Behind Monty Hall's Doors: Puzzle, Debate and Answer?

[chris319]

Even though the die is rolled one time you still can know what to expect, if not the result.

Here's your one roll. What is the outcome? All you know is that you have a 5/6 chance of being wrong.

just because it's a single trial doesn't suddenly make it a 50/50 proposition.

The house odds are 66.7% assuming the contestant switches every time, regardless of the number of trials. If the player doesn't switch every time the odds of winning are 33.3%. The distinction has to be made between the house odds and the player's odds.

It is never an incorrect strategy to switch.

[TLEberle]

Even though the die is rolled one time you still can know what to expect, if not the result.

Here's your one roll. What is the outcome? All you know is that you have a 5/6 chance of being wrong.

Right, but so what? There are n different results for an n-sided die. The outcome is unknown until the event is carried out.

[chris319]

Now to blow the lid off my blowing the lid off.

In all of the circumlocutious explanations of the MHP I've seen, I've never seen it explained in the following simple manner. Suppose the doors are laid out such that the car is behind door 2. Doors 1 and 3 conceal zonks. Now consider the following truth table:

PICK     REVEAL     REMAINING     SWITCH     NO SWITCH

  1         3         1 OR 2      2 (Win)     1 (Lose)

  2         1         2 OR 3      3 (Lose)    2 (Win)

  3         1         2 OR 3      2 (Win)     3 (Lose)

By switching, the only way to lose is to pick the door with the car and switch, giving a 1/3 chance of losing and a 2/3 chance of winning. By not switching, the only way to win is to pick the door with the car, giving a 1/3 chance of winning.

Simple.

/If the contestant picks door 2, either door 1 or 3 can be revealed. It works out the same in either case.

//Headache any better, Brandon?

[dale_grass]

PICK     REVEAL     SWITCH     NO SWITCH

  1         3         2 (Win)     1 (Lose)

  2         1         3 (Lose)    2 (Win)
 
  2         3         1 (Lose)    2 (Win)            

  3         1         2 (Win)     3 (Lose)

There's nothing more fun in a math class than those extension problems.  The table above looks like the probability of winning after a switch is 1/2, but it's still 2/3.  Who wants to take a crack at explaining why?  (This will be on the final.)

[Denials]

Because two of the outcomes are mutually exclusive, right?

[BrandonFG]

/If the contestant picks door 2, either door 1 or 3 can be revealed. It works out the same in either case.

//Headache any better, Brandon?

No. Where's my Advil?

[Kevin Prather]

There's nothing more fun in a math class than those extension problems.  The table above looks like the probability of winning after a switch is 1/2, but it's still 2/3.  Who wants to take a crack at explaining why?  (This will be on the final.)

Let me see if I'm way off base here.

At the very beginning, you have a 1/3 chance of picking the right one, which means you have a 2/3 chance of picking the wrong one. Revealing one of the wrong doors does not change that at all. You're still 1/3 of the time right, 2/3 of the time wrong. The only difference is now you know which one to switch to if you want to switch, and since you're wrong 2/3 of the time at the start, you should always switch.

[chris319]

PICK     REVEAL     SWITCH     NO SWITCH

  1         3         2 (Win)     1 (Lose)

  2         1         3 (Lose)    2 (Win)
 
  2         3         1 (Lose)    2 (Win)            

  3         1         2 (Win)     3 (Lose)

There's nothing more fun in a math class than those extension problems.  The table above looks like the probability of winning after a switch is 1/2, but it's still 2/3.  Who wants to take a crack at explaining why?  (This will be on the final.)

You're double counting the door 2 pick. There are only three options: door 1, 2 or 3. If you want to examine the outcome of picking 2 and revealing 3, here's how to do it:

PICK     REVEAL     REMAINING     SWITCH     NO SWITCH

  1         3         1 OR 2      2 (Win)     1 (Lose)

  2         3         1 OR 2      1 (Lose)    2 (Win)

  3         1         2 OR 3      2 (Win)     3 (Lose)

This is better analyzed as a logic problem than as a math problem.

