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- The Monty Hall problem (or 2 goats and a car)
So there’s the familiar problem that has been around for quite a while now, but personally, I have never been able to get my head around WHY the answer is the case. I’ve seen it written out long ways, short ways, with demonstrations and thought experiments, but never have I been able to bend my mind to accept it.

Problem (in this version, from Wired magazine):

Suppose you’re on a game show and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say Number 1. The host, who knows what’s behind all the doors, opens another; say Number 3, which he knows is hiding a goat. He then asks, “Do you want to pick door Number 2?” Is it to your advantage to switch your choice?

Answer:

Yes. Switching your choice always raises your odds. Originally you had a 1 in 3 chance of guessing the correct door. When Monty shows you the goat behind a door you didn’t choose, the probability of a car being behind that door drops to zero and the chance that there’s a car behind the third door thus rises to 2 in 3. So your odds improve if you switch your choice.

Now, here is where I can’t follow it. Why are the odds after the door has been opened not 1 in 2?

The way I see it, when you are offered the swap, you are effectively given the choice to choose again, between the two remaining doors, which makes it 1 in 2 (surely?). If you choose the same door again, it is the same

*effect*as sticking with the original 1 in 3 odds, but with updated information – making it a 1 in 2. The third door is now irrelevant, which suggests to me that*‘the chance that there’s a car behind the third door thus rises to 2 in 3.’*is bunkem.Anyone care to have a go at explaining it?

Posted 7 years agohmmm

Posted 7 years agoThe easiest way is just to play it out. Consider three doors, A, B and C. A and B have sheep behind, C has the car.

Scenario 1. You pick A. Monty reveals B. To win you need to change to C

Scenario 2. You pick B. Monty reveals A. To win you need to change to C.

Scenario 3. You pick C. Monty reveals A or B. To win you need to stay with C.So, in 2/3 of the scenarios you win by changing, and in 1/3 you win by staying.

Posted 7 years agoIt’s because you could still in theory pick the door he’s already opened. Which is daft. So it’s still a three door problem even though it’s not.

Posted 7 years agor0bh – Member

The easiest way is just to play it out. Consider three doors, A, B and C. A and B have sheep behind, C has the car.

Scenario 1. You pick A. Monty reveals B. To win you need to change to C

Scenario 2. You pick B. Monty reveals A. To win you need to change to C.

Scenario 3. You pick C. Monty reveals A or B. To win you need to stay with C.So, in 2/3 of the scenarios you win by changing, and in 1/3 you win by staying.

you have actually compressed 2 separate scenarios into option 3, so thats a flawed argument.

Posted 7 years agoWhy are the odds after the door has been opened not 1 in 2?

Now you can get into some probability but I find it easier to take a slightly different look at the problem.

You choose one door.

The host then offers you a choice, keep the single door you chose or take

*both*of the other doors.Think about what the probabilities are now.

This is essentially the same choice as previous as you

Posted 7 years ago*know*that one of the doors in pair must include a goat so by showing you this the overall probabilities don’t change. Remember probabilities can only change with new information. Dressing up old information as new doesn’t count.~~r0bh’s explanation is excellent; it answers why it is 2 in 3, and I can follow that through and accept that, but it doesn’t disprove the 1 in 2 idea in my mind (which obviously can’t be ALSO correct if I accept the 2 in 3)~~Edit – ok maybe it doesn’t…For years I’ve had to accept I don’t understand it, and that to accept the 2 in 3 fully I’d have to effectively put faith in the ‘experts’ without being able to prove it to myself – and I can’t do that ðŸ‘¿

Posted 7 years agoyou have actually compressed 2 separate scenarios into option 3, so thats a flawed argument.

No it’s not. The scenarios are only dependent on what door you pick first. Three scenarios for three doors. In scenario 3 the door Monty chooses to reveal is irrelevant.

Posted 7 years agoMilitant_biker – Member

Now, here is where I can’t follow it. Why are the odds after the door has been opened not 1 in 2?

They are. All the rest is just the awesome ability of statistics to prove something that isn’t true.

Posted 7 years agoIt’s also to do with which door the host opens and where the car is.

Posted 7 years agoIf you look at this on wiki its a nightmare of odd symbols and expensive looking maths.

