Turn on thread page Beta
    • Thread Starter
    Offline

    1
    ReputationRep:
    I read "The Curious Incident of the Dog in the Night Time" yesterday, and it was super good. But sadly didn't understand a bit of it - the Monty Hall problem.

    You are on a game show on television. On this game show the idea is to win a car as a prize. The game show host shows you three doors. He says that there is a car behind one of the doors and there are goats behind the other two doors. He asks you to pick a door. You pick a door but the door is not opened. Then the game show host opens one of the doors you didn't pick to show a goat (because he knows what is behind the doors). Then he says that you have one final chance to change your mind before the doors are opened and you get a car or a goat. So he asks you if you want to change your mind and pick the other unopened door instead. What should you do?

    This was published in a magazine and apparantly you should always change and pick the final door because there is a 2/3 chance there is a car behind it. Lots of scientists and mathematicians wrote in to say this is wrong. It then gives you some maths to prove it, but I am not convinced. Surely, knowing where one of the goats is makes the odds 50:50 because your choice is narrowed to two, and one has a car, one has a goat, and the knowledge of where the other goat is isn't given? Also, because you have just the same knowledge about the door you chose, and the one you can choose to switch to, this seems to suggest that had you chosen the other one, the odds would be in favour for you to switch to the one that you might have originally chosen... if that makes sense.

    Someone cleverer than me please explain it to me. I am posting this in the Oxbridge forum because I read a thread about Oxbridge logic a week or so ago and there seemed to be plenty of people who knew a lot about it.

    Emma
    xxx
    Offline

    1
    ReputationRep:
    http://www.io.com/~kmellis/monty.html explains it pretty well... It's quite a well known problem.

    Somewhere there's an applet which proves it but I forget where that is.
    • Thread Starter
    Offline

    1
    ReputationRep:
    I'm still not convinced. I can see where this is coming from, but I think I'm missing something important.

    The probabilities of your initial choice being correct, and the remaining choices have to sum to equal one. Therefore, the probability of the remaining choices have to sum to equal one minus the probability of your initial choice. In this case (with three doors), they have to sum to equal 2/3.
    But, after opening one door, there aren't really three doors involved! Surely the opening of one of the goat doors changes the probability or your initial choice, as well as the probability of the remaining one?
    Offline

    14
    ReputationRep:
    (Original post by covered farm wa)
    I'm still not convinced. I can see where this is coming from, but I think I'm missing something important.



    But, after opening one door, there aren't really three doors involved! Surely the opening of one of the goat doors changes the probability or your initial choice, as well as the probability of the remaining one?

    the logic seems flawed by the fact that monty knows he is opening the door without the prize. If he opened a random one then it would be correct to change if it showed a goat as the other door does adopt the 2/3 probability but as he is opening one he knows not to be a prize then the odds aree equal at 1/2 per door and you have no reason to change.

    Am I correct? if not where have I gone wrong?

    Musicboy
    • Thread Starter
    Offline

    1
    ReputationRep:
    I agree entirely. It seems that once the door is opened, the argument says that the odds of your original choice being right are exactly the same. But to make it add up to one, the other choice becomes more likely. But surely once the door is opened, you should re-evaluate the odds for BOTH doors...?
    Offline

    14
    ReputationRep:
    you should change, and ive got the reason in my head, but whenever i try to convert it to words i lose the reason.

    bah
    Offline

    2
    ReputationRep:
    I think I see the way through this. The argument is based on the fact that the probability of the prize being behind any two given doors is 2/3, and that if you know one of them does not hide the prize, then the 2/3 probability still applies, but only to one door. Seeing why this cannot be right is easier if you consider the problem without picking a door first. Say you have the three doors, each with a 1/3 probability of the prize. One is opened to show a goat. If you pair that one with one of the others, you can say "they originally had a prize-probability of 2/3, which now transfers to the unopened door. It is therefore more likely to contain a prize than the other unopened door." However, try going back a step. What if you had paired the opened door with the other unopened door? The same process applies, and once again you have a 2/3 probability of getting the prize. You have a 2/3 probability for each door, and this adds up to 4/3, which is greater than one and not a valid probability. Therefore, this is not a valid method of determining probability, and the remaining two doors must be assessed on their own.
    Offline

