A fascinating mathematical problem

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Originally posted by Matthew Robinson:
So, answer me this. If you pick door A and I pick door B *and* door C I win 2/3 of the time yes?

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But you can't. One door will be eliminated. So you can pick door B OR C, but never B AND C.

Let's look at it this way. A, B and C each have a probability of 1/3 in the first round. That means A and B have an equal probability of hiding the car.

The host then takes away option C. I can see absolutely no mechanism why option B should now be assigned a greater probability than A, when they had an equal probability in the first round.

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PS I think I might write a program to demonstrate the right answer.

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Do that, I'm looking forward to it.

Jeremy -- Most people assume from the question that the host knows what is behind the door as it's implicit in the question that the host always opens a door with a goat behind it. Besides it doesn't say "the host opens a door and in this case it happens to reveal a goat" so you can argue that it doesn't.

But I agree if the host doesn't know where the goat is then there is no advantage in changing.

Matthew

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Originally posted by Minky:In a game show there are three doors. Behind two of them are goats. Behind the third is a car. You are a contestant trying to win the car. You choose a door and the game show host opens one of the two remaining doors to reveal a goat. Which door is most likely to have the car behind it (should you stick or switch) ?

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Just to check if we're all singing from the same sheet: this is the set-up I was commenting on. The rules are indeed that one door with a goat behind it will be eliminated.

Yet another way of looking at the problem: Does it matter at all if one picks a door in the first round? If we do away with the first round and let the host eliminate a door from the outset, we have two choices (door A and B) with a 50/50 probability of being correct.

With the set-up described above, there is really no difference. Instead of denominating the choice of door A as door A, we give it the moniker of sticking with our choice from round 1. Instead of saying we are choosing door B, we say we switch. But in reality, it is just the description of the process that changes, not the process itself.

I really think that door C is just there to confuse us.

PS I think I might write a program to demonstrate the right answer.

did it a couple of years ago...you can see the results earlier in this post....it confirms that if you stick you win 1/3 of the time....if you change you win 2/3 of the time. (approx for the benefit of PR)

Omer had a very helpful illustration above with his 1,000,000 doors, again posted in the Brain teaser thread a couple of years back. ie if you pick ONE door and stick with it, your chances of picking the car door are a million to one. Not very good. This means the chances of the car being behind one of the other 999,999 doors is .999,999. Almost a dead cert. All YOU have to do is FORCE the Host to open ALL these 999,999 doors. It doesn't matter whether he opens them all at once!! or one at a time!! Surely we can all agree on this !!!!

Assume the host knows where the car is - he is going to open 999,998 "goat" doors and you know this. He is simply going to keep the car-door until last.....it creates a bit of tension (or confusion or boredom!). "Change your mind" on the last door and your chances of winning are .999,999.

Of course you didn't REALLY change your mind, you simply chose 999,999 doors rather than ONE and you forced the Host to open all 999,999.


Same principle applies to 3 doors.

Cheers

Don

PS now assume the Host didn't know where the car was. Chances of him opening 999,998 doors without revealing a car are pretty slim.........

[This message was edited by Don Atkinson on Mon 19 April 2004 at 21:59.]

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To put it another way, when the rules state in advance that one of the three doors will be eliminated by the host, and that this door does not have the car behind it, the real choice only ever is between two doors.

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Sums it up very nicely.

I actually came to the forum because my poor little brain wasn't working well enough to attempt to work on my OU maths courses. So why on Earth did I come to this thread?

I've tried to see how the assumption that the host deliberately chooses a door with a goat might make a difference but I can't, I'm afraid. Maybe it will all make sense in the morning...

Oh! Wait a minute... The goats are indistinguishable in either case but when the host deliberately chooses a door with a goat behind it the doors cease to be indistinguishable (which they were for the problem as originally stated). In this case, we get:

contestant's door..c... g... g
red door (say)......g... c... g
blue door (say).....g... g... c

This time, unlike in my answer to the question as stated, the middle column is also possible because the host chooses the blue door instead of the red door.

So in this case the contestant should switch doors.

[Or I am even more confused than I think]

[This message was edited by JeremyD on Mon 19 April 2004 at 22:12.]

Omer recalls the program I wrote a couple of years back. I think I sent him a copy to verify its authenticity.

I might have kept it somewhere.

Cheers

Don

JerremyD, Markus, jayd, Rasher,

does the 1,000,000 door version help?

Cheers

Don

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Originally posted by Don Atkinson:
JerremyD, Markus, jayd, Rasher,

does the 1,000,000 door version help?

