Quoth Boris B:
Hear hear. What is it about this problem that makes so many so willing to argue, yet unwilling to actually try it out? It’s not as if you need a particle accelerator to set up an experiment. Any yutz can try this out with no more equipment than pencil, paper, and a six-sided die. And the results will be extremely convincing. Even if the logic of the problem isn’t convincing, watching the always-switch strategy win twice as often ought to do the trick.
You were doing great until you got to the second part of step 4 there. Supposing what you said in step four was true, it’s plain to see that there would be only a 2/3 chance that a prize even existed! (2 doors * 1/3 chance of each door winning = 2/3 chance that some door is a winner.) That’s not the game we’re playing here - in our game, there is always (1/1 chance) a prize somewhere. The odds that each unopened door is a winner must add up to 1.
If this is really imporant to you, (step 5 sure seems to be a passionate one) try it out with pencil and paper. Have a friend play Monty, if that helps you.
Or try this: Play the game with 1000 doors. You pick one, and Monty always opens the other 998 that don’t have a prize left behind them, leaving one left. Do you really think there is no advantage to switching? Thinking so is tantamount to believing that you’ll always have a 50/50 chance of picking the right door out of 1000.
The problem can indeed be reduced to, “You can keep the door you picked, or exchange it for ALL the others.” This is the case because out of all the others, Monty eliminates all the incorrect choices for you.
The question gets more interesting if Monty only chooses to open one of the other doors some of the time. Then it becomes a matter of game theory.
And this is where it really gets interesting. At least more so than trying to convince people who post to a public board shooting from the hip, without bothering to put pencil to paper first.
The way the problem is usually stated, you don’t know that Monty will always offer you the choice, or what his motivations might be. If you look at the problem from a game theory perspective and assume that you want to get the good prize and Monty wants you to get a year’s supply of chicken feathers, then if you follow a pattern such as always switching when you have the choice, Monty will start offering the choice only when you’ve picked it right on the first try. Since Monty has the advantage in that he knows where the prize is, the best you can hope for is having a 1/3 probability of winning, which happens if you stick.
If you sometimes stick and sometimes switch, then Monty will offer you the opportunity to switch only when it is advantageous to him, resulting in less than a 1/3 win rate for you. You should always stick with your first choice. And if you always stick, then Monty will not even bother to offer you the choice.
That’s how it would happen in a competitive game. Of course, his real goal is to make the show exciting, so he’d probably offer the switch in a way to make it average out to 50/50.
Note to poor readers: I am in no way suggesting that the probability, assuming he always offers the chance to switch, will be any other answer but switch by 2/3 to 1/3.
Here is another thread on this topic.
http://boards.straightdope.com/ubb/Forum1/HTML/000465.html
Anyone who still doubts the outcome, try it yourself.
Take a deck of cards, have a friend play “Monty”. He gets to see all three cards, and knows which is the ace (car), vs. the two jokers (goats).
You pick a card, then he must reveal a card. (Note: he is not allowed to circumvent the game by revealing the ace.) Thus you know he will reveal a joker.
Or play with the whole (52 card) deck, with the Ace of Spades as the winning card. You pick one, Monty turns over 50 remaining non-ace cards. Do you switch?
Play for money - you’ll learn the lesson better that way.
Hey, Irishman, that’s a good idea. This whole thing kind of reminds me of certain gambling card games. The ones like Blackjack and Stud Poker where you have to guess what cards your opponents have based on the ones you can see, but your guess changes each time you see more.
So I guess five-card stud is a better example. Your opponent has an Ace and a down card showing; you have two Kings. He’ll win is his down card is an Ace, which makes you nervous until more Aces appear in other people’s hands. Did the appearance of those Aces change the identify of the down card? No, it just changed your best guess.
