Monty Hall Problem... For Newbies

[Rolfe]

You shouldn't really need a simulator. It is possible to work out logically that IF Monty ALWAYS opens a door and shows a goat, you will gain twice as often as you will lose by switching.

Most of the argument actually deals with people disputing whether the terms of the puzzle do dictate that he must always open one of the two doors you haven't picked, and reveal a goat. If it is agreed that this is what he always does, it's easy to work out from there. The simulators just show it.

Rolfe.

 

[Professor Yaffle]

Played your link 10 times, got 7 goats and 3 cars....

If someone wants to scroll up (and click back) there's an authorative (or so I supposed) statement on how often you have to play the game to prove it by the practical method...

ETA: 12 goats, 3 cars
EATA: 20 goats, 3 cars - is your link crooked??
EOATA:30 goats, 3 cars. It did say, when I first clicked the link, that it might not work on my browser (chrome) and I should use IE4 (yeah, right). But that should just mean it failed to work at all, not that it would automatically give you goats in Chrome? Come on, I've only just been convinced that of course you should change, and now it's 10/1 in favour of sticking?

It works for me, even though it said it wouldn't work with my browser. Over 1000 runs, when sticking I got 32% cars and 68% goats, when changing i got the reverse. Doing it manually and changing every time for 10 goes, I got 8 cars and 2 goats.

ETA - I wasn't paying any attention to where the car actually was when I played, just picking randomly, and I just noticed that the car is always in the same position. I guess thats why it doesn't work on chrome.

 

[fromdownunder]

I have to say in all honesty that I do not understand the "Monty picks a random door" thingy which has been discussed at some length here. The problem in it's pure form is that Monty knows exactly where the car is and will always pick a goat.

The hundred door example, mentioned earlier several times explains it best. You start with odds of one in 100, and finish up with two doors 50-50 odds. I would go with 50-50 and change rather than assume that my first 100/1 guess was incredibly insightful.

I can't see how this is any different than starting with three doors. Changing increases your chances of winning from 33% to 66%.

Norm

 

[anglolawyer]

It's one of those things I got my head round so long ago I gave up discussing it. Never ceases to amaze me how often it's resurrected though.

It's one of my intellectual boasts, though. Figured out what Peregrinus expectavi pedes meos in cymbalis means, and figured out the Monty Hall problem unaided.

Rolfe.

I want to play!

Peregrinus = Peregrine Falcon
expectavi= expects
pedes= feet
meos= me or my
in = in
cymbalis = cymbals

The Peregrine Falcon expects my feet in cymbals! No problemo. Except is make no sense. Hmm. Take two ...

The Peregrine Falcon usually walks on its tip toes (I know, I know).

Rolfe! What's the answer?

 

[zooterkin]

I want to play!

Peregrinus = Peregrine Falcon
expectavi= expects
pedes= feet
meos= me or my
in = in
cymbalis = cymbals

The Peregrine Falcon expects my feet in cymbals! No problemo. Except is make no sense. Hmm. Take two ...

The Peregrine Falcon usually walks on its tip toes (I know, I know).

Rolfe! What's the answer?

You could try searching the forum, when you'll find this thread.

 

[Caper]

My God are we still at this? I worked this out before I even had a computer. Somebody told me about the puzzle, I think it had been in the Guardian.

Suppose there are 100 doors and one car. And the presenter keeps opening doors and keeps asking you if you want to switch. You keep saying no. Finally you're down to two doors. The one you originally chose and one other.

You still sure you don't want to switch?

Rolfe.

Wow... That is the best explanation yet!!

 

[anglolawyer]

You could try searching the forum, when you'll find this thread.

No, it's OK. I finally worked it out. It means the Peregrine Falcon trod on my cymbals. Obvious really.

 

[Rolfe]

It actually means "I really hate Igor Stravinsky and spit on him as a dirty sell-out to the western powers".

Which reminds me I have a rehearsal tonight for that very piece. It's astounding how it comes over once you realise what it means.

And this is totally OT, please return to your regularly scheduled broadcasting.

Rolfe.

 

[Rolfe]

I have to say in all honesty that I do not understand the "Monty picks a random door" thingy which has been discussed at some length here. The problem in it's pure form is that Monty knows exactly where the car is and will always pick a goat.

Yes and no. It's usually understood to be like that, but it isn't generally made explicitly clear. As you said, when that is made explicitly clear the answer is simple. Most of the arguing, though, relates to people quibbling about whether that is actually the form of the puzzle.

Rolfe.

 

[Mark6]

Wow... That is the best explanation yet!!

Once or twice I saw people (on the web, not in person) who still claimed the chances of the car being behind either of two remaining closed doors (out of 100) are even.

Most of the argument actually deals with people disputing whether the terms of the puzzle do dictate that he must always open one of the two doors you haven't picked, and reveal a goat. If it is agreed that this is what he always does, it's easy to work out from there.

I disagree. Some people just seem to have a mental block about the whole thing even if they know and accept that the host always opens the empty door.

