Earlier in this thread, there was talk about the “two boys” using the concept of “definite female” “chance male”:
D(F) C(F)
D(F) C(M)
C(F) D(F)
etc.
It should be noted that this analysis is complete nonsense.
For it to have any meaning, the groups used must be equal sized and distinct (non-overlapping). Neither trait is true for those grouping (e.g., “D(F) C(F)” would have a large overlap with “C(F) D(F)”)
The reason why some people think it’s 2/3 instead of 1/2 is because the folks who believe it’s 2/3 are NOT recalcuating probability after eliminating a possibility. Here’s a different coin example: What are the odds of flipping a coin and it landing on heads 10 times in a row? 1 over 2^10 right? Well what if you flip a coin 9 times and it lands on heads all 9 times? What are the chances that you get a heads if you flip again? It’s not 1 over2^10, it’s 50%.
After you eliminate possibilities, your odds change. It’s right in the number…you don’t have a 2/3 possibility because you no longer have three choices, you have a 1/2 chance because you now only have two possiblities.
People who say that you start off with a 2/3 chance of the car being in the “other door” are thinking about the problem wrong. They are judging the odds of “the other door” as if you hadn’t eliminated all the other “other doors”
Let’s try this another way: with 52 cards instead of 3 doors.
I’m the host. I have a deck of 52 cards. They’re marked, and you know they’re marked, but I know how to read them and you don’t.
The object of the game is to pick the queen of hearts.
I fan out the cards face down. You pick one randomly and place it face down in front of you. Then I (who, remember, can read all the cards from the back) pick a second card and place it face down in front of me. Then I turn over all of the remaining cards, in order to show you that the queen of hearts is not still in the deck.
So… do you still believe that the odds are 50/50 that the card you chose *randomly *out of 52 is the queen of hearts? Or do you want to switch to the card that I chose *deliberately *out of the 51 remaining?
And the reason for that is that the odds don’t change. The host knows which door has the prize, and the probability is 100% that he’s going to open a door without a prize. The only element of chance in the scenario is your initial choice, which is 1/3. The odds are 2/3 that it’s behind one of the other two doors, and the host is helpfully eliminating one of those choices for you.
If he were to randomly choose which of the two doors to open, and it just happened to be one without a prize, then and only then would the odds change to 1/2. But we’re told at the start that the door he opens will be one without the prize, so there’s nothing random about it.
Bingo. There’s a big difference between “The Monty Hall Problem,” an abstract probability puzzle, and advice for what to do in the real game, because in the real game Monty has more options on what to do and can employ a strategy. In the abstract puzzle, Monty has no choices and can employ no more strategy than a blackjack dealer. (There are other differences, too. It’s not two goats and a car, it’s a goat, a nice washing machine, and a car.)
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You’re right; see above. 2/3 applies to the abstract puzzle, not the show. But I hope you can see the 2/3 answer for the abstract puzzle, thanks to Marylin’s extension to lots of doors.
No, it’s not. Try it, but just use 10 “doors”. 1/10 of the time, you’ll pick the right door the first time. The other 9/10 of the time, that remaining door that you didn’t pick will be the right door. If you’re “playing with a full deck,” you should see the light after only a couple trials. 
Bingo. When I first heard this puzzle (on NPR’s Car Talk), they didn’t explain it well and so I did the correct thing from prob-stat-101, which was to enumerate all the cases and count them, as you’ve done here. Fortunately the number of cases is small and it’s easy to do, and that’s why I’m totally convinced you’re dead right, with regards to the abstract problem.
Later, a friend challenged me to explain it or show my math. On further thought, I came up with the 100 doors explanation, which satisfied him, and finally made sense intuitively to me.
Would this work?
“The Smith’s daughter is in our daughter’s ballet class. Mrs Smith has told me they have two kids.”
I don’t see how the switch from cards to doors helps. If people can’t see the explanation with 100 (or a million) doors, why would 52 cards help?
Of course, if we’re trying it ourselves, manually, cards are an easy way, but just as easy to pull out 9 random cards and the Ace of Spades and do it that way.
It’s easy to figure out if you realize that you are given the choice between the original door you named, and what’s behind BOTH of the other doors. One of the other doors revealed to you will be an empty one (or effectively empty). Two thirds of the the time the other door not revealed to you has the prize, i.e., the prize is behind one of the two doors you didn’t originally name two thirds of the the time. Only one third of the time does the original door you named have the prize. The key is that the door that is revealed is never the grand prize.
Good point.
The funny thing about this puzzle is that there are dozens of ways to explain it, and each person needs a different explanation to finally make it click.
Powers &8^]
I keep trying to hone down the explanation because it’s so simple. I’ll try again here.
First, the rules:
What do you know?
A. You know that the prize will be behind the door you named 1 out of 3 times on average.
B. You know that the prize will be behind one of the other two doors 2 out of 3 times on average.
C. Here’s the good part - you know that the prize is not behind the door that was opened.
Since the prize is behind one of the other doors you did not name 2 out of 3 times, and it’s never behind the one that was opened, it must be behind the other door you did not name and wasn’t opened 2 out of 3 times.