Seems that once the door is open you now have a 50/50 choice on what’s left. I’ll be dipped in dogshit if I can fathom why it’s 2/3, seems to include the open door but you know that’s out so I don’t get it.

Posted 7 years agoNow my brain’s hurting.

Raises eyebrow at Northwind… Do you mean the accepted answer or the ‘answer’ I quoted?

Posted 7 years agoThink of it this way…

You pick one of the three doors.

The odds that you picked the right one are 1 in 3.

The host opens a door and shows you a goat.

At this point, before offering you a choice, would you agree that the odds that you picked the right door are still 1 in 3?

By the host opening the door, we know that the odds of there being a car behind that door have collapsed to 0, while the odds that you picked the correct door to start with is still 1 in 3, you haven’t been asked if you want to change yet.

As the set of all odds has to add up to 1, the other door then must have odds of 2 in 3 that the car is behind it.

Still no choice been allowed here.

At this point you are asked if you want to change. Can you explain to me why the odds would then change to 1 in 2 at this point?

Posted 7 years agonever mind all the maths – has any one ever found out empirically what choice is the best one?

Posted 7 years ago[video]http://www.youtube.com/watch?v=mhlc7peGlGg[/video]

Pathetic voice don’t know why the did that but this help explains it.

Posted 7 years agoLast time this came up someone, GrahamS I think, wrote a script for testing this and proved it empirically. It’s also been done in Excel as well I believe.

The answer is to always swap.

Posted 7 years agoI can’t be bothered to read all the replies, but it did take me a little while to understand the first time around. It makes sense to me like so:

At the beginning, the chance of picking the right door is 1/3. You pick your door. The chance that the car is behind one of the other doors is 2/3. Ignore the host. Whatever he does, since there is no way he’ll pick the car, the odds don’t change. Your door is still 1/3 and the doors you didn’t pick are still 2/3.

It’s just that the other doorS are now just one door.

Posted 7 years agofunkynick – I’m with you up until

At this point, before offering you a choice, would you agree that the odds that you picked the right door are still 1 in 3?

No, once the door has been opened, the odds are no longer 1 in 3. The odds WERE 1 in 3, but once the door is opened the odds of the door STILL being correct are now 1 in 2.

At this point you are asked if you want to change. Can you explain to me why the odds would then change to 1 in 2 at this point?

When asked if you want to change, I view it like this. The opened (or third) door is irrelevant. You have 2 doors in front of you. One contains a goat, or contains a car. You again choose at random (which is 1 in 2). Now half of the time you would choose that same door (which is sticking) and half the time you would choose the other door (which is switching). That’s my logic behind the odds changing to 1 in 2.

Empirically – the solution is correct, switching is successful in 2 in 3 cases. And I’m almost happy with that, but I’d love to know why!

Posted 7 years agoHow about this… if there are 100 doors, 99 goats, one car…

Pick a door, the odds on the car are 1 in 100

Now he opens 98 other doors with goats behind.

Do you contest that at this point, without any other choise being offered, they you have a 50:50 chance of winning the car?

Posted 7 years agoMilliant you need to think back to the start, you had a 33% chance of picking a car and 66% picking a goat.

The hose opens the door and you stick means you still have that 33% chance of picking the car as you didn’t change anything.

You decide to swap that means you now have increased your odds to 66% chance but only if you picked a goat first but you had a 66% chance to do that.

Posted 7 years agobut once the door is opened the odds of the door STILL being correct are now 1 in 2.

Why? What new information do you have about the door that you chose?

The opened (or third) door is irrelevant

No it’s not, as you knew as soon as you chose one door that one of the other two MUST have a goat.

Posted 7 years agoWhen Monty shows you the goat behind a door you didn’t choose, the probability of a car being behind that door drops to zero

That bit it is wrong though it can’t be zero unless there was no car, it’s 33%.

Posted 7 years agoArse I misread that.

Posted 7 years agoConsider three doors, A, B and C. A and B have sheep behind, C has the car.

Scenario 1. You pick A. Monty reveals B. To win you need to change to C

Scenario 2. You pick B. Monty reveals A. To win you need to change to C.

Scenario 3. You pick C. Monty reveals A. To win you need to stay with C.

Scenario 4. You pick C. Monty reveals B. To win you need to stay with C.There are 3 doors, but 4 possible scenarios, its just a logic trick.

