The Monty Hall problem

[humber]

Do you agree with the following:

1) Start with N doors. If I pick a door, and the probability that my door hides the car happens to be p, then the probability that I'll win if I stick with my door is p and the probability that I'll win if I switch to the one door remaining after Monty has revealed N-2 goats is 1 - p.

Yes.

2) If we have N = 100 doors, and each has an equal probability of hiding the car, then each door has a p = 1/N = 1/100 probability of hiding the car.

Yes.

3) If I pick one of the doors, then my door has a p = 1/100 probability of hiding the car.

Yes.

4) Combining 1 and 3: If I stick with my door, I have a p = 1/100 probability of getting the car. If I switch to the one door remaining after Monty has revealed 98 goats, I have a 1 - p = 1 - 1/100 = 99/100 probability of getting the car.

No, why would you think that?

 

[Vorticity]

No, why would you think that?

You got me. I have to admit it.

Even after I figured it out, you still managed to string me along for a few more posts.

Once again, I salute you, sir. Truly, you are a giant amongst gnomes.

 

[humber]

Correct.

What about for the even distribution amongst 100 doors?

If you play the game with one hundred doors and Monty opens one door then obviously there are 99 closed ones. Not the same as 3 doors.
No logical process can isolate one door, so that you can say 1 - p.

As I said 1 -p is justified by a simple binary logic that models Monty.
He does that, and faultlessly.

So, with that in mind, you tell me why the 100 door is offered to validate the 3 door version, unless you think only of the last 3 doors? Which says the same thing as 3 doors.

 

[Vorticity]

By the way, the above few posts remind me strongly of Lewis Carroll's What the Tortoise Said to Achilles (a must read for all enthusiasts of both amphigory and rampant philosophunculism), in which the Tortoise is very much trolling.

 

[humber]

By the way, the above few posts remind me strongly of Lewis Carroll's What the Tortoise Said to Achilles (a must read for all enthusiasts of both amphigory and rampant philosophunculism), in which the Tortoise is very much trolling.

No, you did not pay attention to what I said, and forced your own conclusion.
If I were looking for salutation, I doubt this would be the place.
Never mind.

 

[Vorticity]

If I were looking for salutation, I doubt this would be the place.

And yet you got it anyway!

Awesome show! Great job!

Salutations!

 

[Modified]

So, with that in mind, you tell me why the 100 door is offered to validate the 3 door version, unless you think only of the last 3 doors? Which says the same thing as 3 doors.

It's done because of the way most people think. 1/3 and 2/3 are close enough that they may not see that the answer can not be 1/2. With 100 doors, it becomes intuitively obvious that the result can't be 1/2, even for someone who knows nothing about probability or mathematics. For someone completely lacking in intuition though, this may not be the case.

 

[Jack by the hedge]

As I said 1 -p is justified by a simple binary logic that models Monty.
He does that, and faultlessly.

Yeah, but we were already there on page 1.

So, with that in mind, you tell me why the 100 door is offered to validate the 3 door version, unless you think only of the last 3 doors? Which says the same thing as 3 doors.

The 100 door version is not offered to validate the 3 door version. It is used to emphasize the point which people most commonly miss when they consider the puzzle: that the chances of their original choice being right are not improved by the removal of other choices the host knows to be wrong, so long as he leaves one other door unopened.

The last 3 doors of the 100 door variant do not say the same thing as the 3 door variant unless a crafty troll is keeping up his sleeve a ridiculous plot twist where there is one choice among the 100 doors which arbitrarily has 1/3 chance of winning and the contestant knows which it is.

Of course that would be preposterous, so if we, as reasonable people, stick to a fair 100 door game, the last 3 doors leave us in a position where;

The contestant's original choice has 1% chance of winning,
The two other doors each have 49.5% chance of winning.

So it's not like the 3 door game at all.

(Unless another wicked troll claims that it's a 3 door game where not only does one of the 3 doors have only a 1% chance of winning but the contestant knows which it is...)

 

[Rolfe]

I don't know if he's trolling or not. If he is, it's finestkind.

If he isn't, I realised several pages ago that I have no hope of figuring out what he's talking about, or what reasoning processes he's employing, or even what his point is.

So I lurk, so see if anyone else can figure it out.

Rolfe.

 

[SumDood]

So say you play the 100 door Monty game 100 times. Each time, you pick a door, then Monty reveals 98 goats behind 98 doors leaving the one you picked un-opened and another un-opened door. How many times out of that 100 games would you expect a car to be behind the door you picked?

 

[Jack by the hedge]

So say you play the 100 door Monty game 100 times. Each time, you pick a door, then Monty reveals 98 goats behind 98 doors leaving the one you picked un-opened and another un-opened door. How many times out of that 100 games would you expect a car to be behind the door you picked?

Ah, but you have to pin it down more carefully than that. You have to specify a fair 100 door game, otherwise humber merely declares an arbitrary probability for the initial choice and its complement for the swap.

 

[humber]

I don't know if he's trolling or not. If he is, it's finestkind.

If he isn't, I realised several pages ago that I have no hope of figuring out what he's talking about, or what reasoning processes he's employing, or even what his point is.

So I lurk, so see if anyone else can figure it out.

Rolfe.

OK. Three doors one car. You open one door, no car. You can then say that the other two are 1/2 each. But, when Monty does it, then it's 1/3 for one door and 2/3 for the other. That cannot be the result of probability.
You may express it that way, but probability is not at work in that transformation.
Agree or not?

 

[Subduction Zone]

Oh no, this thread has humbers

Don't worry folks, stand clear. We might have to tent this thread, as you would tent your house for a termite invasion, but with the moderators help we should have all of the humbers out of here and back to the DDWFTTW thread in no time. Just try to ignore him until then. humbers do feed on replies to their posts, so at all costs avoid responding to his nonsense or this thread could have humbers as a permanent feature

 

[69dodge]

Does anything change, though, if after the first switch is made Monty opens another door (that was previously not singled out in any way, say) and then offers me to switch? What if he then offers me to switch back to my original door, specifically?

I'm assuming that when Monty picks his second door to open, he's not allowed to open the door you originally picked, nor is he allowed to open the door you first switched to. In that case, this version of the game is similar to the usual version, in that the probabilities of the doors you picked don't change after Monty opens his doors; the probabilities of the remaining doors increase to accomodate their fewer number. So, if there are n doors, the probability is 1/n that your original door hides the car, the probability for the door you switched to is (n - 1) / [n(n - 2)] , and the probability for each of the remaining n - 4 unopened and unpicked doors is (n - 1)(n - 3) / [n(n - 2)(n - 4)].

If Monty is allowed to choose your original door as the second door he opens, but just happens not to, the problem is somewhat more complicated, and I haven't figured it out yet. (I tried, but I must have made a mistake somewhere because my computer simulation and my pencil-and-paper calculation give different answers.)

 

[SumDood]

Ah, but you have to pin it down more carefully than that. You have to specify a fair 100 door game, otherwise humber merely declares an arbitrary probability for the initial choice and its complement for the swap.

I was hoping by specifying it as the 100 door Monty game, by definition, one door chosen at random has a car behind it and all other 99 have goats.

 

[69dodge]

If Monty is allowed to choose your original door as the second door he opens, but just happens not to, the problem is somewhat more complicated, and I haven't figured it out yet. (I tried, but I must have made a mistake somewhere because my computer simulation and my pencil-and-paper calculation give different answers.)

I found my mistake. Here's the answer.

Suppose there are n doors, where n > 3. Let x = (n - 2)², y = (n - 1)(n - 3), z = (n - 1)(n - 2), and d = x + y + (n - 4)z. Then, the probability is x/d that your original door hides the car, the probability is y/d for the door you switched to, and the probability is z/d for each of the other n - 4 unopened doors.

 

[Rasmus]

I found my mistake. Here's the answer.

Suppose there are n doors, where n > 3. Let x = (n - 2)², y = (n - 1)(n - 3), z = (n - 1)(n - 2), and d = x + y + (n - 4)z. Then, the probability is x/d that your original door hides the car, the probability is y/d for the door you switched to, and the probability is z/d for each of the other n - 4 unopened doors.

My brain will fry if I tr< to understand this any longer .... but thank you for the effort. Maybe I'll have tome over the weekend.

 

[Jack by the hedge]

OK. Three doors one car. You open one door, no car. You can then say that the other two are 1/2 each. But, when Monty does it, then it's 1/3 for one door and 2/3 for the other. That cannot be the result of probability.
You may express it that way, but probability is not at work in that transformation.
Agree or not?

You're saying the calculation of the probabilities cannot itself be the result of probability.

I agree in so far as it's almost meaningless to describe any calculation of probability as being a "result" of probability. It's a bit like saying the calculation of a debt is a result of debt.

"That" (i.e. the change in probabilities) is the result of a change of information available in the two different games, and results in new probabilities.

Since the transformation you are talking about is the comparison of the probabilities in two different games, then I suppose only arithmetic is "at work". In which case, I agree.

 

[Rolfe]

OK. Three doors one car. You open one door, no car. You can then say that the other two are 1/2 each. But, when Monty does it, then it's 1/3 for one door and 2/3 for the other. That cannot be the result of probability.
You may express it that way, but probability is not at work in that transformation.
Agree or not?

I'm going to regret this.

Disagree. Your point would be valid if Monty was choosing his door entirely at random. It is not valid if Monty is barred from choosing the same door you chose.

Rolfe.

 

[Ivor the Engineer]

OK. Three doors one car. You open one door, no car. You can then say that the other two are 1/2 each. But, when Monty does it, then it's 1/3 for one door and 2/3 for the other. That cannot be the result of probability.
You may express it that way, but probability is not at work in that transformation.
Agree or not?

In the classic Monty Hall problem the contestant also has the information that Monty only ever opens one of the two doors the contestant hasn't picked with a goat behind it.

At the start of the game the probability the door the contestant picks has the prize behind it is 1/3 and the probability is 2/3 that it is behind one of the other two doors. Monty cannot open the door the contestant picked and cannot open the door with the prize behind it. This means the probability of the one unopened door of the two doors not picked by the contestant having the prize behind it becomes 2/3 after Monty excludes one of the two unpicked and unopened doors by showing the contestant one of them that has a goat behind it.

If the wording of the puzzle is sloppy or the rules of the game changed then the correct answer may be different or there may be no correct answer.