Talk:Monty Hall problem/Archive 1

message: The problem with this article is that it makes it seem like switching is not really the right answer. This is because of language like "standard" and "classical". It makes it seem like there is some controversy whether switching really is right. And the "Assumptions" section is confusing and irrelevant. If anything it gives the reader false hope that the odds could be 1/2. This is a math problem, there is no POV/NPOV involved. Some people are being really childish here.
:(It's not generally a good idea to stick your new comment at the top of the discussion page, pre-empting the months of diescussion that have taken place before. It's not even tactically a good idea, as most Wikipedia editors will assume that new stuff is added on the end, and may not notice your new comment, especially if you don't sign it with a date-stamp; that is, ~~~~)
:Now, as to the substance of your comment: The reason for the wording about the "classical" answer and so on is that it is not by any means a pure mathematical problem. As explained with great care in the text, the problem as normally stated involves a whole raft of assumptions, without which the answer would be different. If it were stated as a pure mathematical problem, with the conditions outlined properly, then it would indeed be simply a math problem. But that would not be the "standard" or "classical" problem. 
:A person who does not know the practices of the TV show—in fact, the practices as modified for this problem—will not necessarily get the "right" answer no matter how good his math is. That's the reason for the Assumptions section. If you find that section not to be clearly written, by all means edit it.
:(Perhaps you would not so easily conclude that someone is being childish if you read the discussion here before prefixing it with your comment.) Dandrake 08:27, Jul 15, 2004 (UTC)
----
Should you switch?
For the original choice, there are these possibilities:
  123
  CGG
  GCG
  GGC
So whatever door you choose, your chances are 1/3.
For the second choice, call the door you originally
chose 1, and the door Monty opens as 3. Now the
possibilities are:
   123
   CGG
   GCG
(We know that GGC is not a possibility because Monty showed 
us a goat behind door 3.)
So if you choose either 1 or 2, you have a 1/2 chance of
winning. Which means it doesn't improve your chances to
switch.
Now, what's wrong with this argument?
:The two cases CGG and GCG are not equally likely. In fact, GCG is twice as likely as CGG. This is obscured by your relabeling the doors after the fact so that the door Monty opens always gets number 3. You may want to write down all 9 possible equally likely scenarios: the car can be behind one of three doors, the contestant can choose one of three doors, makes 3x3=9. Then analyze how the sticker does (he wins in 3 of the 9 cases) and how the switcher does (he wins in 6 of the 9 cases).
Okay, so let's not renumber. Call the door you originaly pick number 1, and indicate the door Monty opens with parentheses. Then we have the following possibilities:
    1  2  3
    C (G) G
    C  G (G)
    G  C (G)
    G (G) C
So, if you stay with 1, your chance of winning is 1/2. If you
switch to 2 or 3 (whichever remains unopened), your chances of
winning are also 1/2. So it doesn't improve your chances to switch.
Or does it?
:Again, the four cases you list are not equally likely. The first two are equally likely, but they are each
half as likely as the third and fourth case. If the door you originally pick is number 1, then the equally likely
scenarios are CGG, GCG and GGC. Each of them has probability 1/3. The first case splits into the two equally likely
cases C(G)G and CG(G) each with probability 1/6. The two cases GC(G) and G(G)C both keep probability 1/3.''
----
Don't forget that the two goats are distinct.  There are six permutaions for item placement:
      1   2    3
      C   g1   g2
      C   g2   g1
      g1  C    g2
      g1  g2   C
      g2  C    g1
      g2  g1   C
There are three door choices the player can make, yielding 18 possible outcomes:
  1   2    3
  C   g1   g2
  C   g2   g1
  g1  C    g2
  g1  g2   C
  g2  C    g1
  g2  g1   C
  1   2    3
  C   g1   g2
  C   g2   g1
  g1  C    g2
  g1  g2   C
  g2  C    g1
  g2  g1   C
  1   2    3
  C   g1   g2
  C   g2   g1
  g1  C    g2
  g1  g2   C
  g2  C    g1
  g2  g1   C
At this point, the player has "lost" 2/3 of the time.  If the player changed the door
selection now, the choices would be between two doors, like this:
    C  g1
    C  g2
    g1 g2
    g1 C
    g2 g1
    g2 C
Choosing either column 1 or column 2 still gives a 2/3 chance of losing, so switching
doors won't accomplish anything.
