My Statistical Thoughts on Monty Hall Problem (nonREBOL)
[1/12] from: reboler:programmer at: 20Dec2001 2:25
Okay you guys brought this up.
Here is my speculation on the Monty Hall Problem.
[REBOLcontentless discussion about Probability follows]:
Let me see what you guys think about the switch/win probability for the following circumstances.
Monty picks a door randomly from [123].
You pick a door, and tell Monty.
Monty removes the door number from his list that is not yours, and not his.
You are given a choice: stay with your door number or pick the remaining door number.
*** You flip a fair coin: heads  you pick the first remaining number
tails  you pick the second remaining number
Either heads or tails could be keeper or switcher depending on your first choice
You win if your door = Monty's door.
*** Important part follows ***
Your coin flip is COMPLETELY INDEPENDENT (in the true Probability sense) of the first
choice from [123]: the coin flip can have no influence on which remaining door is the
winner, and which remaining door is the winner cannot influence the coin toss. Further,
the coin flip is completely independent of your first choice: your first choice cannot
influence the coin, and the coin had nothing to do with your first choice.
*** Therefore: Heads or Tails both have a 1/2 chance of coming up, and 1/2 chance of
matching Monty's number (winning).
That is the very definition of statistical independence.
Note: In this scenario both you and Monty have the same information as in the "traditional"
scenario (without a coin flip).
Analysis of this suggests that:
 if the Monty Hall Problem gives the switch/win probability of 2/3
 then 
A) choosing a door reduces the chances that the door is a winner (keeper win = 1/3, switcher
win =2/3)
or
B) Monty removing a nonpicked door reduces the chance that a picked door is a winner.
*** If I call heads before a coin is flipped, does that reduce the chances that heads
comes up?***
*** If I call a number on a die, and you then scratch off one of those numbers (not mine)
does that reduce the chance that my number will come up? ***
(A) clearly cannot be the case: choosing a door does not change what is behind it.
(B) clearly cannot be the case: removing a door does not change what is behind the other
doors.
Seems to me that switch/win probability = 2/3 is the same as saying that a fair coin
will come up with heads or tails more often depending on which one is the winner.
Please help me understand the difference between the two scenarios.
My explanation: there are TWO INDEPENDENT CHOICES.
Switch and keep each have a 1/2 half probability because Monty's choice of the door to
open is not RANDOM (i.e. independent of your choice and the winning door).
*** You have essentially made TWO independent choices, whether you switch or keep. The
keeper simply chooses the same door again, the switcher chooses a different door.
Consider: Monty opens the door before you choose:
Monty shows you three doors, and opens one of them (ALWAYS A NONWINNER no matter what
you choose)
You pick a door.
Monty asks you to switch or keep.
Since your choice was between two doors, if doesn't matter if you switch or keep
 all probabilities are 1/2
MONTY IS ALWAYS GOING TO PICK A NONWINNER, YOUR FIRST CHOICE IS TOTALLY IRRELEVANT.
SWITCHING AND KEEPING ARE NONREAL CHOICES, ONLY YOUR SECOND CHOICE COUNTS.
[2/12] from: bga:bugbr at: 19Dec2001 16:48
From: "alan parman" <[rebolerprogrammernet]>
> Seems to me that switch/win probability = 2/3 is the same as saying that a
fair coin will come up with heads or tails
>more often depending on which one is the winner.
>
> Please help me understand the difference between the two scenarios.
Instead of thinking about 3 doors, thing about 10000 doors:
1  You pick one door.
2  The other guy picks 9998 (none of them has the prize) and leaves only
one door.
Would you switch or not?
When you did your first pick, you have 1/10000 chances of winning and the
change that the prize would be in one of the other 9999 doors is 9999/10000.
In other words, the prize *WILL BE* in a door that is not the one you choose
(a 9999/10000 chance is enough to me to say that).
Now, the other guy excluded 9998 doors and the second group only has one
door now... But as we saw above, the probabitily that the prize is in this
second group is 9999/10000. So the probability that the prize is in this
single door is 9999/10000 and in the door you choose is 1/10000. That's why
you *MUST* change to the other door. :)
Bruno
[3/12] from: lgidding:aaa:allianz:au at: 20Dec2001 10:43
Hi,
Finally thought I'd add my two penneth....
Bruno,
I think your logic might be flawed.....
>Would you switch or not?
>When you did your first pick, you have 1/10000 chances of winning and the
>change that the prize would be in one of the other 9999 doors is
9999/10000.
>In other words, the prize *WILL BE* in a door that is not the one you
choose
>(a 9999/10000 chance is enough to me to say that).
>Now, the other guy excluded 9998 doors and the second group only has one
>door now... But as we saw above, the probabitily that the prize is in this
>second group is 9999/10000. So the probability that the prize is in this
>single door is 9999/10000 and in the door you choose is 1/10000. That's
why
>you *MUST* change to the other door. :)
You appear to believe that the original 1/10,000 chance is still valid
after 9998 the doors have been opened.
Surely like the coin which has no knowledge of its previous toss,
the odds have no knowledge of their previous length.
For example:
Imagine a Roulette wheel with 10000 numbers
I pick no 1 and have a 1/10000 chance
The next time the wheel has reduced to 36 numbers
I still have no 1 but now I have a 1/36 chance
Finally the wheel is reduced to just 2 numbers
I still have no 1 but now I have a 1/2 chance
But my odds, if i swap to no 2, cannot possibly be improved
by my knowledge that 9998 PREVIOUSLY losing numbers
have now been dropped from the wheel.
Just a thought
Laurence
[4/12] from: joel:neely:fedex at: 20Dec2001 0:15
Hi, Alan,
alan parman wrote:
> Here is my speculation on the Monty Hall Problem.
>
> [REBOLcontentless discussion about Probability follows]:
... snipped ...
The thread discussing this puzzle has provided many examples
of the difference between "commonsense" reasoning and
Mathematical analysis. In problems like this one, verbal
persuasion counts for naught compared with actual, inarguable
logic. (The same can be said for program proof, but that's
a different flame war... ;)
When all else fails, resort to brute force! In analysis of
probabilities, that means a decision tree which traces the
succession of choices (which door hides the prize, which
door the contestant chooses, which door Monty opens), along
with the joint probabilities of the paths through the tree:
+ m=2 (1/2:1/18) case 1
+ c=1 (1/3:1/9) +
 + m=3 (1/2:1/18) case 2