    0
    ReputationRep:
    If you need convincing, either draw a tree diagram or list all possibilities.
    Offline

    14
    ReputationRep:
    (Original post by theone)
    If you need convincing, either draw a tree diagram or list all possibilities.
    ok so here is my tree diagram:

    getting it right first : 1/3,
    getting it right after 1 door taken away : 1/2, getting it wrong after 1 door taken away :1/2

    getting it wrong first : 2/3
    getting it right after 1 door taken away : 1/2, getting it wrong after 1 door taken away :1/2

    so if you don't swap after getting it right first time you have prob of 1/2 x 1/3 = 1/6
    if you swap after getting it wrong first time then you have prob 2/3 x 1/2 = 1/3

    in both cases you end up correct so your total prob of getting it right if you swap is 1/6 + 1/3 =1/2.

    The same is true for the inverse as the total prob must equal 1. If you swap you have a half chance of being right.

    Musicboy
    Offline

    0
    ReputationRep:
    the million doors example sort of convinced me.

    anyway we had this in year 7 maths with some cards, and the teacher betting a student £1 each time, i tihnk the teacher lost 5 in a row before he gave up.
    Offline

    14
    ReputationRep:
    (Original post by DanMushMan)
    the million doors example sort of convinced me.

    anyway we had this in year 7 maths with some cards, and the teacher betting a student £1 each time, i tihnk the teacher lost 5 in a row before he gave up.
    the million doors example is also wrong. If you had a million doors then your chance of being right first would be 1/1000000 after one door is opened and proved not to have the car your chance is 1/999999 the 1/999998 etc until all the doors but onehave been opened at which point your door has the probability of 1/2 as does the other open door. It does not matter which you pick as you cannot predict.

    Musicboy
    Offline

    0
    ReputationRep:
    (Original post by musicboy)
    the million doors example is also wrong. If you had a million doors then your chance of being right first would be 1/1000000 after one door is opened and proved not to have the car your chance is 1/999999 the 1/999998 etc until all the doors but onehave been opened at which point your door has the probability of 1/2 as does the other open door. It does not matter which you pick as you cannot predict.

    Musicboy
    i think the argument goes something like:
    you pick one door, with 1/1,000,000 chance
    the chance its in another door is 999,999/1,000,000
    every other door except one closes, so there's a 999,999/1,000,000 chance you'll get it if you switch.
    Offline

    14
    ReputationRep:
    (Original post by DanMushMan)
    i think the argument goes something like:
    you pick one door, with 1/1,000,000 chance
    the chance its in another door is 999,999/1,000,000
    every other door except one closes, so there's a 999,999/1,000,000 chance you'll get it if you switch.
    please reread what I have said . if all doors except one open then you must re-assess the probabilities of both the oor you've chosen and the closed one. each with be 1/2.

    MB
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by DanMushMan)
    i think the argument goes something like:
    you pick one door, with 1/1,000,000 chance
    the chance its in another door is 999,999/1,000,000
    every other door except one closes, so there's a 999,999/1,000,000 chance you'll get it if you switch.
    OK, so when there are 1,000,000 doors to choose from, the one you pick has a 1/1,000,000 of being the prize one. But when you eliminate 999,998 others, the probability of it being the prize one surely rises, as it does for the remaining one. Why is the remaining one so much more likely than the one you picked? Really, they're equally likely to start with. If eliminating loads of them makes it really likely that it's that one, then, well surely it does the same for the one you picked. There's always got to be a last door to be picked out of 999,999 but if the one you picked is the right one it doesn't make that one any less a loser.
    Offline

    14
    ReputationRep:
    (Original post by covered farm wa)
    OK, so when there are 1,000,000 doors to choose from, the one you pick has a 1/1,000,000 of being the prize one. But when you eliminate 999,998 others, the probability of it being the prize one surely rises, as it does for the remaining one. Why is the remaining one so much more likely than the one you picked? Really, they're equally likely to start with. If eliminating loads of them makes it really likely that it's that one, then, well surely it does the same for the one you picked. There's always got to be a last door to be picked out of 999,999 but if the one you picked is the right one it doesn't make that one any less a loser.
    correct
    Offline