I thought of another way to explain this, not that Lagged and Manduck and the rest of us haven’t already posted a lot. You guess at a door without opening it; it has a 33% chance of being right. Since there’s exactly one prize, the remaining two doors have a combined 67% chance of being right. Monty opens one, subtracting 0% from final door’s probability. The final door has a 67% chance of being right; your original guess keeps its 33%.
Before anyone says anything, yes the probability of winning is 2/3 if you switch.
However lagged2death wrote <quote>What is it about this problem that makes so many so willing to argue, yet unwilling to actually try it out?</quote> Perhaps that it is that statistical analysis is in no way equivalent to logical proof. This problem is amenable to both, however only one shows that it is correct the other merely suggest that it is probably so.
BTW further kudos to manduck that is one of the simplest, most elegant explanations I have ever seen. Despite prior experience of humans, I still couldn’t believe people were still disagreeing after that.
I agree, Rorschach, that a experimental run will not “prove” the problem in a mathematical sense. But hopefully, it would shake some of the doubters out of the certainty they seemed locked into. Once they see the evidence that switching is the winning strategy, they might consider the logic that backs this up. And if not, at least they’ll have stopped pestering us for awhile.
I’m not trying to convince anyone of anything regarding the Monty Hall problem.
Just play the game for money and you will see that Marilyn was right.
I want to offer another version of the somewhat simliar sibling problem that has a
slight twist which makes clear how important subtle concrete details are
in stating the problem.
Suppose you meet a girl on the street and learn she has one sibling. What is
the probability that the sibling is a boy? 50-50. Why? Here are the possible
scenarios.
On the street, you met either girl-old, girl-young, girl1, or girl2. If you met
girl-young or girl-old, the sibling is a boy. If you met girl1 or girl2, the sibling is
a boy.
I’m sorry for posting so carelessly. Also sorry for beating a dead horse.
i didn’t take a whole lot of statistics at university, but i took enough to have encountered this little scenario before. basically, your opinion on this issue is determined on how you answer the following question:
do you believe in the law of averages?
according to the law of averages, if the first child is female, then the other must be male. if you keep flipping heads, you’re bound to flip tails soon, right?
well… the classic response is to treat each instance as a discrete event and say no. this avoids the classic fallacy of confusing performance and probability… if you keep flipping heads, maybe there’s a greater probability of heads, right?
thing is, as pedestrian as the law of averages is, there is really is some substance to it, even with smaller samples than you would think. just ask a casino. in other words, marilyn is technically right… although in practicality, it can go either way for a while.
This is the toughest part of the problem. Why does it matter if Monty knows what is behind the door?
First, think of the original problem. The initial reaction of most people is that the deliberate opening of an empty door tells you nothing, because no matter what it is always possible to open a empty door.
This is actually very close to being right - it tells you nothing about the door you already chose, because indeed, there was no chance that one of the other doors was not empty, so there was no chance Monty could not choose to open an empty door.
However, if you learned nothing about your chosen door, then the chances that it is correct stay the same. You still have a 1/3 chance of having chosen correctly.
But what if it was completely random? What if any of the three doors could have been opened?
In that case, if an empty door other than the one you chose is opened, you DO learn something about your choice. It becomes more likely to be correct.
This is because, by random chance, 1/3 of the time the winning door will be opened. Since it was not, we must be in one of the other 2 scenarios, 1 of which has your door being correct. A 1/2 chance.
And now the scenario which I don’t think has been brought up in any of these threads before (though I was unable to search for all of them).
What if Monty chooses randomly, but only among the doors you did not choose, and opens an empty door?
It seems to me that in this case, you would actually be better off staying with your original choice.
If one of the other doors was the winner, then half the time Monty would unveil the winner.
But if both the other doors are losers, then every time he will open a loser.
Therefore, 2/3 of the time, if Monty opens an empty door, your original choice was correct!
I don’t think this has been brought up in previous threads, and I hope someone can confirm or disprove this last line of thought.
The reason is that choice 4 becomes more likely, because you can meet either one of the girls.
