Re: The Monty Hall Problem -- again
voice_o...@australia.edu
But again, here's where I get confused....
You said:
> The contestant has a 2/3 chance of choosing a door with nothing, and
> there is always a car behind the other door in this case; the
> contestant has a 1/3 chance of choosing the door with the car behind
> it, and there is always nothing behind the other door in this case.
>
> Chance of winning the car by keeping the initial choice: 1/3.
Why 1/3?
Seems to me, the initial odds were 1:3. But as soon as one of the
doors was opened revealing nothing, his odds just increases to
50:50....as in "maybe the car is behind HIS door...maybe its behind
the OTHER door".
Yes, it is true the the host knows which door has the car. But, the
fact that the host chose to reveal door B rather than door C (from my
example below) tells me nothing about what's behind door A.....and I
cannot see how it affect the odds of what's behind door A either.
On Jan 1, 1:21 am, Don Del Grande <del_grande_n...@earthlink.net>
wrote:
> It helps to know that (a) the host knows which door has the car behind
> it; (b) if the contestant chooses one of the doors with nothing behind
> it, the host will open the other door with nothing; (c) if the
> contestant chooses the door with the car behind it, the host will open
> one of the other two doors.
>
> The contestant has a 2/3 chance of choosing a door with nothing, and
> there is always a car behind the other door in this case; the
> contestant has a 1/3 chance of choosing the door with the car behind
> it, and there is always nothing behind the other door in this case.
>
> Chance of winning the car by keeping the initial choice: 1/3.
> Chance of winning the car by switching = chance of selecting a door
> with nothing initially * chance that the other door has the car behind
> it = 2/3 * 1 = 2/3.
>
> It would be 50-50 if there was a possibility that, if the contestant
> chose a door with nothing, the host would open the door with the car
> behind it.
>
> -- Don
> voice_of_rea...@australia.edu wrote:
> >Greetings:
>
> >I have been pondering the oft discussed "Monty HallProblem" and I
> >still just don't get it. I hope someone can help show me the light.
>
> >In brief --
> >Themonty hallproblem is where a contestant is facing 3 doors. Behind
> >one is a brand new car; the other two: nothing.
>
> >The contestant chooses a door(A). The host then reveals what is behind
> >one of the other doors(B)....nothing.....and offers the contestant a
> >chance to switch to door (C).
>
> >The collective wisdom that has evolved around this problem is that the
> >contestant *SHOULD* switch.
>
> >I don't see why....
>
> >The collective wisdom says that somehow by having been shown that door
> >B is empty, the contestant has gained knowledge that allows them to
> >increase their odds of selecting correctly by switching.
>
> >I don't see why....
>
> >Seems to me that originally, the odds of selecting correctly were 1:3.
> >After seeing that nothing was behind door B the odds are now 1:2.....
>
> >......but it seems to me that its 50:50 whether they switch or stay
> >put.
>
> >So what's the advantage of switching?
>
> >What am I missing??
>
voice_o...@australia.edu
> Chance of winning the car by keeping the initial choice: 1/3.
Suppose we alter the game slightly. Call it the "Senile Monty Hall"
Game.
Monty initially offers a choice of 3 doors. The contestant selects
door (A). Monty reveals that there is no car behind door (B) and
offers a chance to switch to door (C).
According to your statement, the contestant has still has only a 1/3
chance of being correct.
Now, senile Monty Halls looses his cue cards and forgets that he has
already revealed a door. He reveals that there is no car behind door
(C).
Does the contestant still face only a 1/3 chance of being correct?
Seems to me that with each reveal the odds the HIS door contains the
car increase.
Thus after the initial reveal, the chance the his door contains the
car increase from 1/3 to 1/2 -- or 50:50 -- and so there is no
advantage to switching.
Mensanator
> Thank you for you reply.
>
> But again, here's where I get confused....
>
> You said:
>
> > The contestant has a 2/3 chance of choosing a door with nothing, and
> > there is always a car behind the other door in this case; the
> > contestant has a 1/3 chance of choosing the door with the car behind
> > it, and there is always nothing behind the other door in this case.
>
> > Chance of winning the car by keeping the initial choice: 1/3.
>
> Why 1/3?
Because your task was to select 1 of 3 doors.
There are 2 possible outcomes, either you selected the
car or you didn't. But these two outcomes do not have the
same probability, there's a 1/3 chance you picked
correctly and a 2/3 chance you picked incorrectly.
>
> Seems to me, the initial odds were 1:3. But as soon as one of the
> doors was opened revealing nothing, his odds just increases to
> 50:50
Wrong. The collapse of a probability function does NOT change
the odds. Even if ALL doors are revealed giving you complete
information and totally collapsing the probability, that does
NOT alter what the probability was AT THE TIME YOU MADE YOUR
SELECTION.
Probability is not retroactive.
> ....as in "maybe the car is behind HIS door...maybe its behind
> the OTHER door".
Right, just as I said there were two possible outcomes, you
picked correctly or you didn't.
>
> Yes, it is true the the host knows which door has the car. But, the
> fact that the host chose to reveal door B rather than door C (from my
> example below) tells me nothing about what's behind door A.....and I
> cannot see how it affect the odds of what's behind door A either.
Yes, it doesn't effect the odds of what was behind door A at the time
you made the selection. Being given a chance to switch and being
given partial new information does NOT alter the fact that your
initial choice had a probability of 1/3.
It is also assumed that if you were to switch, you would switch
to the other unopened door (unless you WANT the goat for a
Satanic ritual).
By having a guaranteed zonk door revealed, you are guaranteed to get
the car IF YOU HAD ORININALLY PICKED INCORRECTLY. The chance that you
picked incorrectly is greater than the chance you picked correctly.
This also assumes you are ALWAYS given the chance to switch doors.
If Monty only makes the offer when the car is initially picked,
then it would be foolish to switch. But that doesn't make a 50:50
chance either.
> > >What am I missing??- Hide quoted text -
>
> - Show quoted text -
Mensanator
> Follow-up to my previous response......
> On Jan 1, 1:21�am, Don Del Grande <del_grande_n...@earthlink.net>
> wrote:
>
> > Chance of winning the car by keeping the initial choice: 1/3.
>
> Suppose we alter the game slightly. Call it the "Senile Monty Hall"
> Game.
>
> Monty initially offers a choice of 3 doors. The contestant selects
> door (A). Monty reveals that there is no car behind door (B) and
> offers a chance to switch to door (C).