    0
    ReputationRep:
    well in my head its clear - switch, if this was real i would switch. The prob at the beginning that you've picked the right door is only 1 in a million. that stays the same, it doesn't change. you're highly unlikely to have chosen the correct door first time round. Now when all the other doors are open its much more likely the other remaining closed door is the prize one, since the probability you were first right is so small.
    Offline

    1
    ReputationRep:
    The crux of the problem is that they aren't randomly openning doors, they always choose to the ones that doesn't contain the prize.
    This doesn't change the probability of you having picked the correct door in the first place.
    Offline

    0
    ReputationRep:
    alaric and music boy are definitely correct: there's no good reason to switch (although switching won't do any harm: it's 50-50).

    Best way of looking at it is either:

    (i) common sense: Goldman Sachs are giving out free booze at Freud's, Love Bar or Raoul's; you don't know which but can assume the chances are equal for each; you head for Love Bar; your mate rings you to tell you he's at Freud's and there's no free drink there; do you about face and head for Raoul's?; *******s you do; there's still just as much chance it's at Love Bar as at Raoul's.

    (ii) logic: remember that probabilities only hold when premises remain the same. Premises can change when you get further information. Comparing the chances BEFORE you get further info (there's a one in three chance that that door is the right door) with the chances AFTER you get the extra info (three doors but not that one) is not comparing like with like. You have to calculate on the basis of the new premises (ie take account of the further info). New premises (forget that you've made an earlier choice and rather start afresh) = either this door or that: 50-50.

    In any case, with the Monty problem you can incorporate all the information you have at the outset and the answer ought to be plain.

    Premises:
    1. There are three doors.
    2. One contains a prize.
    3. You'll be shown one of the non-prize doors.

    It follows that you're only ever truly choosing between one prize door and one non-prize door. Hence, 50:50.
    Offline

    1
    ReputationRep:
    (Original post by houdini)
    alaric and music boy are definitely correct: there's no good reason to switch (although switching won't do any harm: it's 50-50).
    That's not what I'm saying, I'm saying that it is better to switch.

    It's a very well known 'problem' in probability, there's another page describing it wolfram.com: http://mathworld.wolfram.com/MontyHallProblem.html

    The point being that the door is deliberately chosen, and so the original probability doesn't rise.

    Think about it like this:
    -You pick a door.
    -You theoretical commit to that door. The chances of you picking it correctly are 1/3, the chances that you've got it wrong is 2/3.
    -The prize has a 2/3 chance of being behind one of the two doors you didn't choose.
    -One deliberately non-winning door is openned, reducing the probability of it being behind that door to 0. The probability of it being behind either one of those doors is still 2/3. This gives the remaining door a probability of 2/3.
    -No amount of door openning can change the probability that you'd picked correctly first time, if you'd not seen the empty door you'd still believe you have a 1/3 chance.

    The confusion lies in that Monty can and does pick the door that doesn't have a prize when it does lie behind one of the unchosen doors. That extra insight and choice he has makes the problem confusing.

    Alaric.
    Offline

    0
    ReputationRep:
    alaric is correct. monty is never going to open your door, but the fact that he doesn't pick a certain door other than the one you pick gives you information about that door - it is more likely to be the winning one.

    to try to see this, go back to the million door scenario. *Whichever* door you pick, you know that Monty can open 999,998 doors with nothing behind them. Monty was never going to open your door anyway, but the fact that he didn't open the other closed door makes it hugely likely that it is behind that door.

    the correct application of "the probabilities must add up to one" argument is this. suppose you pick door A. monty opens another door. but ****whichever**** door you pick, monty could have done this. the probability of it being behind door A knowing that it is not behind the door monty opens is 1/3, (the conditional probability of an event knowing a certain event is the probability of that first event). and so the prob. of its being behind the other door is 2/3.

    try http://math.rice.edu/~ddonovan/montyurl.html if you are still not convinced.
 
 
 
Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Updated: February 16, 2004
Poll
Black Friday: Yay or Nay?
Useful resources
Uni match

Applying to uni?

Our tool will help you find the perfect course

Articles:

Debate and current affairs guidelinesDebate and current affairs wiki

Quick link:

Educational debate unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.