Cheers

Don

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Since this post I've revised my previous post, now claiming that the contestant should switch if the host deliberately chooses a door with a goat. However, the question does not appear to say this so I still think "it makes no difference" is the answer to the question as stated.

NB If I wake up tomorrow morning and realise that what I've written is rubbish, I reserve the right to claim I was simply stirring.

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Originally posted by Don Atkinson:
JerremyD, Markus, jayd, Rasher,

does the 1,000,000 door version help?

Cheers

Don

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Not really, because the problem as originally stated is the limiting case of only two options (the minimum necessary to offer a choice).

I wrote a program which does 10,000,000 trials. Here's the output from 3 runs:

Stick Wins=3333339 (0.3333339%), Switch Wins=6666661 (0.6666661%)
Picked A=3333206, B=3332522, C=3334272

Stick Wins=3329695 (0.3329695%), Switch Wins=6670305 (0.6670305%)
Picked A=3333878, B=3335417, C=3330705

Stick Wins=3334789 (0.3334789%), Switch Wins=6665211 (0.6665211%)
Picked A=3333184, B=3330970, C=3335846

You'll find the code here.

Note that a) I made it excessively literal for the sake of readability and b) it worked first time which is so unlikely it means I made a mistake somewhere.

Matthew

Matthew, Don, and other code warriors:

How does the code change if the problem is goat-badger-car? Just curious. I've no doubt that your simulations accurately model the problem as you see it. I still maintain that the flaw is in the definition.

Can anybody explain to me how lines 1 and 3 in John Channing's explanation differ (see p.2 of this thread)? I still maintain that, if you can treat the goat separately in lines 1 and 3, there must necessarily be two versions of line 2 (which puts win/lose at 50/50). And if you can't, well then... you're still at 50/50.

Either goat = goat always, or never. Half the time is cheating.

PR said "he then wages the mathematician that on 50 flicks that he will win money if he can get more than 19 tails or heads. The mathematician believing in 50/50 will jump at the idea, but ultimately lose"

I don't understand this at all. Are you saying that the Financier would offer me a bet at evens that I will win if I can throw at least 20 heads in 50 coin flips?

If so I take the bet every time.

Matthew

jayd,

Matthew, Don, and other code warriors:

ok, my simulation wasn't written in any basic (not Basic) computer code, it was a simple Excel spread-sheet. I don't consider myself to be a computer nerd.

As with any aspect of life, including computer programming, its vital to be precise as to the actual question.

The debate at present is as much about the question as it is about probability. In the various published texts about the Monty Hall Show, the question, or the rules of the Show, seems to be slightly vague.

My UNDERSTANDING is that Monty Hall always knew where the car and the goats were and, he always opened the first door to reveal a goat, never a car.

Under these circumstances, "changing you mind" is your best bet. The run of shows ended once the contestants in later shows began to realise this.

The example with 1,000,000 doors is the same, in principle, to the three doors, and for some people it does help them to appreciate what is actually happening. It doesn't help everybody and obviously you are one of these. Perhaps you could persaude your partner to set up (say) 30 dummy shows and see what the outcome is if you "stick"; then run a further 30 and "change you mind". My spread-sheet did this for me (and also looked at what happened if I "randomly" decided whether to stick or change my mind....my chances then were 50/50).Creating a spread-sheet also saved Mrs D from having to buy two goats for these "dummy" experiments.

Doesn't matter whether you use a goat and a badger or two goats. You want to win the CAR. In the 1,000,000 door game you could put ANY useless articles behind 999,999 of the doors. Pick ANY 999,999 doors out of the 1,000,000 and the chances are that ONE of these 999,999 has the car. All that Monty Hall is going to do is avoid openning the one with car until the end.

Simple

Cheers

Don

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Doesn't matter whether you use a goat and a badger or two goats.

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Well, there's something we agree on:

Pick | Reveal | Action | Outcome
goat | badger | switch | switch wins
goat | badger | stick | stick loses
badger | goat | switch | switch wins
badger | goat | stick | stick loses
car | badger | switch | switch loses
car | badger | stick | stick wins
car | goat | switch | switch loses
car | goat | stick | stick wins

Did I miss any outcomes? (Bearing in mind that we've already ruled out revealing the car as an intermediate step, thus eliminating half the "pick goat first" and half the "pick badger first" outcomes).

Simple, indeed.

JerremyD, Markus, jayd, Rasher,

I agree that the question HAS to be clearly defined. I've tried to do that in my previous post, despite the various published texts being sometimes vague.

Is MY explanation of the question clear?

