in reply to Re^4: Marilyn Vos Savant's Monty Hall problem in thread Marilyn Vos Savant's Monty Hall problem
And that statement, while better than how it is presented the vast majority of the time, still isn't sufficient to make the answer unambiguous. There is still the question of Monty's motivations which could make it either adviseable to switch (up to 2/3 odds of winning) or to stay (up to 100% chance of winning).
An interesting followup on Marilyn. Later she tackled a restricted version of the problem that I presented at Spooky math problem (the restriction being that the two envelopes hold money, one has twice the money of the other) and correctly analyzed an argument for whether you should always switch. But she incorrectly analyzed whether you could do better than even odds. I know a couple of probability theorists who pointed out her mistake to her, but she never admitted to her mistake.
Make of that what you will.
Re^6: Marilyn Vos Savant's Monty Hall problem
by TrekNoid (Pilgrim) on Aug 23, 2004 at 17:45 UTC

And that statement, while better than how it is presented the vast majority of the time, still isn't sufficient to make the answer unambiguous. There is still the question of Monty's motivations which could make it either adviseable to switch (up to 2/3 odds of winning) or to stay (up to 100% chance of winning).
I might be misunderstanding the issue then.
The question, as I understood it, is that I find myself on a game show, and I've picked a door... the host has just opened a door to reveal a goat, and then asked me if I want to change my choice to the other door.
So, in order to make my decision on this *one* event, I have to test it... and in order to test it, I must set up *identical* events, devoid of the host's motivation, to test whether I should change in this one instance.
It would be different if the question was 'in general, what should I do if I don't know the host's motivations', but the question states that the host opened a door with a goat... therefore my examination of the question should assume that fact
I don't know if this is coming across correct in text or not, so I apologize if it isn't...
Put another way, I'm setting up a model to that mimics the original question, which assumes the host opens a goat door
The possibility that the host *might* not offer the choice... or might reveal the prize first... etc... don't enter into the answering of this question, because the question supposes that the opening of the goat door has already happened, and now you have to make a decision based on the probabilities in play.
Trek  [reply] 

I believe that you are stating the problem correctly, but then adding a big assumption before analyzing it. You assume that that, whether or not you have the right choice, the host will always put you in the current situation  you've picked a door, and another has been opened showing a goat. With that assumption then you should switch, and will have 2/3 odds of winning if you do.
But that is a big assumption. In analyzing probabilities you cannot just work with what has happened, you have to work with what could have happened instead. A probability problem is never fully specified until it includes both knowledge of what did happen and what might have happened.
In particular in this case there are models of the host's possible behaviour in which you would be an idiot to switch. (The host is trying to keep you from winning the car.) There are models of the host's behaviour in which it is a good idea to switch. (The host wants to draw the game out.) And the problem statement does not provide enough information to unambiguously decide which model of the host's behaviour is correct. (Saying that the host has enough knowledge to always draw the game out is not saying that the host will choose to do that.)
Therefore the problem is not fully specified. To come up with an answer you have to add an assumption of some sort.
If you want to make it truly unambiguous, you can state the problem like this, Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. The game works like this; you pick a door and the host, who knows what's behind the doors, will open another door and show you a goat. Then the host asks whether you want to switch your choice. You decide, then the host opens the other doors and you get the car if you've chosen it. When you play this game, is it to your advantage to switch your choice of doors?
The wording change is subtle but important. With this wording it is clear that, no matter what, the first time the host opens a door there will always be a goat. Not only have you been told what actually happened in the game, but you've also been told what would've happened no matter what your initial choice was.
 [reply] 

In particular in this case there are models of the host's possible behaviour in which you would be an idiot to switch. (The host is trying to keep you from winning the car.)
You've made this statement more than once now, and it completely loses me. Can you demonstrate it? I don't actually see where motivation comes into it; the stipulation is that you have a choice to switch or not switch. That implies that a goat door was opened by the host. (If the prize door was opened by the host, you no longer need choose.)
And afaict, that a goat door was opened is enough to give you an advantage if you switch.
 [reply] 



But that is a big assumption. In analyzing probabilities you cannot just work with what has happened, you have to work with what could have happened instead. A probability problem is never fully specified until it includes both knowledge of what did happen and what might have happened.
This is where I think we're in disagreement... Not with your statement, because I agree with it if you're trying to predict future behavior.
But that's not what we're trying to do in this case. We're not 'about to go on the show'... and we're not 'about to begin the prize round'.
The problem states that we're *already* in the prize round, chosen a door, and the host has revealed a goat door. That's where we *start*
So we have to calculate the probabilities based on that set of variables. You can't examine the possibility that "The host might not have opened a goat door first" because this question *started* with that stated.
Basically, I guess I'm stating that I agree with you that if we're talking about all possible future behavior, then we need to consider that the host might just open the prize door... or not offer a choice... etc...
But for this specific problem, we already find ourselves faced with the host opening a goat door and offering a switch, so to decide *from here* what to do, we have to assume these as fixed variables in our analysis.
You're beginning before the game started... I'm beginning from the point where two actions have already occured... Which changes things.
It's like turning on the TV and finding a car race in it's last lap and wondering what odds are that the car in the back of the pack could win... You would base your conclusion on how many times you've seen the back car win from the last lap in the past... You wouldn't consider the possibility that he *might* have been in the front or the middle, because he isn't...
If you were going to analyze the probability of his winning the *next* race, all that comes into play... but to analyze the probability that he's going to win *this* race, then you discard those options that can no longer occur.
That's how I see the Monty Hall problem. You're considering the probabilities based on *any* version of the game... I'm considering the probabilities of this *one* game, where the goat door is already open, and a choice has been offered.
And, since someone's mentioned 'arguing' with you... for the record, I've enjoyed this discussion, regardless of the outcome of it, or whether we ever agree on it :)
Trek
 [reply] 

