't, making your odds 1 in 3. Obviously, the average of 1/3 and 2/3 is 1/2.

I really hope this helps.

 

[kranky]

callspread, we got Moonbeam to see the light, so let's get you converted too.

Let's say there's 100 doors and one car (the rest have goats). I'll let you have your choice: you can pick ONE door, or you can pick 99 doors.

If you pick ONE door, I'll show you the goats behind 98 of the other ones and leave one unopened. You can switch to the remaining door if you want.

If you pick 99 doors, I'll show you the goats behind 98 of the doors you picked. You can switch to the remaining door if you want.

By your logic you have a 50/50 chance of getting the car in either case. You know I'm not really revealing anything by showing you 98 goats, since you knew they were there all the time.

Do you really think you have a 50/50 chance whether you pick one door, or 99 doors?

 

[callspread]

Kranky,
The chance of choosing the correct door is represented by the equation:
chance = number of doors chosen/number of doors to choose from.

Since you are allowed to choose only one door the numerator is fixed at one. The chance is therefore 1/number of doors to choose from. Each time one door is eliminated by revealing the goat behind it the number of doors to choose from is reduced by one, causing the chance to tend towards 1.

By showing me a goat you are effectively reducing the number of doors available, thereby reducing the denominator in the equation. When the number of doors reaches 2 the chance is 1/2. This is true whether you start with 100 doors or with 2. At every stage, the starting number of doors is not relevant, only the number remaining is relevant.

 

[Xede]

LOL, this thread is getting pretty interesting. It's becoming a little bit of a crusade. How many people will be converted to switching??

Let me say this: I will bet $1000 vs your $10 that 1) a pure switching strategy has higher probability of winning than a pure stay strategy and 2) the correct odds of winning with a pure switching strategy are 2/3, not 1/2.

I can't explain the correctness of the switching strategy any better than I did already, so I won't clutter up the thread with more explanations.

 

[cxim]

Xede,

I'll take that bet...

Who is going to hold the money ?

I don't want to hurt you too bad so just make it $100. to $10.

even let you do the test yourself with an unbiased judge to observe &amp; verify your results.

 

[Argo]

Hehe. I get a better idea. How about I write a program, that will simulate this situation. It'll prove that chances are the same. Sounds good? I'll do that after I get back from work. I'll also post the source code, so that nobody can acuse me of cheating.

 

[kranky]

callspread, you didn't respond to my post, so I'll go with yours.

<< The chance of choosing the correct door is represented by the equation: chance = number of doors chosen/number of doors to choose from. >>

Agreed.

<< Since you are allowed to choose only one door the numerator is fixed at one. The chance is therefore 1/number of doors to choose from. >>

Agreed.

<< Each time one door is eliminated by revealing the goat behind it the number of doors to choose from is reduced by one, causing the chance to tend towards 1. >>

No, because you chose already. Your odds were fixed based on the number of doors hidden when you chose your door. You said it yourself in the first quote above.

<< By showing me a goat you are effectively reducing the number of doors available, thereby reducing the denominator in the equation. >>

Only true if you haven't chosen yet. But you did. You are saying that your odds of success improve just because I am showing you a goat that you already knew was there.

<< When the number of doors reaches 2 the chance is 1/2. This is true whether you start with 100 doors or with 2. At every stage, the starting number of doors is not relevant, only the number remaining is relevant. >>

Again, only true if you haven't chosen yet. But you did. Use your own explanation against the example I just gave where you can pick either 1 door or 99 doors. I'll reiterate: using your logic, whether you start with one door or 99 doors, you are saying the odds are 50/50
in both cases. Clearly, choosing 99 doors is better than choosing just one. They are not equal.

You pick one door out of 100. Are there (at least) 98 goats behind the other 99 doors? Yes, no question. And if I show you 98 goats - goats you already know are there - how does that make your initial choice a 50/50 chance?

Xede, way to go out on a limb!

 

[br0wn]

Some ppls just can't see the TRUTH, even after you show them
the way

Don't blame them, most of us have our own BLINDSPOT. Sometimes
it also takes me HOURS to understand a simple fact, and when
I finally get it I kick myself for being that stupid .

For those who think that switching doesn't give you 2/3 winning chance
of the car, don't argue anymore....think carefully and read
all the messages from the beginning....there are so many evidences,
proofs and even illustrations for you.
If you still can't get it, oh well....maybe SOMEDAY you will understand it.
Ask your math teacher, ask your math professors, and hopefully
they will enlighten you.

I won't try to convince you anymore, it will be the same as trying
to convince blind person that roses are red (well at least most roses
are red).

edit : Argo, yes write that program, and it will prove to you that
the chance of switching is 2/3 for winning the car and HOPEFULLY
it will open up your mind.
Hatboy and I have written the program and verify this.
Even Hatboy has posted his java program, look somewhere in this thread.