>
> According to your statement, the contestant has still has only a 1/3
> chance of being correct.
Not BEING correct, WAS correct.
>
> Now, senile Monty Halls looses his cue cards and forgets that he has
> already revealed a door. He reveals that there is no car behind door
> (C).
>
> Does the contestant still face only a 1/3 chance of being correct?
Not BEING correct, WAS correct. And even though the contestant now
has complete information, that does NOT retroactively alter what
his initial probability was.
>
> Seems to me that with each reveal the odds the HIS door contains the
> car increase.
But doesn't change what his odds WERE at the time he made his choice.
>
> Thus after the initial reveal, the chance the his door contains the
> car increase from 1/3 to 1/2 -- or 50:50 -- �and so there is no
> advantage to switching.
Wrong. Nobody's saying switching doesn't change his chances of
winning,
but that they change from 1/3 to 2/3, so there IS an advantage to
switching.
You don't have to take our word for it. Write yourself a program
to compare how always switching compares to always standing pat
and see which strategy wins the car more often.
Don Del Grande
>Thank you for you reply.
>
>But again, here's where I get confused....
>
>You said:
>> The contestant has a 2/3 chance of choosing a door with nothing, and
>> there is always a car behind the other door in this case; the
>> contestant has a 1/3 chance of choosing the door with the car behind
>> it, and there is always nothing behind the other door in this case.
>>
>> Chance of winning the car by keeping the initial choice: 1/3.
>
>Why 1/3?
Because, the probability that the host opens a door without the car
behind it is 1 regardless of which door the contestant chose.
Therefore, the probability that the contestant chose the door with the
car is still 1/3, and the probability that the car was behind one of
the other two doors is still 2/3 - and once one of the doors was
opened, the probability that it is behind one of the two unchosen
doors is now equal to the probability that it is behind the unchosen
door that has not been opened.
-- Don
Tim Smith
<89811489-7b7c-47c0...@n33g2000pri.googlegroups.com>,
voice_o...@australia.edu wrote:
> Thus after the initial reveal, the chance the his door contains the
> car increase from 1/3 to 1/2 -- or 50:50 -- and so there is no
> advantage to switching.
Code up a simulation and try it. You will soon see that you are
massively wrong.
Your "reasoning" has the same error that the following "reasoning" has:
The Sun will either rise tomorrow or it won't. Therefore, the
chances that the Sun will rise tomorrow are 50:50.
--
--Tim Smith
Tim Smith
<6fcd35b3-8d21-4925...@40g2000prx.googlegroups.com>,
voice_o...@australia.edu wrote:
> Seems to me, the initial odds were 1:3. But as soon as one of the
> doors was opened revealing nothing, his odds just increases to
> 50:50....as in "maybe the car is behind HIS door...maybe its behind
> the OTHER door".
>
> Yes, it is true the the host knows which door has the car. But, the
> fact that the host chose to reveal door B rather than door C (from my
> example below) tells me nothing about what's behind door A.....and I
> cannot see how it affect the odds of what's behind door A either.
(Assumption: host knows where the car is, and he ALWAYS opens a door
without the car and gives you the chance to switch).
It doesn't affect the odds. You pick a door from the 3. You have a 1/3
chance of having picked the car. Nothing the host does after that, but
before the car is revealed (short of actually moving the car behind the
scenes) changes that. Your door has a 1/3 chance of having the car.
Let's say you take door 1. There is a 1/3 you have the car. There is a
2/3 chance it is behind one of the other two doors.
The host opens door 2 to show no car. At this point, there is still a
1/3 chance you have the car. However, now you know that *IF* the car is
not behind your door (2/3 chance that it is not), *THEN* it must be
behind 3.
One thing you are overlooking is that the host is constrained. If you
picked the car, he can pick which door to open, but if you did NOT pick
the car, his pick of door to open is forced, since he can't open your
door.
Here are two other ways that sometimes help confused people to
understand this puzzle.
1. Let's change the game slightly. Instead of 3 doors, let's play with
1000000 doors. You pick #1. The host opens 999998 other doors, leaving
door #591881 and yours. Still think it is 50:50 that your initial pick
happened to be the one in a million pick that hit the car?
2. Let's try a different change. Instead of opening an empty door and
then offering you the chance to switch, the host does not open any
doors. Instead, he offers you the choice between sticking with your
initial pick, or switching to *both* other doors. You gonna stay or
switch? (This version is mathematically equivalent to the original,
given that there are always two doors with no prize).
As a bonus, I'll give you a third way to realize why switching is
better. At the beginning, you have no knowledge about which door wins.
So, without loss of generality, you can always pick #1.
Now consider if you play the game a bunch of times and you never switch.
When do you win. The answer is simple: you win if and only if the car
was behind door #1 that round. That's 1/3 of the time.
How about the person who always switches? He wins if and only if the
car was *NOT* behind door #1. He wins when it is behind #2 *and* he
wins when it is behind #3.
--
--Tim Smith
voice_o...@australia.edu
> In article
> <89811489-7b7c-47c0-b688-6bfc5d8bd...@n33g2000pri.googlegroups.com>,
>
> Code up a simulation and try it.
>
Unfortunately, the software package I initially chose (statistics101
-- resampling freeware) has an odd quirk that won't allow me to use my
initial coding approach (oddly, you cannot use variables as arguments
for functions....) So I'm either going to have to try doing it in Java
( a language I have not used in a loooooong time) -- or download a C++
compiler and examine some of the simulation available online.
On Jan 3, 1:27 pm, Tim Smith <reply_in_gr...@mouse-potato.com> wrote:
> Here are two other ways that sometimes help confused people to
> understand this puzzle.
>
> 1. Let's change the game slightly. Instead of 3 doors, let's play with
> 1000000 doors. You pick #1. The host opens 999998 other doors, leaving
> door #591881 and yours. Still think it is 50:50 that your initial pick
> happened to be the one in a million pick that hit the car?
Yes.....
Just as I would if he left my door and #449987...or #234987...or #2. I
have a 50:50 chance that he is just trying to psyche me out and get me
to switch away from the door with the car...my original choice door
#1. Lots of numbers makes it "seem" different...but its not.
The same logic applies in Poker.
Technically there's nothing "special" about a Straight. We could just
as easily make the winning hand 2479K. But as Humans, we like to see
numbers in sequence...thus we declare 23456 or 45678 or whatever, a
special hand.