Monty now eliminates an un-chosen goat (if there are two to chose from, it
doesn't matter which one he chooses, so I'll always choose the rightmost).
Note that in 4 out of the 6 cases, Monty is "forced" to give the player a car if
he switches, and this is the key -- Monty's choice is not random.  The
resulting column for a switch is:
    C
    C
    g1
    C
    g2
    C
Here's an expanded chart, where the player's first choice is bolded and the
eliminated goat is replaced with a hyphen.
I've put a "#" where a switch wins, and a "-" where a switch looses.
  1   2    3      win/lose
  C   g1   -         -
  C   g2   -         -
  g1  C    -         #
  g1  -    C         #
  g2  C    -         #
  g2  -    C         #
  1   2    3
  C   g1   -         #
  C   g2   -         #
  g1  C    -         -
  -   g2   C         #
  g2  C    -         -
  -   g1   C         #
  1   2    3
  C   -    g2        #
  C   -    g1        #
  -   C    g2        #
  g1  -    C         -
  -   C    g1        #
  g2  -    C         -
-----
But why count the two goats separately? What if there
weren't any goats, just doors left empty - that wouldn't
change the probabilities, would it?
----
Suppose Monty doesn't open the door, but after your initial
choice he still gives you a chance to "switch", meaning you can
either stay with your original choice, or if you switch you'll
win if the car is behind either of the other two doors.
Obviously you want to switch, because then you have two doors,
with two chances to win, as opposed to only one door, one
chance to win, if you don't switch. But what if Monty opens
one door, with a goat, before you decide. That means one of
the other doors isn't really a chance to win. Doesn't that
change things? Doesn't it mean that it's more likely that the
car is behind each of the other doors than orginally thought? Doesn't it make it more likely that your original choice was a winner? Aren't the odds now 1/2 rather than 1/3?
If Monty opens a door , then yes the chances for that door may change.
It depends on what Monty does when a car is behind the door you first choose, whether he open the two empty doors with equal chance.
Before Monty opens a door and you decide to switch, yes your chances
for all trials are 2/3, but get this:
Suppose a car is placed and you choose door 1. If Monty decides to open door 2 with a certain frequency over door 3 if the car is behind 1 and
makes this determination before he asks for your decision to switch and
before he opens a door, you chances for door 2 open are set before any
door is open. Your chances for door 2 are set, for door 3 are set and
the average is set (2/3).
Now suppose two people play the game one chooses door 1 and one chooses
door 2 and door 3 is open. Aren't the chances for each player to switch equal? No, because one door was chosen that Monty could not open, therefore
the situation is not symmetric. One player will gain, the one that goes
from the protected door to the unprotected door, and the other will lose
chances, the one going from the unprotected to the protected door.
Well, suppose Monty opens both of the other doors, and they're
both goats. Still think you should switch? Doesn't showing both
goats behind the other doors mean that your original choice is
a sure winner? That the odds for not switching are now 1.0, not
0.33? So why wouldn't showing the goat behind one of the other
doors also change the odds?
:The odds can't change because they are locked in when you first choose. When you choose a door with a goat (2/3 odds), Monty must show you the other goat, so the car will be behind the door you didn't pick. So 2/3 of the time, switching will get you the car.
:Here's another way to put it: you win if you pick a goat and then switch. The odds of picking a goat are 2/3, so if you always switch you win 2/3 of the time.
----
Here's another question. Suppose while you're deciding whether to switch, another contestant is brought in and Monty lets them choose one or the other of the two closed doors. Aren't their odds of winning 50-50 for either choice? 
:No. If they choose my original door, they win with 1/3 probability, if they choose the other one they win with
2/3 probability.''
I know you think this, but why? There are just two doors, and
the car is behind one or the other of them. Why doesn't that
make the odds 50-50 for the new contestant? 
:Because the situation is not symmetric. The new contestant knows something, and that information helps.
WRONG!! The second contestant knows nothing but the chances are still
asymmetric anyway. Why?