+ p=1 (1/3:1/3) + c=2 (1/3:1/9)  m=3 (1/1: 1/9) case 3
 
 + c=3 (1/3:1/9)  m=2 (1/1: 1/9) case 4

 + c=1 (1/3:1/9)  m=3 (1/1: 1/9) case 5
 
  + m=1 (1/2:1/18) case 6
+ p=2 (1/3:1/3) + c=2 (1/3:1/9) +
  + m=3 (1/2:1/18) case 7
 
 + c=3 (1/3:1/9)  m=1 (1/1: 1/9) case 8


 + c=1 (1/3:1/9)  m=2 (1/1: 1/9) case 9
 
+ p=3 (1/3:1/3) + c=2 (1/3:1/9)  m=1 (1/1: 1/9) case 10

 + m=1 (1/2:1/18) case 11
+ c=3 (1/3:1/9) +
+ m=2 (1/2:1/18) case 12
The tree above assumes that all choices are made "fairly", in that
all possible outcomes of a single decision are equally likely.
Therefore, to take a couple of specific examples:
* In case 1, the prize is behind door 1 (p=1). This can happen
onethird of the time, so the probability is 1/3. Then the
contestant chooses door 1 (c=1). This can happen onethird
of the time, so the cumulative probability is 1/9 (1/3 * 1/3).
Monty chooses to open door 2 (m=2). Since he could have
opened either 2 or 3 (since the prize is behind door 1 and the
contestant picked door 1), Monty's choice of door 2 happens
half of the time UNDER THESE CIRCUMSTANCES, producing a
cumulative probability of 1/18 for case 1.
* In case 8, the prize is behind door 2 (p=2), which can happen
onethird of the time; net probability is 1/3. Then the
contestant picks door 3 (c=3), which can happen onethird of
the time; net probability 1/9. Monty has no choice but to
open door 1 (m=1), which must happen every time UNDER THESE
CIRCUMSTANCES, producing a net probability of 1/9 for case 8.
For which cases does the "stay with your first choice" strategy
win the prize? In cases 1, 2, 6, 7, 11, and 12 (the ones in
which the contestant picked the right door, after which it makes
no difference which other door Monty opens). Each of those six
cases has a net/cumulative probability of 1/18. Since they are
mutually exclusive, their sum of 6/18 or 1/3 is the probability
that the "first choice" strategy wins.
For which cases does the "always switch AFTER Monty opens a
door" strategy win the prize? In all of the remaining cases,
3, 4, 5, 8, 9, and 10! Each of *these* cases has a net/cumulative
probability of 1/9. Since they are mutually exclusive, their sum
of 6/9 or 2/3 is the probability that the "always switch" strategy
wins the prize.
QED.
Having done this analysis, we ought to be able to see the argument
by symmetry.
Whatever door the contestant picks initially has a 1/3 chance of
being right. Therefore the contestant who always stays with his
first choice has a 1/3 chance of winning.
After Monty opens a door, the contestant knows that the opened
door did not contain the prize. His first pick STILL has only
a 1/3 chance of being right. Since one of the two remaining
choices has now been eliminated by Monty, the (only!) other
unopened door now has a 2/3 chance of being the right one.
Many "common sense" analyses fall down in assuming that all of
these choices are completely independent. In fact, Monty's
option(s) for which door to open will be dependent on both
the prize door and the contestant's choice. Therefore some
knowledge about the state of the game is "leaked" by Monty's
action.
This sort of probabilistic "leakage" is behind a large number
of security and cryptographic attacks, FWIW.
jn

; sub REBOL {}; sub head ($) {@_[0]}
REBOL []
# despam: func [e] [replace replace/all e ":" "." "#" "@"]
; sub despam {my ($e) = @_; $e =~ tr/:#/.@/; return "\n$e"}
print head reverse despam "moc:xedef#yleen:leoj" ;
[5/12] from: bga:bugbr at: 20Dec2001 10:07
From: <[lgiddingaaaallianzcomau]>
> I think your logic might be flawed.....
LOL! :) No, it is not, I assure you.
> >Now, the other guy excluded 9998 doors and the second group only has one
> >door now... But as we saw above, the probabitily that the prize is in
this
> >second group is 9999/10000. So the probability that the prize is in this
> >single door is 9999/10000 and in the door you choose is 1/10000. That's
> >why you *MUST* change to the other door. :)
>
> You appear to believe that the original 1/10,000 chance is still valid
> after 9998 the doors have been opened.
And it *IS* valid. See the other analysis posted to this list.
> Surely like the coin which has no knowledge of its previous toss,
> the odds have no knowledge of their previous length.
No no no no... This is the wrong way to think about it!
> For example:
> Imagine a Roulette wheel with 10000 numbers