>
> 2. Let's try a different change. Instead of opening an empty door and
> then offering you the chance to switch, the host does not open any
> doors. Instead, he offers you the choice between sticking with your
> initial pick, or switching to *both* other doors. You gonna stay or
> switch? (This version is mathematically equivalent to the original,
> given that there are always two doors with no prize).
>
Assuming that I win if the car is behind either door, then of course
I'll take more doors. In *THIS* case, I do have a 2/3 chance of
winning. So how is this equivalent?
> As a bonus, I'll give you a third way to realize why switching is
> better. At the beginning, you have no knowledge about which door wins.
> So, without loss of generality, you can always pick #1.
>
> Now consider if you play the game a bunch of times and you never switch.
> When do you win. The answer is simple: you win if and only if the car
> was behind door #1 that round. That's 1/3 of the time.
>
> How about the person who always switches? He wins if and only if the
> car was *NOT* behind door #1. He wins when it is behind #2 *and* he
> wins when it is behind #3.
No...he wins if the car is behind door #2 *or* door #3...depending on
the round. But for a given round, he only wins if it is behind his ONE
ultimately chosen door.
> Your "reasoning" has the same error that the following "reasoning" has:
>
> The Sun will either rise tomorrow or it won't. Therefore, the
> chances that the Sun will rise tomorrow are 50:50.
>
Not at all. Indeed, its hard to see why it is even a distant
approximation of my reasoning.
If nothing else, the Sun is governed by laws of physics; Monty Hall is
not.
Toni...@yahoo.com
> On Jan 3, 1:00 pm, Tim Smith <reply_in_gr...@mouse-potato.com> wrote:> In article
>
> > Here are two other ways that sometimes help confused people to
> > understand this puzzle.
>
> > 1. Let's change the game slightly. Instead of 3 doors, let's play with
> > 1000000 doors. You pick #1. The host opens 999998 other doors, leaving
> > door #591881 and yours. Still think it is 50:50 that your initial pick
> > happened to be the one in a million pick that hit the car?
>
> Yes.....
>
you REALLY think it is THE SAME probability that you chose from the
beginning the correct door than the other single door that remains
closed is the correct one? If you do think it is the same then maybe
you need to study a litte further all this logic and probability
stuff...!
> Just as I would if he left my door and #449987...or #234987...or #2. I
> have a 50:50 chance that he is just trying to psyche me out and get me
> to switch away from the door with the car...my original choice door
> #1. Lots of numbers makes it "seem" different...but its not.
>
> The same logic applies in Poker.
>
== Dude, keep yourself far...FAR...away from casinos!
Regards
Tonio
grabber
"Tim Smith" <reply_i...@mouse-potato.com> wrote in message
news:reply_in_group-92D...@news.supernews.com...
> In article
> <6fcd35b3-8d21-4925...@40g2000prx.googlegroups.com>,
> voice_o...@australia.edu wrote:
>> Seems to me, the initial odds were 1:3. But as soon as one of the
>> doors was opened revealing nothing, his odds just increases to
>> 50:50....as in "maybe the car is behind HIS door...maybe its behind
>> the OTHER door".
>>
>> Yes, it is true the the host knows which door has the car. But, the
>> fact that the host chose to reveal door B rather than door C (from my
>> example below) tells me nothing about what's behind door A.....and I
>> cannot see how it affect the odds of what's behind door A either.
>
> (Assumption: host knows where the car is, and he ALWAYS opens a door
> without the car and gives you the chance to switch).
Additional necessary assumption: when he has a choice of two doors with no
car, he choses at random which to open.
Dave Seaman
> On Jan 3, 1:00 pm, Tim Smith <reply_in_gr...@mouse-potato.com> wrote:
>> then offering you the chance to switch, the host does not open any
>> initial pick, or switching to *both* other doors. ?You gonna stay or
>> switch? ?(This version is mathematically equivalent to the original,
>> given that there are always two doors with no prize).
> Assuming that I win if the car is behind either door, then of course
> I'll take more doors. In *THIS* case, I do have a 2/3 chance of
> winning. So how is this equivalent?
Does it make a difference how the doors are opened? Let's examine two
scenarios. You agreed to switch to the two unopened doors.
1. The host opens your two doors simultaneously. You win if either of
them has the car.
2. The host opens one door at a time, always opening an empty door
first. You win if the second door has the car.
Since you agreed to switch to the two unchosen doors, you evidently
didn't think it was important to know how the doors would be opened. Do
you still think that?
If you think it makes a difference, then how long a time interval is
needed in order for there to be a difference? One minute? One second?
One nanosecond?
>> As a bonus, I'll give you a third way to realize why switching is
>> better. ?At the beginning, you have no knowledge about which door wins. ?
>> So, without loss of generality, you can always pick #1.
>> Now consider if you play the game a bunch of times and you never switch. ?
>> When do you win. ?The answer is simple: you win if and only if the car
>> was behind door #1 that round. ?That's 1/3 of the time.
>> How about the person who always switches? ?He wins if and only if the
>> car was *NOT* behind door #1. ?He wins when it is behind #2 *and* he
>> wins when it is behind #3.
> No...he wins if the car is behind door #2 *or* door #3...depending on
> the round. But for a given round, he only wins if it is behind his ONE
> ultimately chosen door.
Just like in the scenario above where you take two doors, which you have
already agreed to. If the doors are opened one at a time, then you only
win if the last door opened has the car. But if both doors are opened at
once, then you win if either door has the car.
>> Your "reasoning" has the same error that the following "reasoning" has:
>> The Sun will either rise tomorrow or it won't. Therefore, the
>> chances that the Sun will rise tomorrow are 50:50.
> Not at all. Indeed, its hard to see why it is even a distant
> approximation of my reasoning.
> If nothing else, the Sun is governed by laws of physics; Monty Hall is
> not.
You completely missed the point. The mere fact that there are two
possibilities does not make them equally probable.
--
Dave Seaman
Third Circuit ignores precedent in Mumia Abu-Jamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>
Daryl McCullough
>
>Thank you for you reply.
>
>But again, here's where I get confused....
>
>You said:
>> The contestant has a 2/3 chance of choosing a door with nothing, and
>> there is always a car behind the other door in this case; the
>> contestant has a 1/3 chance of choosing the door with the car behind
>> it, and there is always nothing behind the other door in this case.
>>
>> Chance of winning the car by keeping the initial choice: 1/3.
>
>Why 1/3?
