The Monte Hall Problem.

[Crazy Guggenheim]

Say someone's on "Let's Make a Deal". So they get to choose between Door 1-2-3. Say they choose door #2. Now Monte opens up door #1 and there's cardboard boxes. Now when he offers them x amount of money - no complex conjugates or Non-Linear Functions lol - does it pay to switch? Why?

[democracyindanger]

1. What's behind door #1 reveals no further information

about #2 or #3. So, no.

[Crazy Guggenheim]

2. It does reveal info by the fact that "they opened" a door.

:dilemma:

[AllegroRondo]

3. Yes!

If you stick with your choice, you have a 1 in 3 chance of being right.

If you switch, you now have a 2 in 3 chance of being right, because one of the wrong ones was eliminated.

[Ready4Change]

4. I didn't believe it, but now I know that switching makes sense.

This topic came up in another forum. I SWORE switching would make no difference. So much so that I wrote a little computer simulation in JAVA to prove it. It ran through the scenario 3,000 times. First 1000 were using the switch method, second used a don't switch method, and the last 1000 either switched or not by a random choice.

The results CLEARLY indicated that switching worked best. Switching gave a win about 66% of the time. NOT switching only won about 33%. ANd random switching gave 50/50 odds.

I couldn't believe it. But there it was.

Here's what I've decided is the logic behind the results:

Each door has a 33% chance of being the winner. When you pick one the odds are 66% that the winner is one of the other two. However, it is CERTAIN that one of the other two is a looser. So, when that looser is revealed, the other door is the winner 66% of the time.

But I had to run that program of mine several times before I was convinced.

[Rabrrrrrr]

5. Aye, I couldn't believe it either til I did the math

It seems totally against logic.

But, the numbers (and the computer simulations) don't lie.

[democracyindanger]

7. How did you randomize where the prize was?

Was it behind #2 and #3 equally?

I don't doubt your results, I'm just having trouble seeing how it can be. When you start, you have a 33% chance of picking the right door. That changes to a 50% chance when one door is eliminated. So either switching or not, you've got an equal chance of picking the right door.

[Crazy Guggenheim]

8. When you picked it was 1 out of 3.

[democracyindanger]

9. Okay...I think I'm getting it now

If you stick with the original choice, you're still at 33%. In order to take advantage of the new odds, you have to choose again.

This is why I don't gamble--I just hand over money.

[Ready4Change]

12. If I recall correctly...

On each pass, the program would generate a random number between 1, 2, and 3 as the "winning door." It generated another random number as my selected door.

It would then pick one of the two non-selected doors to reveal. Of course, if one of those doors was the winner it would reveal the other door.

It was a pretty simple sim. Ran real fast and produced pretty consistent results for any number of passes above 30 or so. (Below 30, with only 10 tries with each method, you'd get anomolies that wold otherwise average out with more iterations.)

[Commie Pinko Dirtbag]

13. Are you my lost twin?

http://www.democraticunderground.com/discuss/duboard.php?az=show_topic&forum=105&topic_id=2795493#2795556

[Ready4Change]

14. Programmers think alike, I think.

But being twins is far cooler. ;)

[aePrime]

16. Can be proved with probabilty theory

But this is entertaining.

#include <iostream>
#include <ctime>
using namespace std;

int chooseDoor()
{
  return rand() % 3;
}

int otherDoor(int notMe)
{
  int door = notMe;
  // Efficiency, thy name is randomness (come on, just a geometric rv, which is
  // guaranteed to evaluate to a success as the limit approaches infinity :)).
  while ((door = chooseDoor()) == notMe)
    ;

  return door;
}

int lastDoor(int notMe1, int notMe2)
{
  if (notMe1 == notMe2)
    return otherDoor(notMe1);
  else
    return 3 - (notMe1 + notMe2);
}

int main()
{
  srand(time(0));

  int numRuns               = 0;
  int numCorrectSwitched    = 0;
  int numCorrectNotSwitched = 0;

  for (numRuns; numRuns < 5000; ++numRuns) {
    int prizePosition = chooseDoor();
    int ourChoice = chooseDoor();

    // What door will the host open? The one that we didn't choose, and the one
    // without the prize.
    int hostChoice = lastDoor(prizePosition, ourChoice);

    // Now we switch. Pick the door that's not the host choice, and not the
    // one we had before.
    int newChoice = lastDoor(hostChoice, ourChoice);

    if (ourChoice == prizePosition)
      ++numCorrectNotSwitched;
    if (newChoice == prizePosition)
      ++numCorrectSwitched;
  }

  cout << "Percentage without switching: " << static_cast<double>(numCorrectNotSwitched) / numRuns << endl;
  cout << "Percentage with switching: " << static_cast<double>(numCorrectSwitched) / numRuns << endl;
}

[crispini]

6. Oh, not THIS.

It always pays to switch, but I don't remember why. I think I heard this one on Car Talk.

Damn, now I'm going to have to go look this up!

[Crazy Guggenheim]

10. The person who didn't switch won the goat. But they still won ..

because the person was a Repuke.

:rofl: :thumbsup: :toast:

[Worst Username Ever]

11. Supposedly, yes.

But I don't remember why.

[jmowreader]

15. Well, it all depends...

If you take the money, you know you have the money.

If you DON'T take the money, you have a fifty-percent chance of winning a gorilla.

(An aside: does anyone remember the LMAD episode where the guy won a gorilla, had a chance to trade his gorilla in on a shot at one of the three doors, and kept the gorilla?)

Now here's the REAL question: if you didn't take Monty's money and Door Number 2 had either a gorilla or Officer O'Malley of the FBI (who's normally behind Door Number 1), what would you do with your prize?