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# My Statistical Thoughts on Monty Hall Problem (non-REBOL)

### [2/12] from: bga:bug-br at: 19-Dec-2001 16:48

From: "alan parman" <[reboler--programmer--net]>
> 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: 20-Dec-2001 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: 20-Dec-2001 0:15

Hi, Alan, alan parman wrote:
> Here is my speculation on the Monty Hall Problem. > > [REBOL-content-less discussion about Probability follows]:
... snipped ...
The thread discussing this puzzle has provided many examples of the difference between "common-sense" 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 one-third of the time, so the probability is 1/3. Then the contestant chooses door 1 (c=1). This can happen one-third 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 one-third of the time; net probability is 1/3. Then the contestant picks door 3 (c=3), which can happen one-third 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:bug-br at: 20-Dec-2001 10:07

From: <[lgidding--aaa--allianz--com--au]>
> 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 no-prize 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: 20-Dec-2001 8:02

Hi, Laurence, [lgidding--aaa--allianz--com--au] 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

### [7/12] from: reboler:programmer at: 20-Dec-2001 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 three-door problem 2) This drastically changes the value of switching from your interpretation.

### [8/12] from: bga:bug-br at: 20-Dec-2001 14:07

From: "alan parman" <[reboler--programmer--net]>
> 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 three-door 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: 21-Dec-2001 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: nitsch-lists:netcologne at: 21-Dec-2001 0:12

RE: [REBOL] Re: My Statistical Thoughts on Monty Hall Problem (non-REBOL) [lgidding--aaa--allianz--com--au] 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:bug-br at: 20-Dec-2001 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: 20-Dec-2001 5:05

On Thu, 20 Dec 2001 11:56:48 +1000 [lgidding--aaa--allianz--com--au] 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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