Look, one Box is protected no
matter who is watching. When two doors remain, there is a protected
door and an unprotected door. Monty could never open the protected door,
but could have opened the unprotected one, so the chances are unequal,
NO matter who walks in or out. It doesn't matter what the second 
contestant knows or doesn't know. The fact that the second contestant
"knows nothing" merely means that he will probably pick either door
with equal probablity and half the time get a 33% winning odds and
half the time a 66% door with an average winning percent of 50%
Consider this: Monty has three boxes, in one of them is a hundred dollar bill. You don't know where. But then
Monty says: Ok, I'll help you, point out two hats and I'll gladly combine their contents into a new box. You point out two,
and he does as he said. So now there are two boxes left, the new one and the untouched one, one of them has the bill.
But the odds are not 50-50: you know that the bill is twice as
likely to be in the new box than in the untouched one.''
Suppose they don't
know which of the two doors I chose. Then aren't their chances
50-50? 
:Sure.
BTW, the reason I'm raising these questions is to try to suggest why
the Monty Hall problem is a "problem". It wouldn't be a problem if the answer were obvious...
: Consider the extreme case: there are 1000 doors, one has a car. I pick a door, then Monty opens
998 loser doors. Now you come in. Do you pick my door or the other one?''
Yes, that is the argument that most convinces me that there is an advantage to switching. But how can we be sure that this result
applies to the 3 door case. Certainly the benefit of switching
is very much higher in the 1000 door case that in the 3 door case. It obviously decreases as the number of doors decreases. Maybe the advantage vanishes altogether in the 3 door case?
----
Suppose they choose the door you originally chose. Aren't their chances of winning the same as your chances if you choose not to switch?
:Yes.
----
I think this is the most suprised I've ever been my the output of a program I wrote myself.  I was sure it was going to come up 50/50.  (Empirical Proof)
:Me too, with this PHP script here - very suprised indeed. Now my brain hurts. --Dan Huby 08:57, 15 Jul 2004 (UTC)
replies: text: If you're still not convinced after seeing the results of your program, I have a poker game on Thursday nights you're welcome to join...  --LDC
Here's a similar problem in the game of Texas hold 'em that many people don't believe at first: I offer you your choice among three two-card hands: (1) a pair of fours, (2) an ace and a king of different suits, (3) a ten and a jack of the same suit.  After you choose, I choose one of the remaining hands.  Then we deal five cards face up, and the player who can make the best five-card poker hand with any combination of his two cards plus the five on the board wins.  Which cards do you choose, and what is the expected outcome?  It turns out that whoever chooses first loses.  If you choose the A-K, I choose the 4-4 and have a slight advantage.  If you choose the 4-4, I choose the 10-J and have a slight advantage.  If you choose the 10-J, I choose the A-K, and have an advantage!
----
This is an interesting and helpful article but it should be completely rewritten; it doesn't sound like an encyclopedia article to me.  The link to the Perl program is interesting, too, but it's original research.  What place does that have in an encyclopedia?
 I take the rather abstract view of an encyclopedia as something
 which explains everything.  Thus I find the article actually
 quite accessible and helpful, written in a friendly direct
 way.  The perl code is perhaps less useful to someone wanting
 to understand the problem in an article-reading sort of way,
 but I think of it as sort of a multimedia addition - a working
 Monty Hall Problem machine to tinker with. -J
----
I moved the following explanation from the article:
In short, the reason for the above result is that Monty will ALWAYS show the other goat.  Look at the following situation with C stand for Car, G for Goat, X for the goat that Monty picked, and with the player always pick the first door:
  Initial   Monty     Result
  ==================================================
  CGG       CGX       Sticker wins and switcher lose.
  GCG       GCX       Sticker lose and switcher wins.
  GGC       GXC       Sticker lose and switcher wins.
As can been seen above, the chance for the switcher to win is twice that of the sticker.
I believe this is not a compelling argument; it will only convince people who already believe the result or don't follow closely. If I were to criticize the argumentation, I would point out that the case
  CGG       CXG       Sticker wins and switcher loses
is missing. So this makes two cases in favor of the sticker and two cases in favor of the loser: 50-50. To counter this objection, one would have to argue lengthily that the four listed cases are not equally likely, that in fact the first and the fourth case both have probability 1/6 while the second and the third both have probability 1/3. But why? And so on.