/Brandon: We could analyze this some more until your headache is better.

[chris319]

Let me see if I'm way off base here.

At the very beginning, you have a 1/3 chance of picking the right one, which means you have a 2/3 chance of picking the wrong one. Revealing one of the wrong doors does not change that at all. You're still 1/3 of the time right, 2/3 of the time wrong. The only difference is now you know which one to switch to if you want to switch, and since you're wrong 2/3 of the time at the start, you should always switch.

Without switching your chances of winning the car are 1/3. Switching inverts the logic of the outcomes and gives you the 2/3 odds. Switching gives you the logical NOT (inverse) of the outcomes.

/Need more Advil?

[Matt Ottinger]

At the very beginning, you have a 1/3 chance of picking the right one, which means you have a 2/3 chance of picking the wrong one. Revealing one of the wrong doors does not change that at all. You're still 1/3 of the time right, 2/3 of the time wrong. The only difference is now you know which one to switch to if you want to switch, and since you're wrong 2/3 of the time at the start, you should always switch.

You've got it.

[dale_grass]

You're double counting the door 2 pick.

Not when you count the pick and reveal as the experiment.  Then there are two outcomes associated with picking the correct door: either one of the two remaining doors can be revealed by the host.  Thus, there are 4 elements in the sample space.  The catch is that the two middle rows in my table each have probability of 1/6: 1/3 you'll pick Door 2 and 1/2 Monty will reveal 1 or 3.

[chris319]

You're double counting the door 2 pick.

Not when you count the pick and reveal as the experiment.  Then there are two outcomes associated with picking the correct door: either one of the two remaining doors can be revealed by the host.  Thus, there are 4 elements in the sample space.  The catch is that the two middle rows in my table each have probability of 1/6: 1/3 you'll pick Door 2 and 1/2 Monty will reveal 1 or 3.

I don't see a bright future for you in the fields of statistics or logic.

[dale_grass]

You're double counting the door 2 pick.

Not when you count the pick and reveal as the experiment.  Then there are two outcomes associated with picking the correct door: either one of the two remaining doors can be revealed by the host.  Thus, there are 4 elements in the sample space.  The catch is that the two middle rows in my table each have probability of 1/6: 1/3 you'll pick Door 2 and 1/2 Monty will reveal 1 or 3.

I don't see a bright future for you in the fields of statistics or logic.

I'll pass that along to my graduate advisor in the math department.

[chris319]

You're double counting the door 2 pick.

Not when you count the pick and reveal as the experiment.  Then there are two outcomes associated with picking the correct door: either one of the two remaining doors can be revealed by the host.  Thus, there are 4 elements in the sample space.  The catch is that the two middle rows in my table each have probability of 1/6: 1/3 you'll pick Door 2 and 1/2 Monty will reveal 1 or 3.

I don't see a bright future for you in the fields of statistics or logic.

I'll pass that along to my graduate advisor in the math department.

If I were you, I'd keep it a secret.

[beatlefreak84]

You're double counting the door 2 pick.

Not when you count the pick and reveal as the experiment.  Then there are two outcomes associated with picking the correct door: either one of the two remaining doors can be revealed by the host.  Thus, there are 4 elements in the sample space.  The catch is that the two middle rows in my table each have probability of 1/6: 1/3 you'll pick Door 2 and 1/2 Monty will reveal 1 or 3.

I don't see a bright future for you in the fields of statistics or logic.

I'll pass that along to my graduate advisor in the math department.

If I were you, I'd keep it a secret.

Now hold on a minute here...I've taught the Monty Hall problem in various math classes for years, and, in all three textbooks that I have used that explained how it works, what Dale wrote is the exact explanation they use.  You can even go more general and show that, even if the probability that Monty opens either of the two doors in the "bad" case is not equal, you will still get the same 2/3 probability winning when switching.

It's basically an exercise in conditional probability and/or in using tree diagrams and the "multiplication rule."

Anthony