Re^6: Marilyn Vos Savant's Monty Hall problem
by BrowserUk (Pope) on Aug 23, 2004 at 20:50 UTC

I'm not sure of the relevance of this. Stats was never my strong suit but...
If the following simulation bears any resemblance to reality (of the restated version of the problem), then you only get a greater chance of winning if you always switch your choice after the host opens the first door on the goat.
If at that time you make another choiceeither, whether to switch or not, or a random choice between the two remaining doors (which amounts to the same thing?)then your odds of success remain at 33.33%.
#! perl slw
use strict;
use List::Util qw[ shuffle ];
our $TRIALS = 1_000_000;
my $DIVISOR = $TRIALS / 100;
my( $stick, $switch, $newChoice );
for my $try ( 1 .. $TRIALS ) {
## Randomly hide the prizes behind the doors.
my @doors = shuffle 'car', 'goat', 'goat';
## Pick a door
my $choice = int rand 3;
## The host opens a door that I didn't pick
## and that isn't the car
my $opened = grep{ $_ != $choice and $doors[ $_ ] ne 'car' } 0 ..
+2;
## Count my wins if I stick or switch
$doors[ $choice ] eq 'car' ? $stick++ : $switch++;
## Make a new choice from the remaining two
my $new = ( grep{ $_ != $opened } 0 .. 2 ) [ rand 2 ];
## And if I make a completely new random choice.
$doors[ $new ] eq 'car' and $newChoice++;
}
printf "Odds of
Choose again: %.3f
Win if you don't switch: %.3f
Win if you do switch: %.3f\n",
$newChoice / $DIVISOR, $stick / $DIVISOR, $switch / $DIVISOR;
__END__
P:\test>test
Odds of
Choose again: 33.331
Win if you don't switch: 33.286
Win if you do switch: 66.714
P:\test>test
Odds of
Choose again: 33.370
Win if you don't switch: 33.353
Win if you do switch: 66.647
Assuming that this simulation isn't at fault, I'd love to see a (preferably layman's terms) explaination for why a predetermined strategy (always switch) would have such an effect on the odds of success?
Examine what is said, not who speaks.
"Efficiency is intelligent laziness." David Dunham
"Think for yourself!"  Abigail
"Memory, processor, disk in that order on the hardware side. Algorithm, algorithm, algorithm on the code side."  tachyon
 [reply] [d/l] 

Maybe it's easier to think about like this: if you always switch, you only lose if you picked the door with the car first. That's a 1 in 3 chance to lose. (And a 1 in 3 chance to win if you never switch).
Solo

You said you wanted to be around when I made a mistake; well, this could be it, sweetheart.
 [reply] 

Assuming that this simulation isn't at fault, I'd love to see a (preferably layman's terms) explaination for why a predetermined strategy (always switch) would have such an effect on the odds of success?
Suppose there were 100 doors... and I allowed you to choose one of them, what's the odds you picked the right door? 1%
What's the odds that the prize is behind one of the other 99? 99%
So, if I gave you a choice of switching your choice from Door 1, and allowing you to pick all 99 of the other doors, which would you take? Obviously, the 99.
The torture is that I first open 98 of the goat doors (because I know where the prize is), leaving two... The one you picked, and the one remaining. None of that changes the fact that you're initial guess was only 1% chance of guessing right.
Therefore, switching from your initial choice at the last minute is effectively the same as changing to all 99 doors.
Occassionally this strategy would fail, but only about 1% of the time, since that was the likelihood you chose correctly to begin with.
Same is true with three doors. Your chance of picking correctly initially is 1/3rd... If you were allowed to switch to the other two doors, you'd have a 2/3rds chance of winning.
At least, that's how it was initially explained to me, and it seems pretty reasonable.
Trek
 [reply] 

