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# Monty Hall problem/Talk

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Should you switch?

For the original choice, there are these possibilities:

      123
      CGG
      GCG
      GGC

So whatever door you choose, your chances are 1/3.

For the second choice, call the door you originally chose 1, and the door Monty opens as 3. Now the possibilities are:

       123
       CGG
       GCG

(We know that GGC is not a possibility because Monty showed us a goat behind door 3.)

So if you choose either 1 or 2, you have a 1/2 chance of winning. Which means it doesn't improve your chances to switch.

Now, what's wrong with this argument?

  The two cases CGG and GCG are not equally likely. In fact, GCG is twice as likely as CGG. This is obscured by your relabeling the doors after the fact so that the door Monty opens always gets number 3. You may want to write down all 9 possible equally likely scenarios: the car can be behind one of three doors, the contestant can choose one of three doors, makes 3x3=9. Then analyze how the sticker does (he wins in 3 of the 9 cases) and how the switcher does (he wins in 6 of the 9 cases).

Okay, so let's not renumber. Call the door you originaly pick number 1, and indicate the door Monty opens with parentheses. Then we have the following possibilities:

        1  2  3
        C (G) G
        C  G (G)
        G  C (G)
        G (G) C

So, if you stay with 1, your chance of winning is 1/2. If you switch to 2 or 3 (whichever remains unopened), your chances of winning are also 1/2. So it doesn't improve your chances to switch.

Or does it?

  Again, the four cases you list are not equally likely. The first two are equally likely, but they are each half as likely as the third and fourth case. If the door you originally pick is number 1, then the equally likely scenarios are CGG, GCG and GGC. Each of them has probability 1/3. The first case splits into the two equally likely cases C(G)G and CG(G) each with probability 1/6. The two cases GC(G) and G(G)C both keep probability 1/3.

Don't forget that the two goats are distinct. There are six permutaions for item placement:

          1   2    3

          C   g1   g2
          C   g2   g1
          g1  C    g2
          g1  g2   C
          g2  C    g1
          g2  g1   C

There are three door choices the player can make, yielding 18 possible outcomes:

      1   2    3

      C   g1   g2
      C   g2   g1
      g1  C    g2
      g1  g2   C
      g2  C    g1
      g2  g1   C

      1   2    3

      C   g1   g2
      C   g2   g1
      g1  C    g2
      g1  g2   C
      g2  C    g1
      g2  g1   C

      1   2    3

      C   g1   g2
      C   g2   g1
      g1  C    g2
      g1  g2   C
      g2  C    g1
      g2  g1   C

At this point, the player has "lost" 2/3 of the time. If the player changed the door selection now, the choices would be between two doors, like this:

        C  g1
        C  g2
        g1 g2
        g1 C
        g2 g1
        g2 C

Choosing either column 1 or column 2 still gives a 2/3 chance of losing, so switching doors won't accomplish anything.

Monty now eliminates an un-chosen goat (if there are two to chose from, it doesn't matter which one he chooses, so I'll always choose the rightmost). Note that in 4 out of the 6 cases, Monty is "forced" to give the player a car if he switches, and this is the key -- Monty's choice is not random. The resulting column for a switch is:

        C
        C
        g1
        C
        g2
        C

Here's an expanded chart, where the player's first choice is bolded and the eliminated goat is replaced with a hyphen. I've put a "#" where a switch wins, and a "-" where a switch looses.

      1   2    3      win/lose

      C   g1   -         -
      C   g2   -         -
      g1  C    -         #
      g1  -    C         #
      g2  C    -         #
      g2  -    C         #

      1   2    3

      C   g1   -         #
      C   g2   -         #
      g1  C    -         -
      -   g2   C         #
      g2  C    -         -
      -   g1   C         #

      1   2    3

      C   -    g2        #
      C   -    g1        #
      -   C    g2        #
      g1  -    C         -
      -   C    g1        #
      g2  -    C         -

But why count the two goats separately? What if there weren't any goats, just doors left empty - that wouldn't change the probabilities, would it?