<<quoted lines omitted: 6>>
> by my knowledge that 9998 PREVIOUSLY losing numbers
> have now been dropped from the wheel.
The problem here is that you are ignoring the fact that the other guy told
you about the 9998 noprize entries and this is exactly what make this
problem interesting! When 9998 numbers were excluded, you learned that they
were not the correct ones (and they were part of the original problem).
Initially you had 1/10000 chance of winning and that's why we can be pretty
sure that you didn't and the prize is in the other group. Now, from the
other group, all but *ONE* has been released as not having the prize... the
prize can not move groups and it will still be on the second group
(9999/10000 chance) and the chance the it is the first one you picked is
1/10000.
Bruno
[6/12] from: joel::neely::fedex::com at: 20Dec2001 8:02
Hi, Laurence,
[lgiddingaaaallianzcomau] wrote:
...
> You appear to believe that the original 1/10,000 chance is
> still valid after 9998 the doors have been opened.
>
He does. It is. See below.
> Surely like the coin which has no knowledge of its previous
> toss, the odds have no knowledge of their previous length.
<<quoted lines omitted: 3>>
> The next time the wheel has reduced to 36 numbers
> I still have no 1 but now I have a 1/36 chance
The last sentence above is the key to the flaw in your
analysis/argument. You talked about changing the number
of outcomes from the wheel and then discussed the odds for
the next time
.
There is no "next time" in the Monty problem we've been
discussing. Your argument would only be relevant under
the following scenario:
* One round of the Monty game is played where the prize is
placed behind a door, the contestant picks a door, and
win/loss is declared. I think we all agree that the
probability of a win is 1/3.
* A door is removed from the stage, leaving two doors.
* A "next time" round is played, where the prize is placed
behind a door, the contestant picks a door, and win/loss
is declared, WHERE THE LOCATION OF THE PRIZE AND THE
CONTESTANT'S CHOICE ARE UNCORRELATED TO THOSE OF THE
"FIRST TIME" ROUND. I think we all agree that the
probability of a win is 1/2.
However, in the game we've been discussing, analyzing,
simulating, and argu^H^H^H^Hthinking ;) about, there is
ONLY ONE ROUND, so all of the decisions about which door
Monty can open, and which doors remain available to the
contestant's keeporswitch decision ARE ALL CORRELATED
TO THE UNCHANGING FACTS OF THE TRUE LOCATION OF THE
PRIZE AND THE CONTESTANT'S FIRST CHOICE.
It is precisely those notsoobvioustothecasualobserver
correlations that make this game a good demonstration of
why everyday common sense falls down and skins its knees
so often over mathematical and probabilistic problems.
(Not to mention design and verification of programs! ;)
Let me offer a counterexample to the line of analysis
you proposed. Suppose I have four coins (penny, nickel,
dime, and quarter). I place them in my cupped hands and
shake them, then separate my hands without letting you
see their contents. I tell you (truthfully!) that one
of my hands contains a single coin and the other hand
contains three coins. You have to guess which hand has
only one coin.
If we can assume that my shaking hands are an unbiased
randomizer, and that I am truthful, then I hope we can
agree that your probability of winning is 1/2.
Now, suppose I wiggle the fingers of one hand, say the
left one, and a coin drops to the floor from that hand.
Obviously this action changes neither the original
distribution of coins between my hands, nor your stated
guess as to which hand holds only one coin.
Now, suppose I wiggle the fingers OF THE SAME HAND, and
another coin drops to the floor from that hand. Again,
this action changes neither the original distribution of
coins nor your stated guess.
However, I hope we can agree that you can now state with
absolute certainty which hand originally held one coin.
The "state" of the system (WRT to which hand originally
held one coin and which hand you originally guessed) has
not changed, but YOUR INFORMATION ABOUT THE STATE *HAS*
CHANGED and may be used to guide any future decisions
you make WITHIN THIS SAME ROUND.
Someone may find it interesting to analyze (or simulate)
the probability of success if I allow you to change your
mind after the first coin has dropped.
jn