Cheers

Don

Can anybody explain to me how lines 1 and 3 in John Channing's explanation differ

IF you lable the doors A,B,C(simply for refernce, and the fact that they are there)

In line 1 the car is behind Door C
In line 3 the car is behind Door B

Does this help?

Cheers

Don

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Originally posted by Don Atkinson:
_Can anybody explain to me how lines 1 and 3 in John Channing's explanation differ _

IF you lable the doors A,B,C(simply for refernce, and the fact that they are there)

In line 1 the car is behind Door C
In line 3 the car is behind Door B

Does this help?

Cheers

Don

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Yes. I was afraid you would say that.

Position isn't a factor; the order in which the items occupy the stage has no bearing on the possible outcomes. Ergo, lines 1 and 3 are identical. And we're back to even odds.

Please, let it stop!!!

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Originally posted by jayd:
Position isn't a factor; the order in which the items occupy the stage has no bearing on the possible outcomes. Ergo, lines 1 and 3 are identical. And we're back to even odds.

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If you want to try and argue that they are the same then you might as well say that CGG = GCG = GGC because they all have 2 goats and therefore they all cancel and you either have 0% or 100% chance of picking the car.

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Originally posted by John Sheridan:
Please, let it stop!!!

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Originally posted by jayd:
Position isn't a factor; the order in which the items occupy the stage has no bearing on the possible outcomes. Ergo, lines 1 and 3 are identical. And we're back to even odds.

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If you want to try and argue that they are the same then you might as well say that CGG = GCG = GGC because they all have 2 goats and therefore they all cancel and you either have 0% or 100% chance of picking the car.

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Ha. Whole lotta things wrong with that... but you were only joking.

(right?)

Still waiting to hear which outcomes I've omitted from my scenario. What's missing that makes switching such an obvious choice?
Bueller? Bueller? Anyone?

Or, for the goat=goat crowd, the number of outcomes is {edit: typo} halved, but their distribution is unchanged:

Pick | Reveal | Action | Outcome
goat | goat | switch | switch wins
goat | goat | stick | stick loses
car | goat | switch | switch loses
car | goat | stick | stick wins

And, for nondegenerate goats, it's identical to goat/badger:

Pick | Reveal | Action | Outcome
goat1 | goat2 | switch | switch wins
goat1 | goat2 | stick | stick loses
goat2 | goat1 | switch | switch wins
goat2 | goat1 | stick | stick loses
car | goat1 | switch | switch loses
car | goat1 | stick | stick wins
car | goat2 | switch | switch loses
car | goat2 | stick | stick wins

[This message was edited by jayd on Tue 20 April 2004 at 0:59.]

I think I am finally converted to the 2/3 probability of switching, but only after the info that the host knowingly opens a goat door, rather than always opening door 1.

Matthew - how does your program work? does it pick a random number for the car to be behind and then pick a random number that is your choice and then randomly open one of the other 2 doors?

I really don't understand the logic some people are using here.

Why does the host always opening a door with a goat affect the outcome?

As far as I can see switching or sticking does not make a difference.

Initially each door has 1 in 3 chance, the host opens door with goat, that leaves one of two doors with a car so its 1 in 2 chance! Where am I going wrong here? If I choose to switch I dont gain anything. Please explain - without another problem or some program.

Now I've finally come around, the important point is that the host knowingly opens a door with a goat behind it.

If you know the host will open a door with a goat in it then by picking door 3 you are actually picking doors 1 and 2. After you selection there are 3 things that can happen:

1. If the car is behind door 1, the host will open no 2 and you will switch and win the car.
2. If the car is behind door number 2 then the host will open door number 1 and you wil switch and win.
3. If the car is behind door number 3 then the host will open 1 or 2 and you will switch and lose.

If however the host opens a random door first which could be the car then switching would have no effect.

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Originally posted by Two-Sheds:
Now I've finally come around, the important point is that the host knowingly opens a door with a goat behind it.

If you know the host will open a door with a goat in it then by picking door 3 you are actually picking doors 1 and 2.

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Why?

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After you selection there are 3 things that can happen:

1. If the car is behind door 1, the host will open no 2 and you will switch and win the car.
2. If the car is behind door number 2 then the host will open door number 1 and you wil switch and win.
3. If the car is behind door number 3 then the host will open 1 or 2 and you will switch and lose.

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No 3 above includes two possibilities:
3. If the car is behind door number 3 then the host will open 1 and you will switch and lose.
4. If the car is behind door number 3 then the host will open 2 and you will switch and lose.

Whether you switch or stick makes no difference.