What's Behind Door Number Three?

For detailed explanation :

http://en.wikipedia.org/wiki/Monty_Hall_problem

Or movie explanation:

http://www.youtube.com/watch?v=mhlc7peGlGg

Depending on how deep you want to get into this, you might consider reading a recent book:

Digital Dice: Computational Solutions to Practical Probability Problems, Paul J. Nahin. Princeton University Press.

The so-called Monty Hall problem is one of the problems Nahin shows how to solve using simulations. BTW, he agrees with Marilyn!

Thanks!

Monty

I know this is an old thread, but I've thought of a simple explanation which I've never seen anywhere.

If the player chooses (say) door 1, there is 1/3 probability the car is behind it, 2/3 probability the car is behind one of the other doors. When Monty opens one of the other doors (say door 2) and reveals a goat (which he does because he knows where the car is) there is 2/3 probability the car is behind door 3. So better to switch.

Say there are two players. The first player chooses door A. The second player chooses door B. Monty Hall opens door C and reveals a goat. Should the first player switch to door B? Should the second player switch to door A?

Why would the probability of a car being behind a door change based on its popularity or the act of switching before a reveal?

It seems like the explanation is suggesting the players should each switch, but the true probability can't be different depending on who is looking at the door.

It makes no difference, it's 50:50 whether they switch or not. Because player A or B must have chosen the car - if both had chosen a goat the car is behind door C, so the statement Monty opens door C and reveals a goat is impossible.

But you see the problem, right?

Imagine player one and player two are playing remotely and neither is aware the other is playing. From either players perspective it is indistinguishable from the original problem. While you and I, the observers, realize that one of the players has won the car and the chance is 50/50, each player if we assume the above explanation is correct should switch for a 2/3 chance of winning the car....which of course is impossible.

The fact that another player has chosen the other door doesn't really have a bearing on the chance a car is behind a particular door.

It seems that once Monty opens the third door revealing a goat there are only two doors left. Whether someone has bet on one door or multiple people have bets on each door, the probability should be 50/50, The idea that if Monty opens C then anyone originally betting on A should switch to B while simultaneously anyone originally betting on B should switch to A would only help one group as much as it hurt the other but before finally knowing, it doesn't seem to change anyone's odds.

No, I don't see a problem. In your scenario if I understand it, the players don't have a 50:50 chance of winning, it's possible neither will win. Each player has a 1/3 chance of winning if he doesn't swap, 2/3 if he does.

In the original game, if Monty did not open another door with a goat (which he knows about) but just asked the player if he wants to swap, whether the player swaps or not, when his chosen door is opened there is still a 1/3 chance of the car. It's Monty opening a door with a goat that changes the odds, as in my #4.

No, in my scenario it is not possible that neither player will win once a goat is revealed by Monty. They cannot both have a 2/3 chance of winning by switching places.

Let's do this: assume a scenario wherein no player knows about other players, 3 players, 3 doors, 1 car, 2 goats. 1st player chooses door A, 2nd player chooses door B, 3rd player chooses door C. Monty opens door C revealing a goat and the 3rd player loses and is out.

Now the scenario appears as before to the 1st player and to the 2nd player.

Are you saying that the second player should switch to door A? Are you saying the first player should switch to door B?

Are you saying that switching will increase of their odds of winning?

Are you saying that if they switch it affects the odds of the car being behind a certain door?

Are you saying that there can be more than one door with > 50% odds to have the car behind it?

You - "No, in my scenario it is not possible that neither player will win once a goat is revealed by Monty"

Assume both players are present. If the car is in door 3, player A picks door 1, player B picks door 2, neither wins if neither swaps. But Monty cannot open a door with a goat, which hasn't already been picked, so he can't offer a swap option. So I wouldn't expect Monty to have a 2-person game.

If he does, and both players pick the same door, say door 1, Monty can open door 2 or 3 with a goat. If player 1 doesn't swap, he has a 1/3 chance of winning. If player 2 swaps, he has a 2/3 chance of winning (or vice versa). When the door with the car is opened, one of the players gets it. If both swap, both have a 2/3 chance of winning. That's OK, they'll have to sort out ownership between themselves.

You - "They cannot both have a 2/3 chance of winning by switching places."

They can, as explained above.

If neither player is aware of the other, it's no different from having the game on 2 separate occasions. There must be 2 cars available in case both win it.

You - "Let's do this: assume a scenario wherein no player knows about other player"

As above, it's like having n separate games. There have to be n cars available in case all win.

I don't see why you've brought 3 players into it, let's stick with 1 or 2 for now.

The reason I brought a third player in was in an attempt to help you see that the choice of the door did not affect the odds of that door having a car or goat.

I don't actually understand how you have arrived at your argument. You seem to place limitations without reason on what Monty might be able to do.

Maybe it might become more clear if we take the firat choice away. Assume there are 10 doors and 10 contestants are given a ticket with a number 1 through 10 corresponding to each door such that each door has only one ticket. There is one car behind one door and the rest have goats. Monty opens 8 doors with goats behind them and then asks the only two players left with unopened doors if they want to switch doors.

If they did not have knowledge of the other players then the scenario would look very similar to the original problem to each player (except with even better odds for switching if the original solution is to be believed).

Lets say door 1 and door 2 are the remaining unopened doors and player A and player B hold the ticket to each respectively.

According to the given the original solution player A should switch from door 1 to door 2 while player B should switch from door 2 to door 1, supposedly because there are increased odds of winning if they switch. How can there be greater than a 50% chance for more than one of the remaining two doors?

Before we go any further, do you agree that in the game as normally played, it's better for the player to swap?

I do actually. I'm suggesting these variations because they don't seem to differ by anything that should reasonable affect the probability/outcome, yet I wouldn't think switching should improve odds in these variations.

I'm not trying to dissuade anyone of the accepted answer for the normal set up. I'm having trouble wrapping my head around why the variations could he different though.

OK thanks. I'll get back but busy with the season.

Merry Christmas

Still thinking about this and getting brain ache.

But what I about limitations on what Monty might be able to do, I meant in a 2-person game both players might pick a goat. I make the probability of that 2/3*1/2 = 1/3. In that case he can't open a door to show a goat, and as he needs to have control he wouldn't offer the game.

Rational thinking, 3 doors 2 contestants.
Each Contestant has 3 choices, so 6 potential choices / 2 = 3 so each has 1/3.
Eliminate one door for each contestant we are down to 2 doors for each.
Contestant A chooses door 2, Contestant B chooses door 1.
It is irrespective if one or both contestants swop the door or remain as the original choice, the odds are still 50-50. The odds do not increase or decrease, even if one contestant drops out in the first round the remaining choice for the contestant is still 50-50.
End of the day there was always a "backstage shuffle of the prize" because if a contestant picked the car on the first shot there was no more show !

You are disregarding one element of the activity: The order in which ‘What is behind curtain number XXX?’ is revealed.

The host is aware of the location of the big prize.

At least one contestant (perhaps both of the contestants) has selected a fizzle prize. If both have selected a non-winner curtain, the host must select one contestant to have that fact revealed first. Is this selection by host whimsy, or is a coin flipped backstage and the host prompted to select the flip winner?

OR… is a three face coin flipped and the big prize reveal is selected with this method. Open curtain #1 (a Chrysler Cordoba, with rich Corinthian leather upholstery), and both contestants take home a sheep?

Why assuming scenarios - 2 players, 3 players, 5 players? The main is 1 host and 1 player, and 3 doors. If the player doesn`t change its choice, the odds to win the car is 1/3. If the player want to change his choice, the odds to win the car is 2/3.

Simple!

That's the standard Monty Hall problem, there's no dispute about the odds there. But TINAC has brought up a variation with 2 (or more ) players.

Apparently, not so simple.

It is a bit complex to understand...

I will try this way -

Original Probability is 1/3 that my selection has Car.

Which means 2/3 prob that it doesn't have Car.

Right?

The Compere knows which has car and which doesn't have - so he opens a door, which he KNOWS doesn't have car. Right ?

Now what happened ?

The probablity of 1/3 and 1/3 between each of the two doors now gets merged (since one of them which doesn't have car is opened) - and the left out door has 2/3rd Probability of having car.

This is theory - but it also has a lacuna. I am not very sure whether this had been ever practically carried out on a sufficiently large scale to confirm.

Much like what I said in #4