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Monty Hall - a conundrum within a conundrum

  1. Oct 18, 2006 #1
    Here are two variations on the Monty Hall question - one where the host knows and one where the host doesn't. There is consensus on the former, but I'm interested in the latter.

    I'm using playing cards here, but the principle is identical.

    THE HOST KNOWS​

    The object is to find the Queens.

    There are 6 rows of 3 cards, all face down. In each row there is a Queen and two Jacks.

    From each row you select a card and keep it face down.

    From each remaining pair of cards, a host who knows what they are turns over one of them and reveals a Jack. (So there is no misunderstanding, the host will always do this)

    So now there are the six cards you chose and six cards not eliminated.

    To acquire the most Queens, should you stick or swap? Or are the odds 50/50?

    Monty Hall Answer: Swap - advantage 2/3.

    Reasoning: probability was always that you originally chose 2 Queens and 4 Jacks. You know the other six cards are the opposite of your choice. Therefore, on average, the other cards will most likely have 4 Queens and 2 Jacks.

    Essentially this is standard Monty Hall. I'm assuming we're already agreed on this answer. Now consider the following scenario, which is the same in all respects except for the host's state of mind.

    THE HOST DOES NOT KNOW​

    The object is to find the Queens.

    There are 6 rows of 3 cards, all face down. In each row there is a Queen and two Jacks.

    From each row you select a card and keep it face down.

    From each remaining pair of cards, a host randomly eliminates a Jack. (The host's random elimination is defined as follows: if a Queen is discovered, you simply start again; all the cards in each row are shuffled and you make a new selection; keep doing this until you get the desired outcome from the host - 6 eliminated Jacks.)

    So now there are are the six cards you chose and six cards not eliminated.

    Is the solution the same?


    Simon
     
    Last edited: Oct 18, 2006
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  3. Oct 18, 2006 #2

    HallsofIvy

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    In the Monty Hall problem, in which the host does not know where the prize is you have to assume that he opens a door the prize is not behind by chance and so must take into the account the probability of that happening. It's fairly easy to show that, under these conditions, swapping does NOT improve your chances.
     
  4. Oct 18, 2006 #3
    Monte Hall, Live from Crete

    Monte shows the contestant three doors. Behind one hides a car, and behind the other two stand an honest person and a liar, respectively. The contestant is invited to choose one door, and does. However, before opening that door, Monte reveals another door with one of the persons. They, knowing the contents behind the other doors, will offer (truthfully or not) that information. Is there now a strategy for maximizing the chance of winning the car?
     
  5. Oct 18, 2006 #4

    Doc Al

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    I would say that the host's state of mind is irrelevant and the answer is the same for either scenario: you improve your odds of winning by swapping. Just like in the first scenario, the host always reveals a Jack due to how the game is rigged. The fact that he doesn't know ahead of time is irrelevant, since if he reveals a Queen by chance the game resets and starts over.

    Try it and see!

    (FYI: The business of 6 rows is a distraction. Just use 3 cards and replay as many times as it takes to see the pattern.)
     
  6. Oct 18, 2006 #5
    Thanks for your reply Doc.

    It appears to make sense that the reasoning applied to the knowing host also applies to the random one. In both cases, the odds are 2/3 that you originally chose a Jack. In both cases, the host does eliminate a Jack from the other two cards - whether by design or accident. Therefore in both cases, the odds are 2/3 that the remaining card you didn't choose is the Queen.

    That answer is completely logical - as far as it goes.

    Now for the bombshell: it is wrong!
    HallsofIvy is correct.

    This is counter-intuitive, but the 2/3 advantage in swapping only applies when the host knows. In any game where a Jack is eliminated, but without such knowledge, the odds become 50/50

    Try it and see! :smile:
     
  7. Oct 18, 2006 #6

    Doc Al

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    Sorry, but I don't buy it! :smile:

    The odds of your first pick being a winner: 1/3
    Thus, the odds of the winner being in the other two choices: 2/3
    The odds, if the winner is in the other two, that you will pick it if you switch: 100%! (Monty revealed the loser already.)

    Now, what would be different: Monty asks if you want to switch but reveals nothing. In that case, it is pointless to switch.

    What makes you think the odds become 50-50? Whether Monty "knows" or not should be irrelevant. All that matters is the information Monty reveals.

    To exaggerate this: Have 99 Jacks and one Queen. Pick a card at random. Now Monty reveals 98 Jacks (who cares that it probably takes forever for him to do that since he doesn't know where the Queen is--if he reveals a Queen, just do it over!)--do you pick the one remaining choice? You better!
     
  8. Oct 18, 2006 #7

    Office_Shredder

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    Ok, let's just look at one row, because it's equivalent, and do out all the math.

    First, say you picked a jack. There is a 2/3 chance of this happening. Then Monty reveals a card. 1/2 the time, you reshuffle. 1/2 the time, he reveals the other jack, and you switch to win. So you win 1/3 of the time here, and reshuffle 1/3 of the time.

    Then, suppose you pick a queen instead. There is a 1/3 chance of this happening. Every single time, you lose. So there is a 1/3 chance of you losing

    Doc Al, the problem with your 100 card example is that if he picks 98 jacks in a row, it's probably because he was picking out of a pile of 99 jacks, and you had the queen the entire time (actually, half the time that's true)
     
  9. Oct 18, 2006 #8

    shmoe

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    It still won't matter if you switch or not. Since Monty is picking randomly and you are as well, you can think of this game as a sequence of 99 J's and one Q. The first in the sequence is your pick, the next 98 are Monty's. All 100 sequences are equally likely:

    QJJJJ...JJJ
    JQJJJ...JJJ
    ..
    JJJ...JJJQJ
    JJJ...JJJJQ

    The middle 98 won't get played, only the first and last. Each is equally likely, so it doesn't matter if you switch or not.
     