 

[Meed]

I have to say in all honesty that I do not understand the "Monty picks a random door" thingy which has been discussed at some length here. The problem in it's pure form is that Monty knows exactly where the car is and will always pick a goat.

"Monty picks a random door and happens to reveal a goat" is not the problem in it's pure form. It's a variant. Because it has different odds for switching and sticking than the original problem, it is useful for demonstrating that Monty's "intent" or "strategy" matters in the original.

The hundred door example, mentioned earlier several times explains it best. You start with odds of one in 100, and finish up with two doors 50-50 odds. I would go with 50-50 and change rather than assume that my first 100/1 guess was incredibly insightful.

In the standard Monty you'd have a .99 probability of winning by switching here. But if Monty randomly opened 98 doors and, unlikely as it was, did not happen to reveal the car... then you would have a 50% chance of winning the car regardless of whether you stuck or switched.

 

[ehcks]

In the standard Monty you'd have a .99 probability of winning by switching here. But if Monty randomly opened 98 doors and, unlikely as it was, did not happen to reveal the car... then you would have a 50% chance of winning the car regardless of whether you stuck or switched.

<Quibble>I think that's not quite right. The odds of him randomly picking 98/100 doors without the car behind it are low enough that you should assume you've picked the correct door at the start.

 

[GlennB]

<Quibble>I think that's not quite right. The odds of him randomly picking 98/100 doors without the car behind it are low enough that you should assume you've picked the correct door at the start.

It is right.

Your odds of having picked correctly at the start were, and still are, 1 in 100. Nothing has changed there as long as Monty chose his doors at random.

The point of the classic MH problem is that *sometimes* his hand is forced and what he shows you isn't random.

 

[ehcks]

It is right.

Your odds of having picked correctly at the start were, and still are, 1 in 100. Nothing has changed there as long as Monty chose his doors at random.

The point of the classic MH problem is that *sometimes* his hand is forced and what he shows you isn't random.

The odds of randomly picking 98/99 doors without a car has to be worse than 1/100.

 

[Meed]

It is right.

Your odds of having picked correctly at the start were, and still are, 1 in 100. Nothing has changed there as long as Monty chose his doors at random.

No, they go up to 1/2.

The point of the classic MH problem is that *sometimes* his hand is forced and what he shows you isn't random.

Yeah, in the classic it would stay 1/100.

The odds of randomly picking 98/99 doors without a car has to be worse than 1/100.

Say you randomly chose door 1. So he will not open this one. Then Monty randomly selected door 2 as the door not to open. Then if all the remaining doors happen, by chance, to be goats, that means that one of the two random picks at the start was the car. Each had an initial 1/100 chance of containing the car, reduced to 1/2 with the elimination of the remaining 98 doors. It should work out the same way if he just randomly chooses doors to open until there's one left instead of randomly choosing which one not to open at the start.

 

[GlennB]

The odds of randomly picking 98/99 doors without a car has to be worse than 1/100.

In the version we're discussing it's like Russian Roulette with a 100-chamber revolver. You've only survived this far by pure chance, and we're down to 2 chambers. The one you chose is the final chamber in this analogy.

Do you prefer that final chamber to the penultimate chamber? If so, why?

 

[GlennB]

No, they go up to 1/2.

Yep. Lazy speak here.

 

[Brian-M]

<Quibble>I think that's not quite right. The odds of him randomly picking 98/100 doors without the car behind it are low enough that you should assume you've picked the correct door at the start.

He's not randomly picking 98/100 doors. He's randomly picking one door out of the remaining 99 doors. All the doors he opens are the ones which haven't been picked.

So you each have a one chance in 100 of picking a door with a car behind it, and if no car is seen when all the other doors are opened, then this means that the car is equally likely to be behind either door.

(Yes, he's only picking one door out of 99, which would seem to give him slightly better odds, but there's also a chance that the car isn't behind any of the 99 doors he has to pick from, which brings the odds back to even.)

 

[jsfisher]

The odds of randomly picking 98/99 doors without a car has to be worse than 1/100.

If there be a car behind one of the 99 doors, the chances of picking 98 doors with no car is exactly the same as picking 1 door with the car. (I.e. Pick one door, then open all the rest.) 1 in 99.

[For completeness: There's 1 chance in 100 that Monty will have 99 doors with no car and 99 chances in 100 that the car will be among his 99.

In the former case, his chances of opening 98 doors, all goats is 1 in 1 (since they all have goats behind them, and in the latter case, 1 in 99.

So, is overall odds of revealing only goats while choosing doors at random is:

1/100 * 1 + 99/100 * 1/99 = 2/100 = 1 chance in 50.

And that makes sense if you think of it in terms of the contestant picked one door and Monty picked one door, each at random. 2 chances in 100 that one of those two doors has the car.]

 

[ehcks]

But that means that picking 98 doors wrong is more likely than picking 97 doors wrong, which extends to being more likely than picking one door wrong.