The crucial and convincing argument is the one given in the article: a switcher wins the car if and only if his original choice was a goat. Only if the reader understands that point will they truly be convinced. AxelBoldt, Wednesday, July 3, 2002
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What's the connection with Three card monte? The other seems to be a confidence trick rather than a probability trick? DJ Clayworth 17:08, 27 Aug 2003 (UTC)

message: The article is wrong. Switching will do nothing to change the odds. The past has no effect whatsoever; all one needs to know, is the current state of affairs. That means that there is a 50/50 chance of getting a goat. Furthermore, doors don't know if they are picked; It is only in the human mind that you lay "ownership" to one or the other. So "switching" is in the mind of the human only. This article is a hoax. ChessPlayer 17:18, 8 Mar 2004 (UTC)
replies: text: The article is absolutely correct as written, and your comment demonstrates why the problem is so famous.  It's not a hoax, it's just counterintuitive.  Monty introduces information when he opens a door, because he always opens a door that has a goat.  I would explain further, but this is already rehashed here on the talk page and in the article itself. Isomorphic 17:32, 8 Mar 2004 (UTC)
replies: text:  The situation is, there are two doors; one is a goat. Therefore the odds are fifty fifty, and choosing either door doesn't matter. ChessPlayer 20:14, 8 Mar 2004 (UTC)

text: In the coin example I gave above there are also two possibilities; Things are not always as simple as they seem. DJ Clayworth 14:22, 9 Mar 2004 (UTC)

text: Like everyone, I had trouble wrapping my head around this. So I broke out a deck of cards and played it myself. I used a red King as the car and two black deuces as the goats. I played a few "hands"; each time I picked a random card and turned it over, so unlike the game show I could see whether my initial pick was correct immediately, but that didn't matter since I was always switching. Then I'd find a deuce and treat it as the door that Monty opens, and then I'd switch to the remaining card and see whether I won or lost. Rather quickly I noticed something: The moment I randomly pick the card and flip it over, I could tell instantly whether switching would cause me to win or lose, because when I picked the King (the car), I knew that switching would cause me to lose (because I'd have switch to a deuce, there's only one King). And conversely, whenever I picked a deuce, I knew that switching would cause me to win (because the only card to switch to will be the King after the other deuce is removed by Monty). That's when it hit me: 1. Whenever I pick a deuce, switching will cause me to win. 2. Everyone knows that two out of three times I'm going to pick a deuce. Therefore: Two out of three times I'm going to win by switching!   ...   Interesting side note: In the late sixties my parents were on Monty Hall's game show in Vegas, and they won the car... although my mom knew which curtain to pick because she could just barely see the wheels of the car below the curtain... needless to say she did not switch. =o) - Eisnel 08:28, 15 Jul 2004 (UTC)
thread_title:  Hoax 

message: I was talking to a friend about the Monty Hall problem today and he told me about a similar problem, and I don't think there is a Wikipedia article on it and I'm not sure how to present the solution either, but anyway here is the problem:
You are on a gameshow and the host holds out two envelopes for you to choose from A and B. So you choose an envelope (A) and it's got $2000 in it. The presenter then says that one of the envelopes has twice as much money in it as the other one and offers you the chance to switch. So you think about it this way... "If I switch I will go home with either $4000 or $1000, by not switching I will go home with $2000. There is a 50/50 chance that I will double my money by switching. A normal 50/50 bet results in me either doubling my money or losing it all, whereas here I will only lose half. Therefore this is a better than evens bet so I will make the swap." You are just about to swap envelopes when you think about the problem some more - "Surely this can't be right... ". Mintguy (T) 16:13, 15 Jul 2004 (UTC)
replies: text: Interesting, but not really similar.  In the Monty Hall problem, there's just one possible positive payoff, so the only concern is maximizing your chance of getting it.  This problem is more complicated.  It is true that if you switch, the expected value of the new envelope is (1000*0.5 + 4000*0.5) = 2,500, so if all you care about is the average amount of money you'll take home, you should switch.  However, most people in the real world are risk averse, meaning that they may prefer the sure 2,000. Isomorphic 02:58, 16 Jul 2004 (UTC)
replies: text: Yeah but you're wrong, you see, there is no advantage to switching. How can there be? Hahah. Mintguy (T) 07:41, 16 Jul 2004 (UTC)
thread_title:  two envelope problem