The Monty Hall Problem

Comments

1. 

   January 8, 2005

   DaninVan

   I disagree with the supposedly correct answer for quite different reasons than what were posted. This isn’t really a probability issue but more of a shell game/ human nature scam. If you read the scenario right through, you discover that the show host ALWAYS opens one of the doors and gives the contestant the opportunity to switch choices. Therefore the odds were ALWAYS 1:2 NOT 1:3; no way in hell was s.h. going to open the $$$ door (that ruins the game as there are then obviously only boobyprizes left). Forget math, think human nature. The guys that want to run computer programs and sits with 1000 doors have completely overlooked the obvious, there WERE only 2 doors to choose from, the 3rd is a red herring.

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2. 

   January 9, 2005

   Mick H

   Eh? I’m not sure where to start with that. Of course the odds weren’t “always 1:2” – the odds of the originally chosen door being the correct one are always 1:3. There’s no scam involved. Subsequent to the host revealing one of the others as a dud, the chances of the remaining one being the right one now go up to 2:3. So, it makes sense to switch.

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3. 

   January 9, 2005

   DaninVan

   game’s over OR eliminate one of the dud doors for the player to continue… ie the contestant, in advance, will, before the game even starts, ONLY HAVE A CHOICE OF THE $$$DOOR OR ONE DUD DOOR. As I said ealier the third door is a red herring. Math has nothing to do with it.
   Remember the question?
   “Is it better for the contestant to switch his choice to the remaining door or to stick with his original choice?”

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4. 

   January 9, 2005

   Mick H

   I give up. We’re on different wavelengths.

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5. 

   January 9, 2005

   DaninVan

   Hey, comeon; ‘Dean’ is offering $1,000 to prove him wrong…he’s on your side. That’s gotta be worth pouring yourself a fresh cuppa, finding a comfy cushioned chair, letting your mind mellow, then reconsider this puzzle in the perspective that I’ve laid out. Even Steven Hawking has been able to recant…;)

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6. 

   January 9, 2005

   Mick H

   The reason he’s offering $1000 is that he knows he’s right….

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7. 

   January 10, 2005

   Ian

   n to make it favour one side or another based on past history, exactly like the doors. They don’t rearrange themselves bcause the odds say so, this is of course assuming Monty always opens a door.
   The reason why the question is confusing and misleading is relation to the answer is that the real world instance of the choice is a single go, which is not taken into account in the maths.

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8. 

   January 10, 2005

   Mick H

   Ian – No, the “correct” answer is the correct answer. The probability of a hit, for a single go, if you switch, is 2:3. I don’t follow your distinction between odds and probability.

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9. 

   January 10, 2005

   Ian

   The probability of a hit, for a single go, if you switch, is 2:3.
   Where is the calculation for this 2:3 on a single go ? The ones I see are based on a series of attempts, you don’t get a series, you only get one.
   Are you telling me that if I’d been tossing a coin all day and getting only heads, when you toss the same coin once it is more likely to land on tails ?
   The odds for a single attempt are 50:50.
   I only saying I am skeptical, convince me otherwise 🙂

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10. 

    January 10, 2005

    Mick H

    The coin-tossing analogy isn’t helpful here. When you choose a door to start with, the probability (= odds) of you picking the winning door is 1:3. The probability of the winning door being one of the two you didn’t pick is 2:3. Now Monty Hall opens one of the doors you didn’t pick. The probability of the door you picked being the winning door – of course – doesn’t change, at 1:3. Nor does the probability of one of the other two doors being the winning door, at 2:3. But now you know one of those two definitely isn’t the winner, so the 2:3 probability rests on that remaining door alone – the one you can (and should) switch to.

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11. 