It becomes:
1: 0%
2: 25%
3: 25%
4: 50%
In the original problem it’s:
1: 0%
2: 33%
3: 33%
4: 33%
I’ll do an experiment on the Monty Hall problem. Feel free to try this yourselves if you don’t trust me.
I’ll do the experiment 20 times. First I’ll choose by dice roll which doors are correct.
3, 1, 2, 3, 3, 1, 2, 3, 2, 2, 3, 1, 2, 1, 2, 3, 3, 3, 2, 3
Which door you choose at first doesn’t matter, so I’ll choose it at random by dice roll.
2, 1, 1, 2, 3, 2, 3, 2, 3, 1, 2, 1, 3, 1, 3, 3, 1, 3, 3, 2
Then I’ll start playing. I’ll switch doors every time.
1:
I choose 2, Monty opens 1, I switch to 3. Win
2:
I choose 1, Monty opens 2, I switch to 3. Lose
3:
I choose 2, Monty opens 3, I switch to 2. Win
4:
I choose 2, Monty opens 1, I switch to 3. Win
5:
I choose 3, Monty opens 1, I switch to 2. Lose
6:
I choose 2, Monty opens 3, I switch to 1. Win
7:
I choose 3, Monty opens 1, I switch to 2. Win
8:
I choose 2, Monty opens 1, I switch to 3. Win
9:
I choose 3, Monty opens 1, I switch to 2. Win
10:
I choose 1, Monty opens 3, I switch to 2. Win
11:
I choose 2, Monty opens 1, I switch to 3. Win
12:
I choose 1, Monty opens 2, I switch to 3. Lose
13:
I choose 3, Monty opens 1, I switch to 2. Win
14:
I choose 1, Monty opens 2, I switch to 3. Lose
15:
I choose 3, Monty opens 1, I switch to 2. Win
16:
I choose 3, Monty opens 1, I switch to 2. Lose
17:
I choose 1, Monty opens 2, I switch to 3. Win
18:
I choose 3, Monty opens 1, I switch to 2. Lose
19:
I choose 3, Monty opens 1, I switch to 2. Win
20:
I choose 2, Monty opens 1, I switch to 3. Win
14 out of 20 wins. That’s more than 2/3.
You’ll see that the only times I lost was when the door I chose was the winning door. That was 6 out of 20 times. If you don’t trust my dice rolls, then try it yourself.
Another way to look at it is like this. You can choose door 1 OR you can choose whichever one is correct of door 1 or door 2. You do this by first standing in front of door 3 and let Monty open the false one of door 1&2. Then you switch back to the one that Monty didn’t open. That way you’ll win if the prize was originally behind either door 1 or door 2.
The easiest way to look at it is that you have the choice of picking one door for yourself and leaving Monty 2 to play with or you choose one door for Monty and you open the other 2. Which strategy sounds like it would win 2/3 of the time?
Ahhh, the yearly drudging out of the Monty Hall question. Is it 2004 already?
That is the easiest way, but you have to be careful if you look at it that way.
Remember, it depends on whether or not Monty knew what was behind the door.
If Monty is choosing randomly between the doors you didn’t pick, then it is very different.
In that case, if you choose door number 3, and Monty opens door number 2 showing it to be empty, you have a 2/3 chance of winning if you stay with your original door.
The concept also applies to the lottery problem. Say there are 1000 lottery tickets, one and only one winner, and you take one ticket and leave.
All the rest of the lottery tickets are thrown in a big pile on the floor, and people begin picking them up randomly and seeing if they are winners.
By the time there is only one ticket left on the floor, should you switch to take that ticket?
After all, there was only a 1/1000 chance that your original ticket was right.
But no… you should not switch. In fact, the odds are overwhelmingly in your favor that the ticket you already have is the winner.
Nightime, as near as I can tell, that’s backwards. If the choices are random, then there is no advantage to staying or switching.
Thank you for answering. You are right, I had not thought it through enough.
If the door opening is random, then there is no advantage to staying or switching.