>
>Seems to me, the initial odds were 1:3. But as soon as one of the
>doors was opened revealing nothing, his odds just increases to
>50:50....as in "maybe the car is behind HIS door...maybe its behind
>the OTHER door".
You talk about the odds changing when Monty reveals an empty
door. It's true that odds change when you get new information,
but it's not always as simple as "now there are only two
possibilities, so it's a 50/50 chance".
Suppose that you play the lottery every day. Suppose that
it's a one in a million chance of winning. Every morning,
your friend reads the paper to find out who won yesterday's
lottery. Rather than telling you directly who won, he says
"Well, either you won or Leonard Martin won", or "Either
you won or Tom Clement won", or "Either you won or Sally
Hatfield won". The fact that your friend describes the
situation as two possibilities does *not* make it a
50/50 chance.
--
Daryl McCullough
Ithaca, NY
riderofgiraffes
> soon as one of the doors was opened revealing
> nothing, his odds just increases to 50:50
Let's go through a sequence to see where your beliefs
lie.
You choose a door. Monty knows where the car is
hidden, and he acts as follows:
If you have chosen the door with the car, he
chooses a door at random and opens it.
If you have not chosen the door with the car,
he opens the remaining door-less car.
OK. Before Monty opens the door, you agree that
the probability that the car is behind the door
you chose is 1/3.
Now Monty decides which door to open, but he doesn't
tell anyone.
Is your probability still 1/3?
Monty now tells me, and I don't give it away.
Is your probability still 1/3?
We now tell the entire world, apart from you.
Is your probability still 1/3?
You close you eyes, and we open the door Monty has
chosen.
Is your probability still 1/3?
We reclose the door, and you open your eyes.
Is your probability still 1/3?
You are now allowed to switch to "owning" both
of the other doors.
Is your probability now 2/3?
We point at, but do not open, the door Monty
has chosen.
Is your probability is still 2/3?
When does the probability that your first choice is
right change from 1/3 to 1/2?
Why does the probability of your first choice change?
Monty opening one of the other doors tells you nothing
about the door you've chosen, it can only tell you about
the doors he does, or does not, choose.
And if you're having trouble programming a simulation,
I strongly recommend you download Python for free and
learn that.
Tim Smith
<65c60a7e-4e5b-465b...@a29g2000pra.googlegroups.com>,
voice_o...@australia.edu wrote:
> On Jan 3, 1:00 pm, Tim Smith <reply_in_gr...@mouse-potato.com> wrote:
> > In article
> > <89811489-7b7c-47c0-b688-6bfc5d8bd...@n33g2000pri.googlegroups.com>,
> >
> > Code up a simulation and try it.
> >
> I am in the process of doing just that.
>
> Unfortunately, the software package I initially chose (statistics101
> -- resampling freeware) has an odd quirk that won't allow me to use my
> initial coding approach (oddly, you cannot use variables as arguments
> for functions....) So I'm either going to have to try doing it in Java
> ( a language I have not used in a loooooong time) -- or download a C++
> compiler and examine some of the simulation available online.
Actually, don't bother coding it. You don't need to. See below.
> > Here are two other ways that sometimes help confused people to
> > understand this puzzle.
> >
> > 1. Let's change the game slightly. Instead of 3 doors, let's play with
> > 1000000 doors. You pick #1. The host opens 999998 other doors, leaving
> > door #591881 and yours. Still think it is 50:50 that your initial pick
> > happened to be the one in a million pick that hit the car?
>
> Yes.....
>
> Just as I would if he left my door and #449987...or #234987...or #2. I
> have a 50:50 chance that he is just trying to psyche me out and get me
> to switch away from the door with the car...my original choice door
> #1. Lots of numbers makes it "seem" different...but its not.
Whoa...OK, now I see your real problem. In the standard Monty Hall
problem, Monty has no choice about giving you the chance to switch:
1. The car is put behind one door, at random. Monty is told where it is.
2. You pick a door.
3. From the remaining doors, Monty picks one that he knows does not have
the car (which he can always do), and opens that.
4. Monty gives you the chance to switch.
Instead of trying to code it up, simply play the game yourself, but not
the 3 door version. Do 10 doors, numbered 0 to 9. Always pick 0. Do
it a few times, keeping track of how often switching wins and how often
non-switching wins. Keep doing this until you are convinced that it is
not 50:50. It won't take long.
Here, I'll do it. I will use the digits of e to determine what door has
the car (remember, I am picking door 0). So, here is where the car
is on successive trials:
Car non-switch win switch win
2 Y
7 Y
1 Y
8 Y
2 Y
8 Y
1 Y
8 Y
2 Y
8 Y
4 Y
5 Y
9 Y
0 Y
4 Y
5 Y
2 Y
3 Y
5 Y
3 Y
6 Y
0 Y
2 Y
8 Y
7 Y
4 Y
7 Y
1 Y
3 Y
5 Y
2 Y
6 Y
6 Y
2 Y
4 Y
9 Y
7 Y
7 Y
5 Y
7 Y
2 Y
4 Y
7 Y
0 Y
9 Y
3 Y
6 Y
9 Y
9 Y
9 Y
Note that switcher won 47 out of 50 times. Non-switcher won 3 out of 50
times. Still think it is 50:50?
Try it with pi, or log(2), or cos(1), or sqrt(2), or any other
irrational mathematical constant.
> > 2. Let's try a different change. Instead of opening an empty door and
> > then offering you the chance to switch, the host does not open any
> > doors. Instead, he offers you the choice between sticking with your
> > initial pick, or switching to *both* other doors. You gonna stay or
> > switch? (This version is mathematically equivalent to the original,
> > given that there are always two doors with no prize).
> >
>
> Assuming that I win if the car is behind either door, then of course
> I'll take more doors. In *THIS* case, I do have a 2/3 chance of
> winning. So how is this equivalent?
It's the same game. Here's how to see that. Imagine two people, P1 and
P2, playing with two Monty's, M1 and M2. P1/M1 and P2/M2 play in sync.
P1/M1 play the standard game. P2/M2 play the "stay, or switch to both
doors" variant. P1 and P2 always pick door #1. M1 and M2 always put
the car behind the same door in a given round.
Note that if P1 and P2 make the same decision, they get the same result.
This shows that there is an isomorphism between the standard game P1/M1
play, and the "both doors" variant P2/M2 are playing.
--
--Tim Smith
Tim Smith
<5360604.12309953217...@nitrogen.mathforum.org>,
riderofgiraffes <mathforu...@solipsys.co.uk> wrote:
> And if you're having trouble programming a simulation,
> I strongly recommend you download Python for free and
> learn that.