  10. Oct 18, 2006 #9

    Doc Al

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    Boy, was I wrong!

    OK guys, you got me! My reasoning was full of crap--you guys are correct!

    The flaw in my thinking was forgetting that the entire deck gets reshuffled when Monty reveals a Queen:

    1/3 of the time your initial pick is correct; Monty shows a Jack; you lose when you switch.

    2/3 of the time your initial pick is incorrect, but half of those times don't count; for the half that does count (2/3 * 1/2 = 1/3 of the initial starts), you win when you switch.

    So, it's fifty-fifty!

    Good one.
     
  11. Oct 18, 2006 #10

    Hurkyl

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    NO!

    That's the trick -- in this problem you aren't randomly picking one card out of 100. The game resets every time Monty reveals a queen, something that can only happen if your initial choice is wrong.

    It's easiest to see it from a Bayesian standpoint:

    [tex]
    \def\A{\text{I picked the queen}}
    \def\NA{\text{I picked a jack}}
    \def\M{\text{Monty revealed 98 jacks}}
    \def\b{\, \mid \,}

    \begin{array}{l}
    \frac{P(\A \b \M)}{P(\NA \b \M)} \quad \quad
    \\
    \quad \quad = \frac{P(\M \b \A)}{P(\M \b \NA)}
    \cdot
    \frac{P(\A)}{P(\NA)}
    \end{array}
    [/tex]

    Work it out in the two cases where Monty does and does not know where the queen is.


    Though the problem is much shorter if you can convince yourself that, when Monty doesn't know where the queen is, that it doesn't matter who picks first.


    Enumeration works too -- you always pick the one on the left, and Monty always flips the leftmost card(s) of what's left. What are the possibilities?
     
    Last edited: Oct 18, 2006
  12. Oct 18, 2006 #11

    Doc Al

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    OK, OK, rub it in why don't you? :cry:

    :tongue2:
     
  13. Oct 18, 2006 #12
    Damn, I've just prepared my arguement for you Doc, but you've already thrown in the towel! :smile:

    Nevertheless, here it is!

    Simon would have wrote:
    Oh well! :smile:

    The trick was aimed at those already familiar with the usual Monty Hall problem. Having successfully made the counter-intuitive leap away from 50/50 to the 2/3 advantage, you then have to make the same counter-intuitive leap in reverse. Not everyone can do it immediately, as this has proved.

    Simon
     
  14. Oct 19, 2006 #13
    Ok, we've established that the odds are 50/50 if the Host doesn't know and 2/3 in favour of swapping if the Host does know.

    Lets see if you're ready for this:

    You select one of three cards, as before.

    Enter Host 1 who does not know where the Queen is.
    He randomly reveals a Jack.
    He then turns it back over and leaves.
    Nevertheless, you remember it is a Jack.

    Enter Host 2 who did not witness the above.
    He knows where the Queen is and reveals a Jack.
    You are shown nothing new.
    He has turned over the same card as Host 1.

    A game is cancelled if Host 1 reveals the Queen or if Host 2 reveals a different card.

    For all games that proceed: are the odds 50/50 or 2/3 in favour of swapping?

    Simon
     
  15. Oct 19, 2006 #14
    Does observation affect probability?
     
  16. Oct 19, 2006 #15

    Doc Al

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    The same 50/50 analysis seems to apply. (Though I've been wrong before! :uhh: )

    Assuming I understand the rules this time:

    1/3 of the time your initial pick is correct; Host 1 always a Jack; you lose when you switch.

    2/3 of the time your initial pick is incorrect, but half of those times don't count since Host 1 shows the Queen; for the half that do count (2/3 * 1/2 = 1/3 of the initial starts), Host 2's actions change nothing and you win when you switch.

    So, it's fifty-fifty!
     
  17. Oct 19, 2006 #16
    Yes you do understand the rules. They're exactly as stated.

    This is an even greater counter-intuitive trick than the last, and I'm afraid you fell for it.

    Incredible as this sounds:

    After Host 1 randomly reveals a Jack the odds are 50/50

    After Host 2 turns over the same card, this changes! The odds become 2/3 in favour of swapping - though you are shown nothing you haven't already seen.

    I expect you think this is impossible!
    Others will probably agree with you!
     
    Last edited: Oct 19, 2006
  18. Oct 19, 2006 #17

    Doc Al

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    If you can convince me of that, I'll just quietly slink away. :smile:
     
  19. Oct 19, 2006 #18

    shmoe

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    Only 1/2 the games where you pick a queen first survive host 2's pick. Host 2 comes in and has a 1/2 chance of picking the same jack host 1 did, if he doesn't the game is reset.
     
  20. Oct 19, 2006 #19

    Hurkyl

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    The same three things I mentioned in my last post apply here. :smile:

    (1) The fact the game didn't reset gives you information

    (2) The bayesian approach makes it clear

    (3) You can work out the fact that this game, when it doesn't reset, is equivalent to the original Monty Hall problem! Note that it doesn't matter which order the two hosts choose their card -- it becomes more obvious when you reverse it. The knowledgable host picks a non-queen, then the unknowledgable host wields a 50-50 chance of resetting the game that is completely independent of where the queen actually is.

    (4) You can just enumerate it all. :smile:
     
  21. Oct 19, 2006 #20

    Doc Al

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    Once again, I misread the instructions! I thought Host 2 always picked the same card as Host 1. (Which makes no sense, since he didn't witness Host 1 picking. D'oh!) Got me again! :redface:
     
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