    January 10, 2005

    Ian

    The “2:3 probability” is calculated from repeated attempts, i.e. if you choose 3 times then 2 of them will be the winner.
    You only get one go.
    The coin analogy is sound, the likelihood of 20 coin tosses ending up in 20 heads is small, but any single coin toss the odds are always 50:50, no matter how many heads or tails have preceeded it.
    For the doors, the odds are not 2:3 for one go, it is 50:50, that’s what I claim and I have yet to see proof otherwise that does not mention a series of attempts as opposed to one attempt.
    I stress, that in the show you get one attempt, and that is the situation being presented, my skepicism comes from seeing the wrong math applied.

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12. 

    January 10, 2005

    DaninVan

    Right on, Ian! It’s simple flim-flam, there aren’t 3 doors in any case. The third door (EITHER of the dud doors)is always deleted prior to the only round that counts, the money round. Too many posters have tried to redefine the puzzle and/or the ‘the question’. It’s not even necessary to make a selection in rd.1 The result will always be the same: the $$$door and one dud door to choose between. It makes absolutely NO difference which one the contestant is sitting on. The decision is made by free will; until the contestants says which one, the coin is still in the air. I chatted last night, with my Physics Prof neighbor, about this and he got a pretty good chuckle over the earnestness of the mathaholics and there not being able to sit back and see the sleight of hand.

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13. 

    January 10, 2005

    DaninVan

    Mike; many people over the years have ‘known they were right’, that’s why litigation lawyers are so sucessful.

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14. 

    January 11, 2005

    DaninVan

    Oops, make that’successful’

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15. 

    January 11, 2005

    Mick H

    There’s no arguing with some people. I’ve done my best. If you don’t see it, you don’t see it.

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16. 

    January 11, 2005

    Simon

    OK, I’ll give it a go.
    Ian, you’re confused because you’re trying to compare a coin-toss, which is a random, unpredictable future event, with the Monty Hall problem, which is about Bayesian probability. There is a fundamental difference between the two – that being we have information about what is behind the two remaining doors.
    Specifically, we know that when we chose a door, it had a one in three chance of being the winning door. There is therefore a two in three chance that we got it wrong. Now, after MH has opened an empty door we didn’t choose, the original probabilities don’t change; it was one-in-three that we made the right choice, and it’s still one-in-three now, with the remaining door (the one that neither we chose nor Monty Hall opened) representing the two-in-three chance that we got it wrong, the second empty door having been opened. It’s therefore a two-in-three chance that switching would be the right move.
    The key point to think about in comprehending the problem is that it isn’t like a coin toss, because the jackpot-winning door has already been determined, and Monty Hall’s opening an empty door gives us enough information to know that switching gives us a two-in-three chance of winning.
    I’m not sure why you want to differentiate between repeated attempts and a single attempt (actually I do – I made the same mistake when I was trying to understand the problem. But it doesn’t make any difference). The point is that if switching wins two out of every three times, then the chances of winning on a single go are two in three.
    Daninvan – you miss the point that the door Monty Hall opens is determined by your original choice. If you chose wrong (2/3 chance), Monty Hall only has one option – to open the remaining empty door. It’s therefore a 2/3 chance that the other door, which remains untouched, is the winning door, and so you should switch.

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17. 

    January 11, 2005

    DaninVan

    Simon, one more try, the first two rounds accomplish absolutely zero. The contestant doesn’t have to make his decision till the final round. Any ‘decision’ made in rd 1 is completely reversible (in rd. three) it doesn’t count, it’s a sham, it has no effect. The contest only ever was about two doors. You could let your ex mother-in-law make the first choice. Sheesh

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18. 

    January 11, 2005

    Simon

    No, they don’t accomplish zero. Consider the possibilities for a second. If you picked the right door first off (a one in three chance), Monty has a choice of two empty doors to open, leaving one empty door unchosen, and so if you choose to switch you lose. If you picked the wrong door (two in three), Monty has only one door to choose – the other empty door, leaving the unchosen door as the jackpot winner. It therefore follows that in two out of three cases, the unchosen door is the jackpot winner, meaning that switching gives you a two in three chance of winning the jackpot. Given that it’s likelier you picked the wrong door, it’s therefore likelier that if you switch, you win.
    What you’re missing is that the decision you make in round 1 has an effect on Monty Hall’s decision in round 2 – specifically on whether he gets to choose an empty door to open. The combined effect of your decision and Monty Hall’s decision gives you information which weighs the probability in your favour if you switch.
    A useful way to think of it is to reconsider the problem like this: You have a choice of a thousand doors, one with a jackpot behind it. You choose a door. Monty then opens 998 doors, all empty, leaving just the door you chose and one other door unopened. Which door you choose to win the jackpot? Once you’ve answered that, consider that the logic is exactly the same as with just three doors.

