That extra information is an assumption that is supposed to be based on your knowledge of how game shows work. In fact based on Monty's knowledge and motivations it is possible that you want to switch (Monty knows where the car is and always shows a goat), it doesn't matter (Monty doesn't know where the car is), or you should stick with your choice (Monty knows where the car is and is trying to keep you from getting it).

I've never seen the Monty Hall problem presented with that stipulation explicit.

Here's the exact text of the original question, posed to Marilyn in her column in parade magazine:

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?

http://www.fortunecity.com/victorian/vangogh/111/9.htm

The stipulation that the host knows what's behind the doors, and always opens a door with a goat is a given for this problem.

In fact, the additional information that it was 'Monty Hall' came later... which messed the whole thing up because they interviewed him, and he stated that he sometimes opened the prize door right away... and then everyone forgot the stipulations of the original question

I actually remember when this happened, because my college statistics professor was *consumed* with proving Marilyn wrong, and ended up conceding she was correct only after writing a program to do 10,000 permutations that bore her true

Trek

I've never seen the Monty Hall problem presented with that stipulation explicit.

Funny, I've always seen it explicitly stated that the host knows and will not open the door with the car. If it wasn't this way, he'd open the door with the car 1/3 of the time. A quick google search seems to confirm this, though I only checked the first 4 links that weren't applets. The extra information Mr. Hall provides is what makes the puzzle worth puzzling ;)

Update: strike off the cuff response... By the way (after a quick read and next to no analysis) the spooky math problem looks more like a divide-by-an-infinite-number problem. The probability that you pick a number between two numbers in an infinite continuum is undefined.

For what little it's worth...

People seem to be arguing with tilly about this, but this has been my experience too. I have seen this problem published at least a dozen times; I have never seen the assumptions (e.g. that Monty knows what is behind the doors, etc.) stated clearly before.

buckaduck

No, your chances go from 1/3 to 2/3. For your chances to be 1/2, you'd have to assume no bias in what is behind the two (remaining) doors, which means you'd win as often by picking the same door you previously picked (which you appear to agree isn't the case).

You start with these odds for each door having the car behind it: A=1/3, B=1/3, C=1/3. You pick "A" (it doesn't matter). Then, if the car is in A, the odds that Monty reveals B are 1/6, and that Monty reveals C are 1/6. If the car is in B, then Monty reveals C (1/3) and changing doors means you win. If the car is in C, Monty reveals B (1/3) and changing doors means you win.

So keeping the same door means you win 1/6+1/6 = 1/3 of the time. Switching doors means you win 1/3+1/3 = 2/3 of the time, or twice as likely.

You can't have the choice of "switch doors or not" result in you winning 1/3 of the time one way and 1/2 of the time the other way. You'd have 1/6 of the odds unaccounted for.

The human mind appears not well suited for dealing with probabilities (unlike calculating parabolic trajectories).

A similar probability "paradox" was taught to me:

You know that your stats professor has two children. Once you heard him mention his son. What are the odds that his other child is a girl?

Our text book said that the odds are equal of boy-boy, boy-girl, girl-boy, girl-girl and knowing that one is a boy makes the girl-girl case impossible and therefore we have 1/3 chance that the other child is also a boy and 2/3 chance that the other child is a girl.

Despite this being in a text book and being taught without challenge in my statistics class, I disagreed with this result.

My argument is that if mentioning a boy can make the odds of the girl-girl case go to zero, then why should it not affect the odds of the other cases as well. It was my assertion that the professor is more likely to mention a son when he has two sons (or is relatively more likely to have had two sons by virtue of having mentioned one of them) and so the odds are 50-50 just like instinct tells us.

You can structure the problem a little more strictly as:

You know that your stats professor has two children and research shows that boy and girl children are equally likely and that the sex of one's first child does not have a statistically significant impact on the likelihood of the sex of one's next child. You flip a (fair) coin and if heads, you ask your professor the sex of his younger child, but if tails, you ask your professor the sex of his older child. If your professor (truthfully) answers "male", then what are the odds that the other child is female?

Odds that the first child is a boy 1/2, girl 1/2. Odds that the second child is a boy 1/2, girl 1/2. So the odds are boy-then-boy 1/4, boy-then-girl 1/4, girl-then-boy 1/4, girl-then-girl 1/4. If you flip heads (it doesn't matter), and the professor answers "male", then we have the odds b-b 1/4, b-g 1/4, g-b impossible, g-g impossible. Odds that the other child is a girl: 50%.

But in the restructured problem, you not only know that the professor has a son, but also whether the son in question is the older or younger child. So perhaps this difference changes the odds and we've over-thought the problem to come up with an incorrect answer just to justify our initial, instinctive answer.

There is also a check we can perform.

If we go by the text book's interpretation but don't specify whether the sex that we heard mentioned by the professor was "boy" or "girl", then we have the following matrix of possibilities:

 children's sexes
 ---------------- heard
 b-b b-g g-b g-g  -----
 1/3 1/3 1/3  0   "boy"
  0  1/3 1/3 1/3  "girl"

And since there is no reason to bias for "boy" or "girl" being heard, we assign each of those odds of 1/2 so that combined odds are:

      children's sex
      ---------------  heard
total b-b b-g g-b g-g  -----
 1/2  1/6 1/6 1/6  0   "boy"
 1/2   0  1/6 1/6 1/6  "girl"
  1   1/6 1/3 1/3 1/6  total

which violates our initial common-sense assertion. That is, having heard the sex of one of his children suddenly makes it twice as likely that his children are not the same sex. Sounds rather magical.

So, was my text book wrong?

I think it was, but I'm even more sure that this demonstrates that humans aren't very good a analyzing probabilities.

- tye        

Not exactly true, the simpler flip is to say your first choice was likely wrong. 2/3 of the picks will be Goat to start and thus wrong. Monty showing you a goat doesn't change that your orriginal pick has only 33% chance of being right.

The trap is in thinking that because there are two doors that it's a 50/50 guess again. A trap most people fall into being drilled on elementary statistics about a penny flip. If you flip a penny heads 10 times in a row, what are the odds you'll flip heads again? Yes, still 50/50 but those odds were always 50/50.

To simplify, if instead of opening a door he said to you: "Would you like to trade your one door for the other two doors?" you'd likely agree. You'd go from 1/3 to 2/3 chance of winning. Showing you the goat and making you labour on the two doors is simply a form of entertainment.

TM

BTW, in a discussion of this I rewrote the code for people who need more handholding:

#!/usr/bin/perl
my $wins  = 0;
my $loses = 0;

print "\n";
for (my $i=1;$i<=1000000;$i++)
{
  #all goats
  @doors = [0,0,0];
  #add a car
  $doors[int rand(3)] = 1;

  #contestant choice (cute varname declaration)
  my $choice = int rand(3);

  #monty makes his choice
  my $shown = -1;
  while($shown == -1)
  {
    $shown = int rand(3);
    if
    (
      $shown == $choice    || # he won't show contestants choice
      $doors[$shown] == 1     # or show the car
    ) { $shown = -1 }
  }
  $doors[$shown] = -1; # out of the equasion

  #If your on the car now, a change is a loss
  if($doors[$choice] == 1)
  {
    $loses++;
  }
  # presuming you don't pick the exposed goat, else is
  # a win.
  else
  {
    $wins++;
  }
  printf("\r Wins:%8d Losses:%8d Win \%:%5.3f",
         $wins,$loses,($wins/($i))*100);
}
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