Here's a simulation in Perl. Note that it contains a lot of unnecessary
calculations in order to remain faithful to the problem. More on that
after that code and explanation of the output.
=================================================
#!/usr/bin/perl
use strict;
my($num_doors,$num_trials) = @ARGV;
my $switch_wins = 0;
my $stay_wins = 0;
while ($num_trials-- > 0) {
my $car_behind = int(rand($num_doors));
my $pick = int(rand($num_doors));
# figure out what door Monty leaves unoppened
# If we did not pick the car, it is the door the
# car is behind.
my $monty_leaves = $car_behind;
# If we DID pick the car, Monty picks door to
# leave unopened at random
if ( $pick == $car_behind ) {
$monty_leaves = $pick;
while ( $monty_leaves == $pick ) {
$monty_leaves = int(rand($num_doors));
}
}
$switch_wins++ if $car_behind == $monty_leaves;
$stay_wins++ if $car_behind == $pick;
print "$switch_wins:$stay_wins, C=$car_behind, P=$pick,
L=$monty_leaves\n";
#Note: above line wrapped by my news software. Unwrap before running!
}
=================================================
Run it with two arguments: the number of doors, and the number of
rounds. Here's the last few lines of a run for 3 doors, 100 rounds:
65:29, C=2, P=0, L=2
66:29, C=1, P=2, L=1
66:30, C=0, P=0, L=1
67:30, C=0, P=2, L=0
67:31, C=1, P=1, L=0
67:32, C=0, P=0, L=2
67:33, C=2, P=2, L=0
Take the last line. The 67:33 means that the switcher won 67 times, the
non-switcher won 37 times. The rest of the line is that particular
round. C=2 means the car was behind door 2. P=2 means the player
picked door 2. L=0 means Monty opened all the other doors except door 0.
I've seen people code these simulations before, and most people don't
finish them. That's because they get to this point:
my $car_behind = int(rand($num_doors));
my $pick = int(rand($num_doors));
and then write something like this:
$stay_wins++ if $car_behind == $pick;
$switch_wins++ if
and then realize that the condition for this is $car_behind != $pick, and
then say "Do'h!!!!!", and realize that staying wins with probability
1/$num_doors and switching wins with probability
($num_doors-1)/$num_doors, and there's no point in writing the
simulation.
--
--Tim Smith
spudnik
> One thing you are overlooking is that the host is constrained. If you
> picked the car, he can pick which door to open, but if you did NOT pick
> the car, his pick of door to open is forced, since he can't open your
> door.
thus:
don't blame me for what Schroedinger's cat did
to your cosmic litter box, dood.
actually, what you infer is apparently
waht Galileo also didnt believe,
which was his actual heresy (atomism),
not his nonbelief in Ptolemy's hoax (because,
all semi-educated people knew
about the precession of the equinoxes).
you could effectively say that "because"
of the Uncertainty principle,
there is *only* relative vacuum, although
this has been reified into "zero-point energy"
et al ad vomitorium -- "aim if
for the cat's box!"
> all the universe would be one big physical entity
> in which you cant nake a differnce
> between all the SUB ENTITIES!!
thus:
the only complicity that I have seen,
is the Scty. of Transort Mineta's testimony,
that Cheeny stopped the standard Air Force interceptions;
the plane-bombs were probably adequate incendiaries.
thus:
any congruence surd that repeats across the decimal point,
is equal to zero, such as ...9999.9999...; so,
why do you say that 0.349999... is not equal
to 0.350000...? because, the whole import of Hensel,
was to show that integers have the same properties
as decimals, properly oriented -- except that
he didn't use the commonsense ordering of the digits!
thus:
now, isn't that what the key method
of "remote viewers?" well, it depends;
the purveyors of it obviously have to resort
to some sort of hidden special knowledge,
as with magicians, to show some "results...."
anyway, that's the impression I got
from Ghost to Ghost BC radio;
there's a reason, they're called, Spooks!
thus:
no, no, no;
Fermat's unstated proof was not qualified
as to whether or not it applied to n=4, or
he thought at first that it did; clearly,
he wouldn't have needed to prove that case
with his infinite descent contradiction, if
he had already "had" the unstated proof. but,
isn't it clear that n=4 is very special,
after a while of consideration?
thus:
shouldn't it be clear that
photons are an artifact of the idea
of point-particles?... of course, iff
they exist, then they would have to be *the* particles
that actually were zero-dimensional points, but,
since they are waves, as shown by Young, Huyghens et al,
there is really no need for them,
except in the Pauli matrix formalism
of statistical bosons; eh?
Schroedinger's cat is dead --
long-live Schoredinger's cat!
--only 24 hours to impeach Trickier Dick from the N.Admin,
metaphorically typing, or Cheeny & Zbiggy, fo'mo' years;
Good Morning, Afghanistan!
... Good Afternoon, Sudan!
http://tarpley.net/bush12.htm
http://wlym.com/campaigner/8011.pdf -- Brits hate Shakes, Why?
http://www.wlym.com/~seattle/dynamis/
http://www.21stcenturysciencetech.com/current.html
http://www.rwgrayprojects.com/synergetics/plates/plates.html
http://mathdl.maa.org/mathDL/46/?pa=content&sa=viewDocument&nodeId=3163
http://wlym.com/campaigner/8011.pdf -- English, not! (see Psalms 46)
spudnik
this sentence is a lying piece of poop?, or
this sentence is not *necessarily* good poop?
> "Well, either you won or Leonard Martin won", or "Either
> you won or Tom Clement won", or "Either you won or Sally
thus:
that is only dependent upon which base one uses
for his "decimals," because the ambiguity
that's involved with denominators that are prime
to the base, was covered by Stevin;
a huge chunk of classical numbertheory comes out
of this simple realization,
which Simon Stevin codified in "The Decimals" --
imagine that!
> That's still true. The only complaint with 1/3 or 2/3 is when you
> represent it as a decimal. It is not 0.333....
> On the other hand 1/2 is truly equal to 0.5000.....
thus:
That is the question.