; sub REBOL {}; sub head ($) {@_[0]}
REBOL []
# despam: func [e] [replace replace/all e ":" "." "#" "@"]
; sub despam {my ($e) = @_; $e =~ tr/:#/.@/; return "\n$e"}
print head reverse despam "moc:xedef#yleen:leoj" ;
[7/12] from: reboler:programmer at: 20Dec2001 23:11
To: Bruno G. Albuquerque
You are assuming the Monty Hall Problem is:
Monty picks one door as winning door.
You pick one door.
MONTY OPENS ALL OTHER DOORS.
Monty asks you to switch or keep.
I assumed:
Monty picks one door as winning door.
You pick one door.
MONTY OPENS _ONE_ UNPICKED DOOR.
Monty asks you to switch or keep.
I think you'll agree that:
1) This a valid interpretation of the threedoor problem
2) This drastically changes the value of switching from your interpretation.
[8/12] from: bga:bugbr at: 20Dec2001 14:07
From: "alan parman" <[rebolerprogrammernet]>
> To: Bruno G. Albuquerque
>
> You are assuming the Monty Hall Problem is:
>
> Monty picks one door as winning door.
> You pick one door.
> MONTY OPENS ALL OTHER DOORS.
> Monty asks you to switch or keep.
Hmmm? No, I am not. I am assuming that:
1  You pick one door.
2  Monty opens all but one door and shows that all doors he opened has no
prizes in it. In the classic case, "all other doors but one" results in
Monty opening one door.
3  You ara asked if you want to switch or not.
> I assumed:
> Monty picks one door as winning door.
<<quoted lines omitted: 4>>
> 1) This a valid interpretation of the threedoor problem
> 2) This drastically changes the value of switching from your
interpretation.
In the classic case monty opens one door, yes. I just gave an example to try
to show intuitivelly that the other door is the one with the prize, not
yours (yours had a chance of 1/10000 of being the correct one in my example,
while the chance of the other  remaining  door being the correct one was
9999/10000
Bruno
[9/12] from: lgidding:aaa:allianz:au at: 21Dec2001 9:39
Hi,
Thanks Joel,
If my original position could be termed as having picked door 1....
Your last two posts
>When all else fails, resort to brute force!
and the coins in the hand analogy
>Suppose I have four coins (penny, nickel, dime, and quarter)
>I place them in my cupped hands and shake them,
have between them convinced me,
of the advantages of taking up Monty's offer to swap doors!
So my apologies Bruno.......[ for doubting your logic :) ]
.......and please consider my choice to now be door 2.
 Laurence
[10/12] from: nitschlists:netcologne at: 21Dec2001 0:12
RE: [REBOL] Re: My Statistical Thoughts on Monty Hall Problem (nonREBOL)
[lgiddingaaaallianzcomau] wrote:
> Hi,
> Thanks Joel,
<<quoted lines omitted: 8>>
> So my apologies Bruno.......[ for doubting your logic :) ]
> ........and please consider my choice to now be door 2.
That should depend on Monty's answer :)
>  Laurence
>
Volker
[11/12] from: bga:bugbr at: 20Dec2001 21:35
> So my apologies Bruno.......[ for doubting your logic :) ]
>
> .......and please consider my choice to now be door 2.
LOL! No problem at all. :)
Bruno