#! perl slw
use strict;
use List::Util qw[ shuffle ];
my( $stick, $switch, $skip_goat, $skip_car ) = (0, 0, 0, 0);
for ( 1 .. 100_000 ) {
## Randomly hide the prizes behind the doors.
my @doors = shuffle 'car', 'goat', 'goat';
## Pick a door
my $choice = int rand 3;
####
## Uncomment the option you want here to see different scenarios.
###
## Option 1: The host opens a door that I didn't pick
## and that isn't the car
#my @available = grep{ $_ != $choice and $doors[ $_ ] ne 'car' } 0
+ .. 2;
## Option 2: The host opens a random door
#my @available = grep{ $_ != $choice } 0 .. 2;
## Option 3: The host tries to be malicious
#my @available = grep{ $_ != $choice and $doors[ $_ ] eq 'car' } 0
+ .. 2;
#@available = grep{ $_ != $choice } 0..2 if not @available;
####
## End of options
####
# Monty chooses which door to open from the choices
# that he might make.
my $opened = $available[rand(@available)];
if ($doors[$opened] eq 'car') {
$doors[ $choice ] eq 'car' ? $skip_car++ : $skip_goat++;
next;
}
## Count my wins if I stick or switch
$doors[ $choice ] eq 'car' ? $stick++ : $switch++;
}
printf "Odds of
Not getting here if you were originally right: %.3f
Not getting here if you were originally wrong: %.3f
Win if you don't switch: %.3f
Win if you do switch: %.3f\n",
$skip_car / (( $stick + $switch + $skip_goat + $skip_car) / 100 ),
$skip_goat / (( $stick + $switch + $skip_goat + $skip_car) / 100 )
+,
$stick / (( $stick + $switch) / 100 ),
$switch / (( $stick + $switch) / 100 )
Now that said, let me explain why the odds are what they are for each option.
In Option 1, you pick a door and have 1/3 odds of being right. There are 100% odds that you'll see a goat, so the fact that you saw one tells you nothing. Therefore your odds of being right remain 1/3. Since switching makes you right if you were wrong, your odds if you switch are 2/3  so you want to switch.
In Option 2, you pick a door and have 1/3 initial odds of being right. However if you were right, then you're guaranteed to see a goat next, while if you're wrong, you have only even odds of seeing a goat next. Therefore the knowledge that you actually saw a goat conveys information  if you do the math just enough information to tell you that you now have even odds of being right.
In Option 3, you again pick a door and have 1/3 initial odds of being right. However the fact that you saw a goat gives you considerable information  it literally tells you that you must be right. Since you're right, you don't want to switch.  [reply] [d/l] 

Although your three scenarios may be pedantically valid, only the first one makes any real sense.
If there are goats behind two of the doors, then Monty opening the door with the car behind it would make for a really stupid game show. He might as well just open the door you chose since both actions tell you that you picked a goat door.
I suppose you could argue that the problem as stated doesn't make it absolutely clear that you *know* that there are two goat doors (or that there are *always* two goat doors).
But as stated, the only reasonable interpretation is the first one (that makes sense for a game show, that fits common sense, not that fits a mathematician's standard for precision).
As for the real Monty saying that he sometimes opened the "car" door, this is true. And that didn't make the real game show really stupid (yes, I'm sure many regarded the show as plenty stupid, but I'm talking stupid beyond sense, not just a matter of taste) because the real show always (or almost always, I haven't personally reviewed every single episode) had one "goat" door (a joke "prize", and no, it wasn't always a goat), one "car" door (a major prize), and one "appliance" door (a minor prize).
That makes for a more complex statistics problem (as it should, being a real game show). But it has little more to do with the problem as stated other than being the likely inspiration for it and, later, the inspiration for its retitling as well.
Also note that Monty had other options besides opening one of the doors. He could have a prize wheeled out and offer you that in exchange for the door. He often pulled cash out of his pocket and offered you that (perhaps in addition to the prize he just wheeled out). It wouldn't have made much sense to call the show "Let's make a deal" without such negotiations. (:
 [reply] 


my @available = grep {
$door[ $choice ] eq 'car' ? ( $_ != $choice ) : ( $doors[ $_ ] eq
+'car' )
} 0 .. 2;
Hmm.. I guess that's not much clearer either.
Update: so I rewrote the code, and I think this is clearer:
for ( 1 .. 100_000 ) {
my @door = 0 .. 2;
my $car_door = int rand @door;
my $my_pick = int rand @door;
@door = grep { $_ != $my_pick } @door;
## Option 1: The host reveals a goat
#my @available = grep { $_ != $car_door } @door;
# Option 2: The host opens a random unopened door
#my @available = @door;
## Option 3: The host tries to be malicious
#my @available = @door;
#@available = grep { $_ == $car_door } @door if $my_pick != $car_d
+oor;
my $opened = $available[ rand @available ];
if( $opened == $car_door ) {
$my_pick == $car_door ? $skip_car++ : $skip_goat++;
next;
}
## Count my wins if I stick or switch
$my_pick == $car_door ? $stick++ : $switch++;
}
Makeshifts last the longest.
 [reply] [d/l] [select] 



Your simulation is bad.  Nice catch.
Since switching makes you right if you were wrong, your odds if you switch are 2/3  so you want to switch.  That's the explaination that will stick. Thanks.
Examine what is said, not who speaks.
"Efficiency is intelligent laziness." David Dunham
"Think for yourself!"  Abigail
"Memory, processor, disk in that order on the hardware side. Algorithm, algorithm, algorithm on the code side."  tachyon
 [reply] [d/l] 