> If you picked the car, he can pick which door to open, but
> if you did NOT pick the car, his pick of door
> to open is forced, since he can't open your door.
thus:
don't blame me for what Schroedinger's cat did
to your cosmic litter box, dood.
actually, what you infer is apparently
waht Galileo also didnt believe,
which was his actual heresy (atomism),
not his nonbelief in Ptolemy's hoax (because,
all semi-educated people knew
about the precession of the equinoxes).
you could effectively say that "because"
of the Uncertainty principle,
there is *only* relative vacuum, although
this has been reified into "zero-point energy"
et al ad vomitorium -- "aim if
for the cat's box!"
thus:
voice_o...@australia.edu
> On Jan 3, 12:50 pm, voice_of_rea...@australia.edu wrote:
>
> > > 1. Let's change the game slightly. Instead of 3 doors, let's play with
> > > 1000000 doors. You pick #1. The host opens 999998 other doors, leaving
> > > door #591881 and yours. Still think it is 50:50 that your initial pick
> > > happened to be the one in a million pick that hit the car?
>
> > Yes.....
>
> == Are you kidding??
No.
>
> > The same logic applies in Poker.
>
> == Dude, keep yourself far...FAR...away from casinos!
>
Is it your contention then that a poker hand of 42365 has a high
significance probablity-wise than 3578J?
Toni...@yahoo.com
**************************************************************
No, my contention is :
(1) if you think that after you chose a door out of 100,000, then
99,998 other doors with nothing behind them were opened, the events "I
chose the correct door" and "I chose an incorrect door" are equally
probably (0.5 - 0.5 , as you said), and
2) if you think your above "logical reasoning" equally applies to
poker,
then you better don't play poker or else you'll get bitterly
disappointed...and pennyless.
I recommend you the book "Duelling idiots and other probability
puzzlers", by Paul H. Nahin (pag. 192-193) where this Monty Hall
problem is discussed and even a MatLab program (monty.m) is given
there to simulate it.
Also, the next site has a nice discussion on this problem and a
simulation http://www.cut-the-knot.org/hall.shtml
Regards
Tonio
Tim Smith
<e34c5930-a9d2-4c25...@w24g2000prd.googlegroups.com>,
voice_o...@australia.edu wrote:
> On Jan 3, 7:15 pm, Tonic...@yahoo.com wrote:
> > On Jan 3, 12:50 pm, voice_of_rea...@australia.edu wrote:
> >
> > > > 1. Let's change the game slightly. Instead of 3 doors, let's play with
> > > > 1000000 doors. You pick #1. The host opens 999998 other doors, leaving
> > > > door #591881 and yours. Still think it is 50:50 that your initial pick
> > > > happened to be the one in a million pick that hit the car?
> >
> > > Yes.....
> >
> > == Are you kidding??
>
> No.
So you *really* think that if there are a *MILLION* doors, and only one
has a prize, and you have no idea which one has the prize, you have a
50:50 chance of picking that door when you pick one of those million
doors????????
> >
> > > The same logic applies in Poker.
> >
> > == Dude, keep yourself far...FAR...away from casinos!
That is the best advice you are ever going to receive in your life.
--
--Tim Smith
Daryl McCullough
>
>Follow-up to my previous response......
>wrote:
>
>> Chance of winning the car by keeping the initial choice: 1/3.
>
>Suppose we alter the game slightly. Call it the "Senile Monty Hall"
>Game.
>
>Monty initially offers a choice of 3 doors. The contestant selects
>door (A). Monty reveals that there is no car behind door (B) and
>offers a chance to switch to door (C).
>
>According to your statement, the contestant has still has only a 1/3
>chance of being correct.
>
>Now, senile Monty Halls looses his cue cards and forgets that he has
>already revealed a door. He reveals that there is no car behind door
>(C).
>
>Does the contestant still face only a 1/3 chance of being correct?
No, now it is certain. The difference is that now you know something
more about the doors that you didn't pick: You know that both contain
no car. You didn't know that earlier, and this additional information
changes the odds of the car being behind your door.
In the original Monty Hall problem, when Monty reveals a door with
no car behind it, you haven't learned anything significant. You
already *knew* that there was a door with no car behind it. You
just didn't know whether it was door (B) or door (C).
Look, after your initial pick of door A, you *know* that
one of the following two events will happen:
1. Monty will reveal no car behind door B.
2. Monty will reveal no car behind door C.
Let's give names to various probabilities. Let P1 be
the probability that event 1 happens. Let P2 be the
probability that event 2 happens. Let P1(A) be the
probability, calculated after event 1, that the car
is behind door A. Let P2(A) be the probability,
calculated after event 2, that the car is behind
door A. Let P(A) be the probability that the car
is behind door A, calculated *before* either event
1 or event 2 happens.
Since you *know* that either event 1 will happen
or event 2 will happen (and not both), you can
compute P(A) in terms of the other probabilities:
P(A) = P1(A) * P1 + P2(A) * P2
Obviously, by symmetry (there is no difference between
door B and door C), we have:
P1(A) = P2(A)
P1 = P2
So our formula simplifies to
P(A) = 2 P1(A) P1
Now, since either event 1 or event 2 will happen (and not
both), we know P1 + P2 = 1. So P1 = 1/2. So our formula
simplifies further:
P(A) = 2 P1(A) 1/2
= P1(A)
So the initial probability, P(A) is exactly the same as
the probability after event 1. Event 1 does *not* change
the odds (assuming that it is certain that either event
1 or event 2 happens, but not both).
Now, you change the situation by saying that it is possible
that *both* events 1 and 2 happen. (Monty might open *both*
doors). That changes the calculations. It's no longer true
that P1 and P2 add up to 1.
David Bernier
> On Jan 3, 1:00 pm, Tim Smith <reply_in_gr...@mouse-potato.com> wrote:
>> In article
>> <89811489-7b7c-47c0-b688-6bfc5d8bd...@n33g2000pri.googlegroups.com>,
>>
>> Code up a simulation and try it.
>>
> I am in the process of doing just that.
>
> Unfortunately, the software package I initially chose (statistics101
> -- resampling freeware) has an odd quirk that won't allow me to use my
> initial coding approach (oddly, you cannot use variables as arguments
> for functions....) So I'm either going to have to try doing it in Java
> ( a language I have not used in a loooooong time) -- or download a C++
> compiler and examine some of the simulation available online.
>
> On Jan 3, 1:27 pm, Tim Smith <reply_in_gr...@mouse-potato.com> wrote:
>
>> Here are two other ways that sometimes help confused people to
>> understand this puzzle.
>>
>> 1. Let's change the game slightly. Instead of 3 doors, let's play with
>> 1000000 doors. You pick #1. The host opens 999998 other doors, leaving
>> door #591881 and yours. Still think it is 50:50 that your initial pick
>> happened to be the one in a million pick that hit the car?