Posted 7 years agoDrac.. no it’s not, you know for certain that there is no car behind that door as you can see a goat behind it, therefore the probability of it being the car is 0.

Posted 7 years agoWot funkynick said.

Extrapolate the number of doors upwards and it’s then obvious that you should choose again.

Posted 7 years agofunkynick

*Do you contest that at this point, without any other choise being offered, they you have a 50:50 chance of winning the car?*

I would say that, intuitively, it’s a 50:50 chance, because there are only 2 doors left. I fail to see how the other 98 doors (or 1 door) can have any bearing on the choice. I see it as having ANOTHER choice based on the 2 doors, which points to 50:50.drac. I don’t accept that you still have that 33% chance. Thats my problem!

In all of this, I know empirically that I’m wrong, I’d just love to be able to convince myself of the right answer!

Posted 7 years agonever mind all the maths – has any one ever found out empirically what choice is the best one

It’s very similar to something called ‘The principle of restricted choice’ in Bridge. [Missing two equivalent cards such as the J and Q, if one opponent shows the Q, say, on trick 1, then the odds now favour them not having the J and you should play the other opponent for it.] Sort of meaningless if you don’t play Bridge I guess, but the point is it’s a proven principle that’s used to determine strategy in the game.

If you google ‘principle of restricted choice’ there’s all sorts of similarly impenetrable proofs out there as for the Monty Hall problem

Posted 7 years agoMSP… You are correct that there are 4 possible scenarios, but scenarios 3 and 4 have only half the probability of 1 and 2.

See Drac piccie earlier in the thread.

Posted 7 years agogonefishin –

*Why? What new information do you have about the door that you chose?*I have no new information about the door I chose, but I now know that the door that was opened is a goat (and can be discarded?)

My logic means I flush the cache on what I used to know. I can see 2 doors closed, one of them I know holds a goat, one of them I know holds a car.

Posted 7 years agoMy problem is not in accepting the 2 in 3, which can be proven, it is in disproving the 1 in 2 theory. It obviously cannot be both.

The only way I can tell myself it’s not 1 in 2, is the fact that it’s 2 in 3 – ARGH!

Posted 7 years agoIt’s not 1 in 2 as there’s still 3 doors but you know the answer to one so your not going to pick that which increases you odds to 2/3 as you it’s like getting a chance to go again.

Posted 7 years agoThe whole point of this puzzle is that it’s supposed to highlight that our intuition is wrong (similar to the puzzle someone posted a couple of weeks back).

On the first choice you have a one in three chance of making a correct choice, and a two in three chance of goating out.

When the host asks if you want to change your mind, what’s essentially happening is that you’re being offered to swap the one door you have for

*both of the other doors.*You both know one of the remaining two doors contains a goat, by opening a door (with his insider knowledge) the host simply confirms something you knew anyway. Changing your mind nets you both doors, the opening of one of them is a bit of very effective misdirection.

Posted 7 years ago*You both know one of the remaining two doors contains a goat, by opening a door (with his insider knowledge) the host simply confirms something you knew anyway. Changing your mind nets you both doors, the opening of one of them is a bit of very effective misdirection.*Ah ha… this begins to make a little more sense.

Posted 7 years agoSay you choose door 1, and supposing the host gives you the opportunity to keep the contents of door 1, or to swap and keep the contents of BOTH other doors. Obviously you’d swap, because that gives you a 2/3 chance of winning. The host then says I’ll save you the bother of opening both doors, I’ll open one of the doors for you.

The only difference in the actual game is that the host opens one of the doors before asking if you want to swap instead of after, but that’s irrelevant.

Posted 7 years agothe only way to switch and LOSE, is if you picked the correct door first time (probility 1/3)

If switch-and-lose is 1/3 then switch-and-win must be 2/3

Posted 7 years agoI can see 2 doors closed, one of them I know holds a goat, one of them I know holds a car.

Just because there are two outcomes doesn’t mean that the probabilities are equal. For example there are two outcomes with a lottery ticket, it will either win the jackpot or it won’t, but the odds aren’t 50:50

Posted 7 years agoFor example there are two outcomes with a lottery ticket, it will either win the jackpot or it won’t, but the odds aren’t 50:50

Ah! I get this!

sas – your explanation works for me too

Thanks!

Posted 7 years agoReading this thread was worth my time just for the phrase ‘goating out’

Posted 7 years ago

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