There is no change in the odds. There is no switch. You are not picking the door with the prize in rule 2 above, you are simply dividing the doors into two sets. A set with one door, and a set with two doors. Obviously the prize will be behind the set with two doors 2 out of 3 times. And you know which one of those two doors does not have the prize because it was opened, and you don’t even have the option of selecting the opened door for the answer. The first door you named will have the prize 1 out of 3 times, the other unopened door will have the prize 2 out of 3 times.
I always understood the reasons argued why it’s supposed to 2/3, but it’s so unintuitive that I never really believed it. I guess I’m just a real skeptical person. You did a good job explaining it…but the only way I started to believe it is when I actually practiced it with an actual demonstration. I guess it’s one of those things where there’s no combination of words in the English language that can do what “practice it yourself” is able to do.
The whole theory rests on not only a cognizant host, but a benevolent host who will always select the goat door, you’re right. That is the key factor that is so easily forgettable. What’s the opposite of a red herring?
A blue french horn.
I have puzzled over the Monty Hall problem literally for years. Finally I came across an analysis that I found both simple and persuasive:
[ul]
[li]If you DON’T switch, the only way you’ll win is if you initially chose the correct door, which obviously has a probability of 1/3. [/li]
[li]If you DO switch, the only way you’ll LOSE is if you initially chose the correct door, which has that same 1/3 probability. Thus your probability of winning is 2/3. [/li]
[li]So you’re twice as likely to win by switching. [/li][/ul]
I can’t find anything wrong with this line of reasoning.
Try it with 5 doors. The correct answer is that the remaining door, which you didn’t pick, has a 4/5 probability of being the correct one. I can’t arrive at that using your approach, but maybe I’m doing it wrong.
If I understand you correctly, you’re saying that with 5 doors (and Monty Hall still opening just one), my argument gives you a 4/5 probability of winning if you switch.
Not so. Where it breaks down is that this statement is not true with 5 doors: “The only way you’ll lose is if you initially picked the correct door.” After Monty opens one door, there are still THREE doors left that could be the right one, not just a single door. So you have three ways to lose if you initially picked the right door and then switched.
It looks like this gives you a 1/3 probability of winning if you switch (and didn’t initially pick the right door). More explicitly:
If you don’t switch, you obviously have a 1/5 (20%) probability of winning.
If you do switch, the probability of winning is:
–the probability that you initially picked the wrong door (4/5)
TIMES
–the probability that you pick the right door out of the remaining three (1/3)
The result is 4/15, or about 27%. Still better than not switching, but nowhere near 4/5.
If Learjeff is saying you have five doors, you select one, and three of the other doors are opened, then everything works out. Your explanation is very good GRob, and should suffice, but I’ve seen a need to explain a little further once people head down the wrong track with this problem.
Consider 100 doors. When you pick a door, the probability is 99% that you’re wrong. So the winning door is almost certainly somewhere in that group of 99 doors you didn’t pick. If Monty then eliminates 98 of them, then his last unopened door is very likely (p = 0.99) the winner.
In general, for n doors, your chance of winning is 1/n if you don’t switch, and (n-1)/n if you do. This seems pretty intuitive for large n, but when n is only 3 it’s not so obvious.
Ack! All I wanted was to have a nice, comfortable evening, but no. I had to find this little brain-teaser tonight, and it’s been driving me nuts, but I think I figured out the issue between the two camps. I hope this comes out formatted correctly…
Door:…1…2…3
Option 1: X O O
Option 2: O X O
Option 3: O O X
So, these are the three options regarding the prize (X marks the spot.) I pick Door 1, with a 1/3 chance of winning. Monty picks Door 3, which is empty (O). As the contestant, I can now rule out Option 3. which leaves me with a 50/50 chance to pick the correct door! Here’s where I think folks are getting confused, and please correct me if I’m wrong here: If I didn’t know what door Monty had picked, then my odds would indeed be 2/3 to switch.
An example of the 2/3 camp: Imagine a bag with three rocks, one black, the other two white, and whomever gets the black rock wins. The options are the same as above, just replace “door” with “rock”. So, I grab one rock without looking at it (lets arbitrarily call it Rock 1), then my friend looks in and pulls a white rock out and sets it in front of me. The odds are 2/3 that the black rock is still in the bag. The difference is, I don’t know whether my friend picked Option 2 or Option 3; all three are still possible.
Or take the example of meeting the two siblings. The options are: FF, MM, MF, FM. The first child I greet is a girl. If I don’t know the order they’re greeting me, then I can only rule out Option 2, and there’s a 2/3 chance that the other child will be a boy. If I know that the girl I met was the elder child, I can rule out Options 2 and 3, leaving me a 50/50 chance for the gender of the younger sibling.
Just changing a few numbers gives
[ul]
[li]If you DON’T switch, the only way you’ll win is if you initially chose the correct door, which obviously has a probability of 1/5. [/li]
[li]If you DO switch, the only way you’ll LOSE is if you initially chose the correct door, which has that same 1/5 probability. Thus your probability of winning is 4/5. [/li][/ul]Which is what you expected.
No, in that case it’s 50/50. It doesn’t matter if the order of your options are older,younger or first met,last met. You’ve still learned the gender of one of them, and nothing of the other of them.
The key factor is your act of picking a door is not the same as Monty’s act of picking a door. You don’t know what’s behind the doors, so you’re choosing at random. Monty, however, knows what’s behind all three doors and can make his choice accordingly. He will never open a door with the prize behind it in Step 2 of the game.