>
> Yes.....
>
> Just as I would if he left my door and #449987...or #234987...or #2. I
> have a 50:50 chance that he is just trying to psyche me out and get me
> to switch away from the door with the car...my original choice door
> #1. Lots of numbers makes it "seem" different...but its not.
The "real Monty" will never try to psyche you out. He will always
make a switch offer, whether your original choice has a car behind it
or not. You could imagine putting it in writing before playing:
a legal, binding contract.
In real life, there could be a mixture of "real Monties" and "fake Monties".
The "fake Monties" would sometimes not offer a switch when the chosen
door doesn't hide a car.
The problem assumes that we are dealing with a "real Monty" and that we
know that. But in real life, real and fake Monties could look alike
outwardly ...
David Bernier
Denis Feldmann
extremely bad understanding of the issues would make you an easy prey
for casinos, poker players, etc.
Bart Goddard
0aad36...@w24g2000prd.googlegroups.com:
> On Jan 3, 7:15 pm, Tonic...@yahoo.com wrote:
>> On Jan 3, 12:50 pm, voice_of_rea...@australia.edu wrote:
>>
>> > > 1. Let's change the game slightly. Instead of 3 doors, let's play
> with
>> > > 1000000 doors. You pick #1. The host opens 999998 other doors,
> leaving
>> > > door #591881 and yours. Still think it is 50:50 that your initial
> pick
>> > > happened to be the one in a million pick that hit the car?
>>
>> > Yes.....
>>
>> == Are you kidding??
>
> No.
I wonder if you realize that "the odds" depend on the observer.
In this problem, there are 3 perspectives, and each perspective
has different odds. First, there is Monty. He knows where
the car is, so if you pick Door 1, _to Monty_ the probability
you have chosen the car is either 1 or 0.
Second, there is you, who chose a door and then watched Monty
open another door. _To you_, the probability you chose the
right door is 1/3.
Here's where I think folks get confused. There's a third person,
who walks in late. He sees two closed doors and knows one
has a car. _To him_ the probability that the car is behind
Door 1 is 1/2 (because he has less information than you.)
There really is no such thing as "THE odds", since each
person has different knowledge, and therefore has his own
personal "odds". In your explanations, you assert that the
odds change. What really happens is that the player's odds
stay the same, but you are changing your perspective to that
of the late-comer. But the late-comer isn't playing; you are.
Bart
--
Cheerfully resisting change since 1959.
kibby
> On Jan 3, 7:15 pm, Tonic...@yahoo.com wrote:
>
> > On Jan 3, 12:50 pm, voice_of_rea...@australia.edu wrote:
>
> > > > 1. Let's change the game slightly. Instead of 3 doors, let's play with
> > > > 1000000 doors. You pick #1. The host opens 999998 other doors, leaving
> > > > door #591881 and yours. Still think it is 50:50 that your initial pick
> > > > happened to be the one in a million pick that hit the car?
>
> > > Yes.....
>
> > == Are you kidding??
>
> No.
For what it's worth, the Monty Hall problem is a well-established
problem that is not debatable. We're all trying to help you see the
correct answer, not being combattant.
First, your Poker analogy fails because the probability of having a
straight is way lower than having a non-straight. But, having 5
specific cards has the same probability regardless of the cards. Those
are different things.
Here was the point of the response:
With 3 doors, it's hard to see why switching would make a big
difference. But with 100,000 doors, it becomes a little more apparent.
The host knows where the car is, but you don't. The odds of you
initially picking the car were 1/100,000. Good luck.
So the host, knowing the car is behind door #43,512 says, "Well it's a
good thing you didn't pick door #2! Or 3! Or 4! ... Or 43,510! Or
43,511!" Then he skips over 43,512 and keeps going: "Or 43,513! Or
43,514! ..."
Intuitively, you would feel like odds are pretty gargantuan that
switching once he finally finishes would be justified. After all, why
did he conspicuously leave out 43,512? In this case, your intuition
would actually be correct! This is because before he opened anything,
there was a 99,999/100,000 probability that the car was behind ONE of
those many, many other doors. The host opening doors once you made
your pick didn't change that. There's still a 99,999/100,000
probability that it's behind one of those many doors, but now there's
only one to pick from. I'd take the 99.999% chance.
Arguing the point is certainly worthwhile in that it helps you
understand what's going on, but please read some of these responses a
bit closer. The odds are 2/3 of winning if you switch, not 1/2.
Virgil
It certainly has a better chance, probability-wise, of being a winning
hand in any deal.
Virgil
Bart Goddard <godd...@netscape.net> wrote:
In a very real sense, the third person has MORE , not less, information,
at least after Monty opens a door.
He (or she) knows that the car is not behind the opened door.
So that that increase in information, provided Monty is reliable about
always opening a door, changes the expectations.
spudnik
you flipped a coin to decide which of the two?
> In a very real sense, the third person has MORE , not less, information,
> at least after Monty opens a door.
thus:
I never read muc of the correspondence with Bernoulli
on the brachistochrone/tautochrone, but
hd did say that it was the same as Galileo's cycloid. so,
perhaps they were refering to any such epicycloid
that was "rolling" on Earth,
with respect to a source of light in Earth's orbit,
viz Moon ... if that can be made to work,
mechanically/graphically.
> if it went around twice before landing, hemicycloidal?...
> what would it be for three?
thus:
now, y'talking!
> See:http://mysite.verizon.net/cjnelson9/index.htm
> Does anybody measure area and volume by anything but squares and cubes
> with coordinates?
thus:
yes. of course, Wiles was working on it in his basement office
for seven years, I think, sheletering himself
> Wasn't there Serre's epsilon-conjecture, that a Frey elliptic
> curve would not be modular? And that that was eventually proved by
> Ribet? NOVA (PBS network, US) had an episode "The Proof".
thus:
I don't recall if Deffeyes is Dutch, but
he did work with Hubbert at Shell, which is half British;
between the "free market" in the City of London, and
the massive ports of Rotterdam, the cartel is anglo-dutch. well,
that's probably not quite what he says, and
he seems to believe in the mainstream paradigm as
to the ultimate source of oil, based on that funny word, and
that funny dinosaur in the Mobil commercial,
which he dysavows.
just wish I'd read his first book, when it came out, and
it might have had most of what is in this one.
the most important thing, so far, is that, although
heat engines are restrained absolutely by the Carnot cycle,
fuel cells & such are not. still,
they ain't the panacea, either ...
that is probably your God-am what's-her-name,
that you started this funky item with.
thus:
although they have no other simple valuation,
any congruence surd that repeats across the decimal point,
is equal to zero, such as ...9999.9999...; so,
the whole import of Hensel,
was to show that integers have the same properties
as decimals, properly oriented -- except that
he didn't use the commonsense ordering of the digits!
thus:
now, isn't that what the key method
of "remote viewers?" well, it depends....
Pink Pig
I don't know where to put this message, so I just picked one of the
answers to reply to. The Monty Hall Problem is fascinating, because it
can be proved in many ways that the odds in favor of switching are
2:1, but many people are unable to follow the logic of some of the
proofs. My personal preference is to use "restricted choice", an
argument frequently used by contract bridge players to settle similar
problems. The argument goes: If your original choice was correct, then
Monty could have opened either other door, and the probability that he
chooses a specific door is therefore 1/2; if, on the other hand, your
original choice was incorrect, then Monty is left with no choice as to
which door to open (i.e. his choice is restricted), so the probability
that he chooses a specific door is 100%, i.e. twice as likely as the
alternative.
There's a fairly clear explanation in Wikipedia:
http://en.wikipedia.org/wiki/Monty_hall_problem .
Another way of looking at it: the original probability that you picked
the correct door is 1/3, and nothing has happened that will change
that. On the other hand, Monty has eliminated one of the other
choices, leaving a single alternative. It follows that the probability
that the prize is behind the 3rd door must be 1 - 1/3 = 2/3. This is
because Monty is not allowed to open the door you originally chose. He
must open one of the other two doors, and the prize cannot be behind
it.
spudnik
is that, so?
thus:
little thanks for small things;
as far as i can tell,
there is no where in AP-adics, perhaps
only because "..." is an infinite string
of some thing ... a-ha;
cosmological super-APadic-strings!
> i describe AP-adics with polysigned and p-adics,
> not p-adics with polysigned or vice versa.
thus:
you really know that the sky is falling, iff
you know in what direction....
I'm sure that Andrew was not the only one,
who asked Doctor Spock, if that was entirely logical,
Captain Crunch!
thus:
Ribet'd be a great escrow agent; stipulate,
he keeps the money, if you're wrong:
he's at UCBerkeley, I think.
(1990), http://www.numdam.org/numdam-bin/item?id=AFST_1990_5_11_1_116_0
thus:
there was an arbitrary shift of temperament
of the entire scale to "A=44 Hertz" in the late '30s
-- therein lies a tale -- for the sake of "brilliance;"
it hurts....
classically, more-or-less if not an adopted standard
per se, Verdi's tuning of middle-C = 256 cycles/second is a kind
of nice. so, why is is that "C" is the cannonical major scale --
so that Am is the cannonical minor scale?
> Indeed. Yes a good composer will never allow a refrain
> no matter how many times it's repeated. to become stale.
thus:
I think it is fairly safe to say,
Man could always count *with* coconuts;
just don't hang-out underneath even one!
> man could not even count coconuts, and there are many open questions
thus: truly,
they were the Guinsees-Book-obvious-ever-Druids-award kind
of Druids, but that doesn't mean, they were perfectly fake, and
lots of kids have used better-quality 2-by-9s, since
they retired -- a-hem; farmer's are good business men, two....
D&D, not the world's greatest carpenters!
> I will guess that Wookiepoopya has no rational explanation
> for cropcircles, other than "Doug and Dave
> with *the world's smallest pile* of 2by9s."
thus:
groovier than a pile of rocks,
just by comparison with the Face and Genitals ... a-hem;
Stranger in "a" Strange Land territory.
thus:
dood was totally quanta; I mean, if
you're going to say "it's a singleton -- yeah;
that's what it is!" well, OK;
what spin does it partake of -- is it a fermion?
I know, there are those of you who'd jump
on the answer, "weellll, if there's only one?" -- and
so would I, myself & what's-his-name; me, not just so as
to not fall into the Many Universes default; except,
when I'm "@Nancago.edU of." -- death to the triangle!
any way, Stanton F. is higher than a pile of rocks, iff
you listen to Art Bell et sequentia Al Nouri;
is his first name really Albert?... does Bell parlay
of distance at a spooky action, huge media conglome,
semiotic highway signage?... oh, my;
they are actually better than cropcircles, if
you can just get the slant on the "correct side;"
even a mobius strip has at least one!
thus:
did Copernicus expose Ptolemy's hoax?... and, if not,
how'd you know that it was wrong --
did you know that the sky is round?
spudnik
relating to the ages of siblings,
in which she seems to have mistakenly
taken them to be doors. well,
y'make a better floor, than a door, Contestant!
thus:
however, the *shape* of teh coat-
hangar is *extremely* critical.
> Yo, G, reading problems? He agrees it is a crock.
thus:
photon is nonsequiteroxymoronmisnomer,
Moon hits your eye *like* a big-a piizza pie-a;
look at the structure of the "rods & cones...."
an oscilliscopic trace of electronic current
in a wire is Minkowsian timespace (1+1)d,
often callibrated using a Clifford algebra, but
it is actually a spatial phenomenon
of the wire; perhaps the oscillisc. trace is good
as a projection of that.
> > >http://superstruny.aspweb.cz/images/fyzika/foton.gif
> > Compare withhttp://users.accesscomm.ca/john/PHOTON.GIF
spudnik
I immediately found afterward a dyscussion
of this very problem, which is sort of a generalization;
I have not read past the first screen, to dyscern if
she had actually erred or not, as I had thought, but
*if* she had, it was probably because of Mister Hall and
his little problem.
http://www.marilynvossavant.com/forum/viewtopic.php?t=704
CC MvS thus:
it makes a better floor than a door dept.:
well, the wiki Bayesian entry has what I was referring to,
in the Mister Hall item, as the "Boy-Girl problem." now,
there are two basic versions, and it seems,
upon my just-now reading
of them, that Savant confused the two -- perhaps,
rather, with the Mister Hall problem --
thus giving an answer in thirds, as I recall. whereas,
I do recall that it was the problem involving the ages
of the two siblings, in which case I was correct:
If the older of two siblings is a boy,
what's the probabiliy of the younger, being a girl?
I had meant to write to her, over the years.
interestingly, the Mister Hall Problem has an earlier variant,
the Three Prisoners of Game Theory, but
"your" choice only conditions the probability
of the other guy, getting pardoned.
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox