Probably...

[Barry Shnikov]

There was a game show in the States with a car as the BIG PRIZE.

Each week, the finalist is presented with three doors. He picks one of the doors, and the quizmaster then opens one of the other two doors to show that the car was not behind that door. The contestant is then given the chance to change his mind, and pick the other door, or stick with his original choice.

Imagine you are the finalist. Three doors in front of you: Your Door, the Open Door, and the Third Door. The car is behind Your Door, or the Third Door. You have two options: option 1) stick with Your Door, or option 2) pick the Third Door.
Which of the following is correct: to maximise your chance of winning the car, you should:

a) pick option 1

b) pick option 2

c) it doesn't matter, the chance of winning is the same either way.

(I'd like to have done this as a poll, but the preamble is too long, I theenk...)

[ducasi]

I hope those people, like myself, who have heard this puzzle before and know the answer can keep quiet and let those that haven't have a go... :flower:

[ChrisA]

I hope those people, like myself, who have heard this puzzle before and know the answer can keep quiet and let those that haven't have a go... :flower:
Yeah, but can we post something irrelevant or blatantly wrong that nevertheless appears to be insightful?

Not that I have anything like that, but I just thought I'd check...

[ducasi]

Yeah, but can we post something irrelevant or blatantly wrong that nevertheless appears to be insightful? Well this is the CerocScotland forum isn't it?

We've never managed to stop anyone before now... :flower:

[Jazz_Shoes (Ash)]

Well this is the CerocScotland forum isn't it?

We've never managed to stop anyone before now... :flower:

Well looks like Chris is offline so guess we'll need to wait until tomorrow for his answer.

[frodo]

There was a game show in the States with a car as the BIG PRIZE.

Each week, the finalist is presented with three doors. He picks one of the doors, and the quizmaster then opens one of the other two doors to show that the car was not behind that door. The contestant is given two options: option 1) stick with the door he chose, or option 2) pick the other unopened door.

Which of the following is correct: to maximise his chance of winning the car, he should
a) pick option 1
b) pick option 2
c) it doesn't matter, the chance of winning is the same either way.

The obvious answer is either option b) or c). However if you take into account the fact you might have analysed it incorrectly picking option 2 maximises his chances.

Thinking about it a little more there is a 1/3 chance of picking correctly the first time so there must be a 2/3 chance of getting it correct by changing.

But knowing it is probably a trick question I'll go for option 1.

[Dreadful Scathe]

statistically you'd best pick the other door - but you'd hate yourself if you were right in the first place if you did. Personally I'd stick where I am, Im an optimist.

[ShinyWeeStar]

Thinking about it a little more there is a 1/3 chance of picking correctly the first time so there must be a 2/3 chance of getting it correct by changing.
Yeah, but surely you have a 50/50 chance of getting it right whether you change doors or not? Once the first door has been opened that removes it from the equation; you're down to 2 doors and the car's behind one of them...that's a 1 in 2 chance regardless of whether you stick with the door you picked first time round or not, surely? :confused:

In practice though I'd probably choose option 1) and stick with my original decision; sod's law says if I changed my mind I'd find I'd lost the car! :rolleyes:

[El Salsero Gringo]

The answer is, you should *always* change doors - but the challenge is to explain why...

[ShinyWeeStar]

The answer is, you should *always* change doors - but the challenge is to explain why...

Why, do they move the car if you choose the right door? Sneaky! :eek:

[Barry Shnikov]

The woman with the highest IQ in America has a newspaper column and she put this poser in some years ago.
When she published the correct result, she was deluged with post demanding she retract it, because she was wrong, and much of the post came from professional academics dealing with probability and statistics.
So the answer is not easy - that's why I posted it:devil:

[ChrisA]

The answer is, you should *always* change doors - but the challenge is to explain why...
That's only a small part of the challenge. Explaining why you should change doors becomes trivial as soon as you rephrase the question along the lines of

"What would happen if you did this experiment 30 times - how many cars would you win if you a) always changed or b) always didn't change?"

The rest of the challenge is to also explain why the obvious justification for "it makes no difference whether you change or not" is wrong, if you're only doing the experiment once.

[TiggsTours]

Yeah, but surely you have a 50/50 chance of getting it right whether you change doors or not? Once the first door has been opened that removes it from the equation; you're down to 2 doors and the car's behind one of them...that's a 1 in 2 chance regardless of whether you stick with the door you picked first time round or not, surely? :confused:

In practice though I'd probably choose option 1) and stick with my original decision; sod's law says if I changed my mind I'd find I'd lost the car! :rolleyes:
:yeah: to all of this!

[El Salsero Gringo]

The rest of the challenge is to also explain why the obvious justification for "it makes no difference whether you change or not" is wrong, if you're only doing the experiment once.I don't consider that one has properly explained "why" until one has fully justified one's answer, which includes explaining why the wrong but 'obvious' justifications/answers/whatever *are* wrong. So, as I said, the challenge is to explain why...

[El Salsero Gringo]

I had an interesting chat with a lady at Hammersmith before the new year, when there was a raffle. She told me that she and some friends had "discovered" that the common practice of drawing the fifth-prize-winning ticket, followed by the fourth, then third etc - and leaving the top prize to last - was unfair. This, she told me, was because it meant that whoever won the fifth prize (a box of chocolates) had been cheated out of their chance of winning the top prize (something big and shiny). The only fair way to it - allegedly - was to draw the top prize first, then everyone had an equal chance at it.

She was quite adamant that the draw as conducted was unfair. But how would you have explained the truth to her?

[Barry Shnikov]

I had an interesting chat with a lady at Hammersmith before the new year, when there was a raffle. She told me that she and some friends had "discovered" that the common practice of drawing the fifth-prize-winning ticket, followed by the fourth, then third etc - and leaving the top prize to last - was unfair. This, she told me, was because it meant that whoever won the fifth prize (a box of chocolates) had been cheated out of their chance of winning the top prize (something big and shiny). The only fair way to it - allegedly - was to draw the top prize first, then everyone had an equal chance at it.

She was quite adamant that the draw as conducted was unfair. But how would you have explained the truth to her?

Let’s suppose there are 500 entrants with one ticket each, and there are five prizes. Before the draw, your chance of being picked for one of the prizes is 1%. Your chance of being picked for any particular prize is 0.2%. Therefore your chance of getting first prize is 0.2%.

If your ticket isn’t picked for the fifth prize, your chances of winning first prize become 0.2004% (1/499); if it is, your chances of winning first prize become zero. However, you have already won a prize, so your chance of winning one of the prizes is now 100%. Everybody else, at that stage, still has a 0.802% (4/499) chance of winning one of the prizes.

If the first prize is a BMW and the fifth prize is a £10 record token (record? What’s one of those then?) you might feel a little peeved at this point.

But it’s vital to remember that there is no real unfairness; all 500 people start off with exactly the same chance of winning the first prize, and of winning the fifth prize. After all the prizes have been drawn, you are just one of the 499 people who didn’t win the first prize, but by a slightly different route than those people who won nothing at all.

Um. I think that’s clear…

By the way, Gringo, I don't believe a man of your calibrance couldn't work that out yourself...

[El Salsero Gringo]

By the way, Gringo, I don't believe a man of your calibrance couldn't work that out yourself...I did...

What I said was, imagine the draw was done behind a curtain by a machine. and the results announced together after. Did she still think it made any difference to her own chances of winning top prize whether the machine went 5-4-3-2-1 or 1-2-3-4-5?

[Barry Shnikov]

I did...

Oh, it was a test?


What I said was, imagine the draw was done behind a curtain by a machine. and the results announced together after. Did she still think it made any difference to her own chances of winning top prize whether the machine went 5-4-3-2-1 or 1-2-3-4-5?

Simpler than mine!

[El Salsero Gringo]

Oh, it was a test?Well, no, not a test. She was obviously mistaken, but I was wondering what ways there were to explain why. You thought of a different way to me which was interesting.

[DavidY]

Back to the gameshow... I think one of the things that helps me understand this intuitively (although I don't think it affects the maths) was knowing that the game show host knows where the car is.

So when he opens the door after the initial choice, he's giving the finalist more information. If he didn't know, then 1/3 of the time, he'd open a door with the car and the finalist would walk away empty-handed. However this doesn't happen.

I think the maths is something like this:

There are effectively two initial states:
A: finalist picked car first time - with probability 1/3
B: finalist didn't pick car first time - with probability 2/3

If you happen to be in state A, then once the host opens the empty door, the car is still behind the door you first picked, so if you swap you lose, if you stick you win.

If you happen to be in state B, then once the host opens the empty door, the car is behind the other door, so if you swap you win, if you stick you lose.

So if your strategy is to swap then 2/3 of the time you're in state B and win, and 1/3 of the time you're in A and you lose.

Conversely if your strategy is to stick, then 2/3 of the time you're in state B and lose, but 1/3 of the time you're in A and you win.

So a strategy of "Swap" means you win 2 times out of 3.

In state B, if the host didn't know where the car was then random chance would mean that there would be some cases where he would open the door with the car behind it. However, because he does know, this case doesn't happen and removes some of the possible outcomes, so the odds are no longer intuitive.

..er I think, anyway.:)

[David Franklin]

Back to the gameshow... I think one of the things that helps me understand this intuitively (although I don't think it affects the maths) was knowing that the game show host knows where the car is. How do you know that the host knows where the car is? What would happen if he didn't know?

[DavidY]

From the initial description (and from the few times I've seen the Monty Hall problem described elsewhere) the host knows - he opens the door to show the finalist an empty door.

As I said before I don't think it affects the maths once you know he's opened an empty door.

But if everything was random chance, then 2/6 (?) of the time the host would open a door with a car and the finalist loses. But this doesn't happen (for whatever reason)- so this outcome is eliminated and it means the odds of the remaining possibilities aren't so intuitive (that's how I understand it, at least).

[ChrisA]

I don't consider that one has properly explained "why" until one has fully justified one's answer, which includes explaining why the wrong but 'obvious' justifications/answers/whatever *are* wrong. So, as I said, the challenge is to explain why...
Ok, fair point.

Let me try, then, to articulate the part of the question that is hard to answer. I'll have a stab at answering it, but although the sums are easy, I don't really understand the crucial bit. Hopefully someone will tell me if the reasoning is wrong.

Let's imagine a superficially similar experiment, where the contestant is initially shown two closed doors and an open one - the car is behind one of the closed ones.

He's invited to choose one of the doors, and is then given the option to change his choice if he wants.

Clearly, it makes no difference whether he changes or doesn't - half the time he'll have chosen the car so he'll get it if he always doesn't change; half the time he won't so if he always changes, he'll get the car the other half of the time.

So the question is:

Why is this situation different from the situation in the original problem, after the host has opened the door?

Because then, it's still true that:

- the contestant has either chosen the car or not,
- it's behind one of the two closed doors.

And it's a new choice he's making between one of the two closed doors. Or is it?

Going back to the original game, let's suppose the contestant makes his initial choice, and the host opens one of the doors. The contestant is about to switch.

Now imagine bringing someone else in, who hasn’t seen the show up to that point. If he’s invited to choose between the two remaining doors at this point, he'll get the car half the time. This looks like the same choice as the contestant has, but if we believe the maths, it can't be.

What's different about the new person's choice? It's independent of what went before, whereas the original contestant's choice isn't. But this is where I get stuck.

[David Franklin]

What's different about the new person's choice? It's independent of what went before, whereas the original contestant's choice isn't. But this is where I get stuck.Basically, the difference is the original contestant knows what door he chose first, and that there is unlikely to be a car behind it.

[El Salsero Gringo]

Now imagine bringing someone else in, who hasn’t seen the show up to that point. If he’s invited to choose between the two remaining doors at this point, he'll get the car half the time. This looks like the same choice as the contestant has, but if we believe the maths, it can't be.If you mean that he doesn't know which door the contestant originally chose - then yes, his odds are 50/50.

The difference comes about because of the difference between the "true" odds - and the newcomer's best estimator for those odds.

*We* know that the "real" odds are 2/3 - 1/3. But the ignorant newcomer (not knowing which door the contestant had originally picked) doesn't have the information to say which way round those odds go (he might say to himself, I know it's 2/3 - 1/3 in favour of *one* of the two doors, but it's 50/50 as to which way round...)

Similarly, for the people running the show - the odds aren't even 2/3 - 1/3. Having put the car behind one of the doors themselves, *their* odds are 1-0.


So is there any such thing as "true" odds? Well, yes, but only for a truly as-yet unsampled random process. The positioning of the car behind one of the doors isn't really random - because if it was, the host wouldn't be able to open a guaranteed-empty door. If it were really random then the game couldn't be played as stated.

[El Salsero Gringo]

Back to the gameshow... I think one of the things that helps me understand this intuitively (although I don't think it affects the maths) was knowing that the game show host knows where the car is.

So when he opens the door after the initial choice, he's giving the finalist more information. ...and that's the salient point.

Example:

Q: I have two children. One is a boy. What is the probability that that the elder is a girl?

A: 1/3 (possibilities are B-B, B-G, G-B)

Q: For those same two children: The younger one is a boy. What is the probability that the elder is a girl?

A: 1/2 (possibilities are B-G, B-B)

If I ask you the first question, then reveal more information (by telling you the second one) then it's not surprising the odds change.

[David Franklin]

I think one of the things that helps me understand this intuitively (although I don't think it affects the maths) was knowing that the game show host knows where the car is.
So, suppose the host doesn't know, and picks randomly, but we don't televise the shows where he does reveal a car.

Now imagine we're watching a random episode; the contestant has picked his door, and the host has just opened another door to reveal it's not the car. Which of the remaining two doors is more likely to have the car now?

[El Salsero Gringo]

So, suppose the host doesn't know, and picks randomly, but we don't televise the shows where he does reveal a car.

Now imagine we're watching a random episode; the contestant has picked his door, and the host has just opened another door to reveal it's not the car. Which of the remaining two doors is more likely to have the car now?It doesn't make any difference. Because by not televising the shows where he does reveal a car - i.e. for all the shows that you see - the host effectively *does* know where car is.

[David Franklin]

It doesn't make any difference. Because by not televising the shows where he does reveal a car - i.e. for all the shows that you see - the host effectively *does* know where car is.Actually, I think it does make a difference.

By not televising the shows where he reveals a car, we eliminate a proportion of the probability space. But we don't eliminate any of the cases where the original door holds the car (because Monty can't reveal a car by accident in this case). Conversely, we elminate exactly half of the cases where the original door is car-free (because there's a 1 in 2 chance Monty reveals a car by accident). So we have 1:3 (car original), 1:3 (car not original, Monty doesn't reveal car), 1:3 (car not original, Monty does reveal car). After discarding the last case (not televised), we end up with a 50:50 chance that in a televised show the original pick is the car.

Loosely speaking, a show is much more likely to end up being televised if the original pick is the door with a car, and this affects the odds.

[El Salsero Gringo]

Actually, I think it does make a difference.
{stuff}
Fair enough.

Or, for instance, if one chooses only to televise shows on which the _contestant_ reveals the car, then suddenly his 'odds' of winning have jumped to 100%.

But the change in odds is not a feature of what the host knows, or the contestant knows - that's because of a biased sampling of a subset of all the shows.


Loosely speaking, a show is much more likely to end up being televised if the original pick is the door with a car, and this affects the odds.That changes the odds as percieved by the audience - but not as percieved by the contestant!

[David Franklin]

But the change in odds is not a feature of what the host knows, or the contestant knows - that's because of a biased sampling of a subset of all the shows.But I was fairly explicit in explaining how I was sampling, and (it seems) you didn't realise it lead to bias. Even though I think you know I wouldn't have asked the question without a reason, and my rephrasing indicated I was probably "up to something"! The point is that lots of ways of biasing a sample space are not obvious. In your "children" example, most people don't realise saying "one child is a boy" biases the chance of the eldest child being male by excluding the subset "both children are girls".


That changes the odds as percieved by the audience - but not as perceived by the contestant!Again, I tried to make it fairly explicit - "we're watching a random episode" - that I was talking about the odds as seen by the audience.

In fact in the scenario I described, I tried to make it as clear as possible what the assumptions were and how they differed from the original.

Now suppose someone asks:


You win a contest. To pick your prize, you are presented with three doors - there is a car behind only one of them, the other two are empty. You pick one door, and the master of ceremonies picks another door and opens it. A car is not behind that door. You are given the chance to change your mind, and pick the other door, or stick with your original choice.

What should you do?

It might look similar, but this time I've made the problem intentionally vaguer in a couple of significant aspects. And now you can justify all three answers (switch, don't switch, makes no difference) depending on your assumptions...

[LMC]

I've got lost somewhere :(

Back to basics for me: I reckon that it makes no difference.

When you pick from 3, you have a 33.3recurring% chance of getting the right door. When you pick from 2, you have a 50% chance. So intuitively, you should change. But since the doors don't have memories, the probability doesn't "accrue" (probably the wrong word but I only got a C at Maths O'Level :blush: ). So do what you like :shrug:

[El Salsero Gringo]

But I was fairly explicit in explaining how I was sampling, and (it seems) you didn't realise it lead to bias.I know, I know, I was being dumb. Again. It's tough being a donkey.
Now suppose someone asks:
You win a contest. To pick your prize, you are presented with three doors - there is a car behind only one of them, the other two are empty. You pick one door, and the master of ceremonies picks another door and opens it. A car is not behind that door. You are given the chance to change your mind, and pick the other door, or stick with your original choice.

What should you do?It might look similar, but this time I've made the problem intentionally vaguer in a couple of significant aspects. How did you make it vaguer? I'm being dumb again.

[El Salsero Gringo]

I've got lost somewhere :(

Back to basics for me: I reckon that it makes no difference.

Let's see if the problem can be recast to make it more intuitive.

Suppose I have a hundred envelopes, and I tell you that a fifty pound note is in one of them.

You pick and take one. Now I have 99 envelopes, and you have 1. It's 99 to 1 that I'm still holding the money.

I burn 98 of the ones I'm holding, ones that I know are empty. It's *still* 99 to 1 that I'm holding the money - and if I am, it must be in what is now the only envelope in my hand.

I offer you the chance to swap your envelope for mine. Would you take it, or keep your original choice?

If that doesn't convince you, what about if, instead of burning the 98 empty envelopes, I empty them all into a bag, and offer to swap you the bag for your envelope. Or, even more obviously, what if I just offer to swap all 99 envelopes that I'm holding for your 1? It all amounts to the same thing.

If that *still* doesn't do it for you, how about a game of poker?

[David Franklin]

I know, I know, I was being dumb. Again. It's tough being a donkey.How did you make it vaguer? I'm being dumb again.It's not explicitly stated whether or not the MC knows where the car is. If he doesn't, and just opens a door at random, we get into the earlier case I described (the exclusion of some states has already "happened" due to the fact we don't see a car behind the door) - so it makes no difference what we do. (N.B. making all the standard Bayesian assumptions that I can't be bothered to spell out).

If he does, we get to the second ambiguity. Does he always open a door to reveal an empty space? If so, we're in the original problem and should switch.

But it might be that he only offers us the chance to switch if we've already picked the door with a car ('cos he's a meanie!). In which case we should never switch.

[El Salsero Gringo]

It's not explicitly stated whether or not the MC knows where the car is. If he doesn't, and just opens a door at random, we get into the earlier case I described (the exclusion of some states has already "happened" due to the fact we don't see a car behind the door) - so it makes no difference what we do. (N.B. making all the standard Bayesian assumptions that I can't be bothered to spell out).

If he does, we get to the second ambiguity. Does he always open a door to reveal an empty space? If so, we're in the original problem and should switch.

But it might be that he only offers us the chance to switch if we've already picked the door with a car ('cos he's a meanie!). In which case we should never switch.So to sum up, your (new) ambiguity is because now we have to guess the rules as we go along (or infer them from the MC's behaviour)? We don't know that he would *always* (and in fact might not) show us an empty door, or even offer us the chance to switch at all?

I think you're (worthily) pointing out that it's very difficult to make any kind of judgement about risk without some a-priori knowledge of the behaviour of the system. A friend of mine was on holiday. On the first night, he saw cockroach crawl out from under his bed. What was the probability that he'd see a cockroach on the second night? What was the probability that he'd see a millipede on the second night? Kind of hard to say. By the end of the two weeks, when he had some statistics to go on, he was better placed to estimate.

[LMC]

(snips lots of stuff which I think amounts to "it makes no difference whether you switch or not" - which is what I said, although I appreciate that I got the probability calculations wrong)

If that *still* doesn't do it for you, how about a game of poker?
If I still misunderstand you then I'll pass thanks (and not wonder any more why I got cleaned out on New Year's Eve :rolleyes: :blush: )

Now, would anyone like this box, which may or may not contain a cat...

EDIT due to cross-post:

I think you're (worthily) pointing out that it's very difficult to make any kind of judgement about risk without some a-priori knowledge of the behaviour of the system.

^^^ That is what I meant by the doors' (system) lack of memory.

For your next trick, how about a nice argument on justification of lottery number choices? :devil:

[El Salsero Gringo]

If I still misunderstand youTexas Hold'em OK for you?

[LMC]

*promptly folds*

I'll quit while I'm vaguely even, thanks

EDIT: should say that it's all t'rifficly educational thanks guys - I'll carry on sittin ere in the corner and watching if you don't mind (even if you do :na: )

[David Franklin]

So to sum up, your (new) ambiguity is because now we have to guess the rules as we go along (or infer them from the MC's behaviour)? We don't know that he would *always* (and in fact might not) show us an empty door, or even offer us the chance to switch at all?Exactly.


I think you're (worthily) pointing out that it's very difficult to make any kind of judgement about risk without some a-priori knowledge of the behaviour of the system. In the particular case of the Monty Hall problem, a lot of the mathematicians who got it wrong made the "chooses door at random" assumption, from what I've gathered. I think there was a resemblance to a fairly common "brainteaser" going the rounds in math circles at the time, and in that case 1/2 was the correct answer.

[Barry Shnikov]

because of a biased sampling of a subset of all the shows

Oo, it gives me goosepimples when you talk like that...

[Barry Shnikov]

If that *still* doesn't do it for you, how about a game of poker?

Ooops! Mistake-o.

Better try a game with more chance in it - like baccarat.:waycool:

[Asif]

Wow - what a brainteaser. I have seen something similar to this done by Derren Brown on one of his TV programs.

Bascially, he had three identical jewellery boxes and put a diamond ring into one of them and then mixed them around (out of site of the public). He then asked a member of the public to take one (at this point the person had a 1/3 chance of picking the ring).

He then opened one of his boxes and showed it to be empty. This is where the clever stuff happened. It was all down to how he then spoke to the person and the words he chose - he must have known which box the ring was in, in order to decide how he was going to talk to the person.

Example 1) - Person has picked the ring, so DB has to convince them to do the swap. He would say something like ......

"OK, so before i eliminated that box, you had a 33% percent chance having the ring, which means i had a 66% percent chance (having two boxes). Seeing as i have removed one of these boxes, this other box that i've got has now got the 66% chance against your box which only has 33%. What do you want to do?"

This would make the person think that, by DB saying that his box had more chance than their own box, he was telling them to change but at the same time would have his own bluff worked out and in fact did not want them to change, so they would try to double bluff him and would change. The result ...... he triple-bluffed them and got them to give him the box with the ring.

Example 2) - Person has not picked the ring, so DB had to convince them not to do the swap. He starts of with the same banter but then would add something like ....

"But think about it, if i have a 66% chance and you have a 33% percent chance, do you really think i would have told you about this - i would have said something else. So, what you need to do know is work out if i told you this just because i wanted you to change or not."

Again, this would make the person think that, by DB saying that his box had more chance than their own box, he was telling them to change but at the same time would have his own bluff worked out and in fact did not want them to change. However, as there was a greater probabilty that he did have the ring to begin with, he was trying to get them to actually do the swap. They would then decide to stay with their box. The result ..... again he bluffed them and got them to stick with the box that didn't have the ring.

Something like that .... I think!!!!

It basically came down to the fact that he new from the outset where the ring was and, by his very clever wordplay, was able to get the person to either give him the ring or stick with an empty box.

I wouldn't imagine that this gameshow host to be as clever and for the TV company to employ such method, otherwise none of the contesants would ever win.

Hang on .... i seemed to have lost the thread myself. Basically, at the beginning, the player has a 33% percent chance of winning. However, by the time they have to choose there are only 2 choices left, then they have a 50% chance. Probability wise, they orginally have less chance of guessing right, so, given the choice, i would always swap.

My head hurts - i'm going to lie down now! :sick:

[Barry Shnikov]

It's not explicitly stated whether or not the MC knows where the car is. If he doesn't, and just opens a door at random, we get into the earlier case I described (the exclusion of some states has already "happened" due to the fact we don't see a car behind the door) - so it makes no difference what we do. (N.B. making all the standard Bayesian assumptions that I can't be bothered to spell out).

If he does, we get to the second ambiguity. Does he always open a door to reveal an empty space? If so, we're in the original problem and should switch.

But it might be that he only offers us the chance to switch if we've already picked the door with a car ('cos he's a meanie!). In which case we should never switch.

I think there's a fallacy here.

If the MC opens a second door and there is no car behind it, you are in the situation already described. The only other possibility is if he opens the door and there is a car behind it. In which case, probability doesn't enter into it any more - you the contestant pick the obvious door.

Thus it doesn't matter whether the MC knows where the car is or not - if you the contestant don't see the car, you're in the same position regardless.

um...that's right, isn't it?

[Barry Shnikov]

Now, would anyone like this box, which may or may not contain a cat...


I think it definitely contains a cat; the question is, is it alive or dead?

[David Franklin]

I think there's a fallacy here.

If the MC opens a second door and there is no car behind it, you are in the situation already described. The only other possibility is if he opens the door and there is a car behind it. In which case, probability doesn't enter into it any more - you the contestant pick the obvious door.

Thus it doesn't matter whether the MC knows where the car is or not - if you the contestant don't see the car, you're in the same position regardless.

um...that's right, isn't it?No. Let A be the event that the first door picked holds the car, and B be the event that the random door picked by the MC shows an empty door. Write Ac and Bc for A complement and B complement as I can't do superscripts here.

Then P(A) = 1/3, P(B|A) = 1, P(Ac) = 2/3, P(B|Ac) = 1/2.

What we want is P(A|B). By Bayes theorem, this is:

P(A|B) = P(A and B) / P(B) = (1/3) / (P(B|A)P(A)+P(B|Ac)P(Ac)) = (1/3)/(1/3 + 2/3 x 1/2) = (1/3)/(2/3) = 1/2.

In other words, given the MC shows an open door, the probability we picked the car is 1/2.

[Barry Shnikov]

No. Let A be the event that the first door picked holds the car, and B be the event that the random door picked by the MC shows an empty door. Write Ac and Bc for A complement and B complement as I can't do superscripts here.

Then P(A) = 1/3, P(B|A) = 1, P(Ac) = 2/3, P(B|Ac) = 1/2.

What we want is P(A|B). By Bayes theorem, this is:

P(A|B) = P(A and B) / P(B) = (1/3) / (P(B|A)P(A)+P(B|Ac)P(Ac)) = (1/3)/(1/3 + 2/3 x 1/2) = (1/3)/(2/3) = 1/2.

In other words, given the MC shows an open door, the probability we picked the car is 1/2.

Ri-i-ght >he whimpers, holding his head< but if the first door picked holds the car you don't proceed any further, because there's no point in it. The conditional probability doesn't arise.

[David Franklin]

Ri-i-ght >he whimpers, holding his head< but if the first door picked holds the car you don't proceed any further, because there's no point in it. The conditional probability doesn't arise.No, the first door is the one the contestant picks. So you don't know what's behind it until after you proceed.

The point is the fact that the MC, picking at random from the doors you haven't picked, has chosen an empty door, tells you something about the chances that the car is behind the door you did pick.

[El Salsero Gringo]

No, the first door is the one the contestant picks. So you don't know what's behind it until after you proceed.

The point is the fact that the MC, picking at random from the doors you haven't picked, has chosen an empty door, tells you something about the chances that the car is behind the door you did pick....because, if the MC, picking at random, chooses an empty door, you can deduce that it's more likely that you have the car. Right?

[David Franklin]

...because, if the MC, picking at random, chooses an empty door, you can deduce that it's more likely that you have the car. Right?Yes.

Here's a "sharper" variation.

Imagine you are playing Russian roulette with the MC, and there are 2 guns, one with no bullets, and one with 5 bullets. Suppose the MC goes first, takes a gun, and "spins" the barrel and fires - 'click'! He's alive! He does this 20 times in a row (he's a Deerhunter fan!).

If you assume the MC knows and can select the empty chamber in the gun with 5 bullets, this tells you absolutely nothing about which gun he's got - exactly the same thing would have happened regardless of which gun he had. So you have no rational reason to prefer his gun over your own.

But if he is spinning randomly, the chances of him having the loaded gun are less than 1 in a billion, so you would be crazy not to pick his gun if you could.

[Barry Shnikov]

No, the first door is the one the contestant picks. So you don't know what's behind it until after you proceed.

The point is the fact that the MC, picking at random from the doors you haven't picked, has chosen an empty door, tells you something about the chances that the car is behind the door you did pick.

OK, gotcha. Much more instructive than the algebraic version...

[Dreadful Scathe]

No no no. Opening the first box instead of the second door which doesnt contain the cat gives you a 66% chance of not getting the original 33% chance that you had when you originally came on the tv show in your underwear. If you then open the door on the car with the boot also open you'll find Darren Brown behind you going "wooooooooo" and yapping on about boxes.

[ducasi]

The challenge in the puzzle isn't working it out (which is difficult to start with if you've never heard it or anything similar before,) but understanding the solution intuitively once you know it. (At least, it was for me.)

[El Salsero Gringo]

... you'll find Darren Brown behind you ...Oh for goodness sakes, it's Derren Brown - not Darren. From you, DS, I expected so much better.

[Dreadful Scathe]

Oh for goodness sakes, it's Derren Brown - not Darren. From you, DS, I expected so much better.
The extra E makes all the difference - especially at a dull Ceroc event. :eek: :)

[bigdjiver]

My way of explaining it simply:
Describe the doors any way you like, say left, right, middle.
Pick a door, say the left one.
If you swap, and the car is behind the left door you lose. If it is behind the right door or centre door you win. If you swap you win two times out of three.
If you do not swap you only win if the car is behind the left door, and that is one time out of three.

[Bombay Sapphire]

More of these please.

I liked the problem,
and understood the solution fairy quickly,
but then wasted 10 minutes reading geek like
posts about the theory of probability -

Why not start a logic puzzle thread?

:)

[Dreadful Scathe]

like this thread (http://www.cerocscotland.com/forum/showthread.php?t=741) ? :)

[Bombay Sapphire]

Thanks Scathe.

Last post 2003! No new ones since then?

[WittyBird]

The extra E makes all the difference - especially at a dull Ceroc event. :eek: :)
:rofl: