A fascinating mathematical problem
Posted by: Minky on 18 April 2004
I've just finished reading "The curious incident of the dog in the night-time" by Mark Haddon. The book has a brief passage on something called "The Monty Hall" problem. Marilyn vos Savant, the smartest person in the world by virtue of her Guinness record IQ, was asked the following question (my words) :
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) ?
If you have never encountered this problem before, have a think about it.
When Marilyn vos Savant gave her answer she was quite soundly chastised by some very heavy-hitting brainiacs, and this is at least as fascinating as the problem itself, because (as I have also found) this baby seems to have an almost supernatural ability to turn otherwise mild-mannered people into red-eyed monsters
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) ?
If you have never encountered this problem before, have a think about it.
When Marilyn vos Savant gave her answer she was quite soundly chastised by some very heavy-hitting brainiacs, and this is at least as fascinating as the problem itself, because (as I have also found) this baby seems to have an almost supernatural ability to turn otherwise mild-mannered people into red-eyed monsters
Posted on: 19 April 2004 by jayd
quote:
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.
Sums it up very nicely.
Posted on: 19 April 2004 by JeremyD
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.]
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.]
Posted on: 19 April 2004 by Don Atkinson
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
I might have kept it somewhere.
Cheers
Don
Posted on: 19 April 2004 by Don Atkinson
JerremyD, Markus, jayd, Rasher,
does the 1,000,000 door version help?
Cheers
Don
does the 1,000,000 door version help?
Cheers
Don
Posted on: 19 April 2004 by Former member
Hehe Don, I just reread the Brain Teaser thread and saw I was very hard to convince that it's 1/3 and 2/3, I was also sure it's 50%...
I don't have a copy of that spreadsheet, I can only testify it was neat
Omer.
I don't have a copy of that spreadsheet, I can only testify it was neat
Omer.
Posted on: 19 April 2004 by Former member
Rasher
"probability changes as the situation progresses."
Yes!
A good example is in flicking a coin for head and tails. The finite, almost Newtonian law is 50/50, but the universe is asymmetric and varable as the situation progresses. So you start with 50 flicks, and end up with either 25/25 (fluke) or something like 22/28, then again 23/27, then again 26/24, then again 21/29, and maybe if you are lucky you get 25/25.
Now, a financier waging a bet with a mathematician will calculate the element of volatility, which lets say is an approximate standard deviation of 4 points (he may even use iterations as with Equity Options), and then add 2 for costs and profit. So his margin is 6 points. 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 his socks, because the financier has allowed for a margin of error.
As the number of flicks increase from 50 to 100, then 1000, then 10000, 100000, and finally 1000000, the financier can reduce his margin’s as the volatility smoothens (situation progresses) with the greater number of flicks. But ‘never the twain shall meet’ at 50/50 odds again, without a miracle or act of God.
Any creationists out there?
"probability changes as the situation progresses."
Yes!
A good example is in flicking a coin for head and tails. The finite, almost Newtonian law is 50/50, but the universe is asymmetric and varable as the situation progresses. So you start with 50 flicks, and end up with either 25/25 (fluke) or something like 22/28, then again 23/27, then again 26/24, then again 21/29, and maybe if you are lucky you get 25/25.
Now, a financier waging a bet with a mathematician will calculate the element of volatility, which lets say is an approximate standard deviation of 4 points (he may even use iterations as with Equity Options), and then add 2 for costs and profit. So his margin is 6 points. 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 his socks, because the financier has allowed for a margin of error.
As the number of flicks increase from 50 to 100, then 1000, then 10000, 100000, and finally 1000000, the financier can reduce his margin’s as the volatility smoothens (situation progresses) with the greater number of flicks. But ‘never the twain shall meet’ at 50/50 odds again, without a miracle or act of God.
Any creationists out there?
Posted on: 19 April 2004 by JeremyD
quote: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.
Originally posted by Don Atkinson:
JerremyD, Markus, jayd, Rasher,
does the 1,000,000 door version help?
Cheers
Don
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.
Posted on: 19 April 2004 by jayd
quote:
Originally posted by Don Atkinson:
JerremyD, Markus, jayd, Rasher,
does the 1,000,000 door version help?
Cheers
Don
Not really, because the problem as originally stated is the limiting case of only two options (the minimum necessary to offer a choice).
Posted on: 19 April 2004 by matthewr
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
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
Posted on: 19 April 2004 by Former member
I keep wondering off...
If there are 3 doors and one door is opened by the host to reveal a pig (sorry) your original odds have been increased to 2/3, but your position has moved on to a 50/50 chance to keep things simple.
If there are 3 doors and one door is opened by the host to reveal a pig (sorry) your original odds have been increased to 2/3, but your position has moved on to a 50/50 chance to keep things simple.
Posted on: 19 April 2004 by jayd
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.
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.
Posted on: 19 April 2004 by matthewr
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
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
Posted on: 19 April 2004 by Don Atkinson
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
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
Posted on: 19 April 2004 by jayd
quote:
Doesn't matter whether you use a goat and a badger or two goats.
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.
Posted on: 19 April 2004 by Don Atkinson
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
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
Posted on: 19 April 2004 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
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
Posted on: 19 April 2004 by jayd
quote:
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
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.
Posted on: 19 April 2004 by John Sheridan
Please, let it stop!!!
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.
quote:
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.
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.
Posted on: 19 April 2004 by jayd
quote:
Originally posted by John Sheridan:
Please, let it stop!!!quote:
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.
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.
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?
Posted on: 19 April 2004 by jayd
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.]
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.]
Posted on: 19 April 2004 by Two-Sheds
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?
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?
Posted on: 19 April 2004 by Haroon
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.
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.
Posted on: 19 April 2004 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. 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.
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.
Posted on: 19 April 2004 by Haroon
quote:
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.
Why?
quote:
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.
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.
Posted on: 19 April 2004 by Two-Sheds
quote:
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.
Why?
I am also assuming that you know the host will open a door with a goat behind it.
The 3 possible combinations to start with are:
goat, goat, car
goat, car, goat
car, goat, goat
I assume you agree that you have a 2/3 chance of picking a goat.
Now then the host will open one of the other doors and reveal a goat. You know the host has picked a goat on purpose. If you are currently on a goat and switch then you will definitely win (since the host has eliminate the other one) and you have a 2/3 chance of being on a goat from the prior to the door being opened so thus by switching you now have a 2/3 chance of being on a car.