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19. 

    January 12, 2005

    DaninVan

    Ok, I get your point re 1,000 doors but I’m not convinced that applies in the written scenario. Excuse me while I run a few sets with a King, 2 and 3 and a less than enthusiastic wife (my contestant).

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20. 

    January 12, 2005

    Ian

    he question was “is it better to choose a door and switch or choose a door and stay”, then perhaps the math is more appropriate, but it is not represented that way.
    Again, I can toss a coin 49 times and they all come up heads, then I turn to you and ask, the mathematical probability of the single choice is 50:50, regardless of what the previous results have been, I am only asking you about the final coin toss.
    The Monty Hall problem concerns how math is applied wrongly or how the question is being used to confuse.

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21. 

    January 12, 2005

    Mick H

    Ian – you persist in the coin-tossing analogy, but as we keep saying, it doesn’t apply. It’s not the same. When faced with the choice of whether to switch or not, it’s not a choice between two equal possibilities, as it is with a coin toss. That’s because you have some information about one of the doors: you know that one of them was the door that Monty didn’t choose. That tells you something about it. That’s why that door has twice the probability of being the correct door than the one you originally picked.

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22. 

    January 12, 2005

    DaninVan

    Mea culpa, Mick and Simon. I played a bunch of sets using a K,2 and 3 (K being the $$$). You’re right I’m wrong…(grovel, grovel).
    Ian, try it with cards, a dealer and a player. it becomes obvious REAL fast!
    ……………..mumble, mumble, sulk

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23. 

    January 27, 2005

    Adam N

    The smart person would always switch. Why you might ask, because 2/3 of the time the initial guess would be wrong.

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24. 

    February 25, 2005

    michael

    This isn’t a problem concerning what will happen – we already know that, try a 10,000 game experiment. This is a problem in communication. Look at the second stage as a separate occurance, the only connexion with the 1st is that being that one door has been chosen.
    Assume the car is behind door 1 with a goat behind door 2 and another goat behind door 3. This doesn’t change; it remains in force whilst we look at the possibilities of choosing door 1 followed by choosing door 2 followed by choosing door 3.
    There are 4 possibilities (A, B, C and D).
    A. If door one is chosen, door 2 is open showing a goat and door 3 is closed.
    B. Again, if door 1 is chosen, door 3 is open showing a goat and door 2 is closed.
    C. If door 2 is chosen, door 3 is open showing a goat and door 1 is closed.
    D. If door 3 is chosen, door 2 is open showing a goat and door 1 is closed.
    There are no other possibilites. Therefore, the contestant, is looking at one of four possibilities. If he switches his choice, on the first two he looses and on the second two he wins.
    If he doesn’t switch, on the first two he wins and on the second two he looses.
    That, paradoxically, is Bayesian and would therefore determine that switching or not switching is irrelevant to the outcome.
    The challenge here, is not to show why switching is correct, but why the above is flawed.

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25. 

    February 25, 2005

    michael

    Simon – If there were a 1000 doors and the game show host opened 998 – you’d simply HAVE to stick! He’s missed it 998 times – it’s a good bet you got it in the first place!

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26. 

    March 18, 2005

    mntyhll

    math.ucsd.edu/~crypto/Monty/monty.html

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27. 

    March 18, 2005

    jcri

    do you assume that the host knows witch door the car is behind

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28. 

    March 18, 2005

    Mick H

    Yes

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