Suppose Monty doesn't open the door, but after your initial choice he still gives you a chance to "switch", meaning you can either stay with your original choice, or if you switch you'll win if the car is behind either of the other two doors. Obviously you want to switch, because then you have two doors, with two chances to win, as opposed to only one door, one chance to win, if you don't switch. But what if Monty opens one door, with a goat, before you decide. That means one of the other doors isn't really a chance to win. Doesn't that change things? Doesn't it mean that it's more likely that the car is behind each of the other doors than orginally thought? Doesn't it make it more likely that your original choice was a winner? Aren't the odds now 1/2 rather than 1/3?

Well, suppose Monty opens both of the other doors, and they're both goats. Still think you should switch? Doesn't showing both goats behind the other doors mean that your original choice is a sure winner? That the odds for not switching are now 1.0, not 0.33? So why wouldn't showing the goat behind one of the other doors also change the odds?

Here's another question. Suppose while you're deciding whether to switch, another contestant is brought in and Monty lets them choose one or the other of the two closed doors. Aren't their odds of winning 50-50 for either choice?

  No. If they choose my original door, they win with 1/3 probability, if they choose the other one they win with 2/3 probability.

I know you think this, but why? There are just two doors, and the car is behind one or the other of them. Why doesn't that make the odds 50-50 for the new contestant?

  Because the situation is not symmetric. The new contestant knows something, and that information helps. Consider this: Monty has three boxes, in one of them is a hundred dollar bill. You don't know where. But then Monty says: Ok, I'll help you, point out two hats and I'll gladly combine their contents into a new box. You point out two, and he does as he said. So now there are two boxes left, the new one and the untouched one, one of them has the bill. But the odds are not 50-50: you know that the bill is twice as likely to be in the new box than in the untouched one.

Suppose they don't know which of the two doors I chose. Then aren't their chances 50-50?

  Sure.

BTW, the reason I'm raising these questions is to try to suggest why the Monty Hall problem is a "problem". It wouldn't be a problem if the answer were obvious...

  Consider the extreme case: there are 1000 doors, one has a car. I pick a door, then Monty opens 998 loser doors. Now you come in. Do you pick my door or the other one?

Yes, that is the argument that most convinces me that there is an advantage to switching. But how can we be sure that this result applies to the 3 door case. Certainly the benefit of switching is very much higher in the 1000 door case that in the 3 door case. It obviously decreases as the number of doors decreases. Maybe the advantage vanishes altogether in the 3 door case?

Suppose they choose the door you originally chose. Aren't their chances of winning the same as your chances if you choose not to switch?

  Yes.

I think this is the most suprised I've ever been my the output of a program I wrote myself. I was sure it was going to come up 50/50. (/Empirical Proof)

  If you're still not convinced after seeing the results of your program, I have a poker game on Thursday nights you're welcome to join... --LDC

Here's a similar problem in the game of Texas hold'em that many people don't believe at first: I offer you your choice among three two-card hands: (1) a pair of fours, (2) an ace and a king of different suits, (3) a ten and a jack of the same suit. After you choose, I choose one of the remaining hands. Then we deal five cards face up, and the player who can make the best five-card poker hand with any combination of his two cards plus the five on the board wins. Which cards do you choose, and what is the expected outcome? It turns out that whoever chooses first loses. If you choose the A-K, I choose the 4-4 and have a slight advantage. If you choose the 4-4, I choose the 10-J and have a slight advantage. If you choose the 10-J, I choose the A-K, and have an advantage!

This is an interesting and helpful article but it should be completely rewritten; it doesn't sound like an encyclopedia article to me. The link to the Perl program is interesting, too, but it's original research. What place does that have in an encyclopedia?

     I take the rather abstract view of an encyclopedia as something
     which explains everything.  Thus I find the article actually
     quite accessible and helpful, written in a friendly direct
     way.  The perl code is perhaps less useful to someone wanting
     to understand the problem in an article-reading sort of way,
     but I think of it as sort of a multimedia addition - a working
     Monty Hall Problem machine to tinker with. -J

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Last edited August 12, 2001 11:02 pm (diff)
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