Fortune Cookie Says:
No problem is so large it can't be fit in somewhere.
[12/12] from: jofo6117:student:uu:se at: 20Dec2001 5:05
On Thu, 20 Dec 2001 11:56:48 +1000
[lgiddingaaaallianzcomau] wrote:
> Bruno,
>
> I think your logic might be flawed.....
>
> >Would you switch or not?
>
> >When you did your first pick, you have 1/10000 chances of winning and
the
> >change that the prize would be in one of the other 9999 doors is
> 9999/10000.
> >In other words, the prize *WILL BE* in a door that is not the one you
> choose
> >(a 9999/10000 chance is enough to me to say that).
>
> >Now, the other guy excluded 9998 doors and the second group only has
one
> >door now... But as we saw above, the probabitily that the prize is in
this
> >second group is 9999/10000. So the probability that the prize is in
this
> >single door is 9999/10000 and in the door you choose is 1/10000. That's
> why
<<quoted lines omitted: 13>>
> by my knowledge that 9998 PREVIOUSLY losing numbers
> have now been dropped from the wheel.
While realising that this thread is losing it's rebol related theme fast,
I still find it fascinating enough to reply. My take on this problem, and
I'm fairly sure that the validity of this line of reasoning has been
proven
empirically (I think I even did it myself when this problem was first
presented to me, which was probably ten years ago or so):
Everyone agrees that the first choice will have 1/3 probability of being
a winner (we're back at the original problem involving three doors).
Since,
after Monty's elimination of one empty door, there's only one other door
left, we can be sure that if the car is not behind the door we chose
first,
it's behind the other door. Therefore, the whole situation should be
equivalent to this situation: instead of Monty following the usual
routine,
he gives us the choice of either A) sticking with our first choice or B)
looking behind *both* the other doors. We know that the chance of the car
being behind any one door is 1/3, therefore we have 2/3 chance of winning
if we choose the two other doors.
Basically, I think the whole confusion stems from the idea that the two
choices are completely unconnected. But they are not; making the second
choice we *know* what the odds are that we've already found the car. If
we were presented with just the choice of two doors, with a car behind one
of them, it wouldn't matter to us if we knew that the car was *probably*
behind "door A" if we didn't know which one door A was. All information
available is that the car is with probability 1 behind one door and with
probability 0 behind the other door. So the chance of us picking the right
one would simply be 1/2.
Presented with two doors *and* the additional information that the car is
with probability 2/3 behind, say, the left door, I think the choice would
be relatively simple. And in the original problem, this is precisely the
kind of information we have.

Johan Forsberg
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