Lets Make Another Deal!!

Factual Questions
1

I submitted this to cecil, and he did not understand my question…

ok, i admit my original question lacks fluidity… :slight_smile:

I will try again…

Monty starts with three people. Each of these people choose a different door.
So you have three people all choosing a different door. Monty then chooses a LOSING door at random. (not that he randomly chooses any door, he picks one from the two losing doors). So in essence, its the same as if you were the only one playing, and he randomly chooses a loosing door. Based on how I understand the original problem, switching doors increases your odds of winning to 2/3. Now lets say that the two remaining people (one picked the winning door, and the other picked a losing door) read “The Straight Dope”. They know if they switch, they can increase their odds too. If the two people switch, how can both people have odds of 2/3. Is it possible to have a total of 4/3 chance?

for the love of all things holy, please answer me!!!

slowy losing friends over this!!
Here is my original post:
Now that I have lost a few days and probably some friendships, I think I understand the answer of whether or not to choose the other door in “Lets Make a Deal.” Instead of being content, I thought up a small dilema that I cannot figure out. Lets say you start with three people, all of which choose a different door. Then Monty RANDOMLY chooses a losing door, and opens it. That person goes home. So you have 2 people with different doors. Since they are avid readers of the Straight Dope, they know to switch. My question(s) is this. How can both people have a 2/3 chance of winning. Is it possible to have odds greater than 3/3 for someone to win? Or is it 50/50 since they both have the same odds? Or, is this a completly different problem, and I am crazy?

REPLY BACK ----------------------

Your question doesn’t make any sense. One of the two of them will win, whether there’s anything behind a door or not. At this point, the chances for either one of them is 50/50.

In the initial case, there are two scenarios: the person wins (2/3 times) or the person doesn’t win (1/3 times.) Once you have three people, one per door, the game is different – one of those people will win 100% of the time.

In the initial case, Monty doesn’t open a door at random. He KNOWS where the money is, so he opens a door deliberately that he KNOWS doesn’t have the money. In your case, if Monty opens a door randomly, he might open the door and get the winner, and say “You win!” and the game is over. THe other two can switch doors or not, they both lose.

THe difference (and a somewhat subtle one) in the problem is whether Monty knows where the money is and deliberately opens one door that doesn’t have it, or whether Monty is just opening at random.

Imagine the situation slightly different. Instead of three doors, imagine 52 cards, face down, and you win if you pick the ace of spades. Scenario 1: there’s one contestant, you pick a card. Monty turns over 50 othe cards, none of them the ace of spades; you have your card (picked at 1/52 odds) and there’s one other card, do you stick or switch? Obviously, you switch.

Now your scenario: There are 52 cards and 52 people, each one picks a card. Monty starts turning over cards randomly. The chances are, he’ll hit the ace of spades somewhere along the line and there’ll be a winner.

OK?

If you need more discussion, you might go to our website at www.straightdope.com, to the section called “Message Boards”, to ask your question.It might stir up an interesting discussion, anyhow.

2

The short answer is that the presence of other players is irrelevant to the player who is offered the chance to switch. They end up with a door–doesn’t matter whether they chose or got leftovers–and there’s a 1/3 chance that the prize is behind it. Monty opens another door and offers one player the chance to switch. At this point the other player will switch if and only if the chooser switches, so there’s no extra information from knowing that the other player is present.

What I think you’re asking is whether it’s to the advantage of the other player to switch. Assuming that Monty acts so that the switcher has an advantage, it’s to the disadvantage of the other player. Each wins if and only if the other loses, so they’re complementary events, and their probabilities have to sum to 1. So if the probability that the switcher wins by switching is 2/3, the probability that they will lose (and the other player wins) is 1/3.

Is that at all clear?

3

There is no advantage to switching in this case.

The difference is that in this scenario when Monty is picking a door to open, he may pick YOUR door. The possibility of that happening changes the odds.

Forget the other two players. Imagine just you and Monty and the three doors. You pick one and then Monty randomly selects one of the losing doors to open. Unlike the way the problem is normally stated, there’s a chance he’ll pick your door.

Your initial pick:

1/3 chance you picked the prize
2/3 chance you picked the goat

Monty randomly opens a goat door:

1/3 chance you picked the prize on your first pick – you’re safe
1/3 chance you picked the goat but Monty doesn’t pick your door
1/3 chance you picked the goat and Monty picks your door – you’re out

If you’re NOT eliminated at this point the odds that you picked the prize on your initial guess are the same as the odds that you picked the goat on your initial guess. There’s no advantage to switching.

Basically, when Monty picks without including your door, he’s giving you more information that changes the odds. When he includes your door, he’s not.

4

For those following along at home, Pochacco and I are both right. I’m analyzing the game from the perspective of the player who is offered the chance to switch, and he’s analyzing the game from the perspective of a particular player. They aren’t necessarily the same.

5

The difference with three people picking three separate doors is that their choices are not independant. Player 1 picking the red door prevents players 2 or 3 from picking the red door. If they were each allowed to independantly choose a door, there would be 27 distinct scenarios. Instead, there are 6 scenarios. In each scenario, there are two different doors that Monty could pick, giving 12 different ways that Monty could select a given door. For player 1, if Monty picks his own door, he’s out of luck. If Monty doesn’t pick his door, there are 8 remaining scenarios; four where his current door wins, and four where it loses. Thus, his chances of winning if he doesn’t switch doors is 50%, just like the other remaining player.

6

The way I understand it.

The first player to choose gets a choice of 3 doors. (Even if it’s random)
The second player to choose only gets a choice of 2 doors. (Even if it’s random)
The third player doesn’t get a choice of doors.

This complicates the issue somewhat, since you can wind up in 3 equally likely possible final states:

a) Player 1 lost, b) Player 2 lost, or c) Player 3 lost when the losing door was opened.

a) Player 2 had a choice of 2 doors and their initial choice had a 1/2 chance of picking a winning door. They’ve gained no choice and no information. It is not advantageous to switch for either Player 2 or Player 3

b) Player 1 had a choice of 3 doors and their initial choice had a 1/3 chance of picking a winning door. Then it is advantageous for player 1 to switch with player 3, and since player 3 had no choice to begin with but got stuck with the good choice.

c) Player 1 had a choice of 3 doors and player 2 had a choice of 2 doors (one of which turned out to be losing). Player 1 had 1/3rd of a chance of picking the winning door and Player 2 had 1/2nd, so Player 2 is stuck with the good choice and it’s again advantageous for Player 1 to switch.
The answer is: If you are player 1, you always switch. If you are not player 1 you always stay.
Did I get it right?

7

No. Even for player 1 it doesn’t help to switch.

The key is that Monty may pick your original door to open in this version of the problem.

In the normal version of the problem Monty never picks your original door.

That detail changes the final odds from 33/66 to 50/50.

8

No.

Player 1 has a 1/3 chance to pick the right door, and 2/3 chance to pick the wrong door. In the “standard” Monty Hall problem, Monty will never pick the door chosen by the player. In this situation, Monty will choose player 1’s door 50% of the time if player 1 chose incorrectly. Therefore, if Monty doesn’t open his door, there is a 50% chance that he originally chose the correct door.