The Monty Hall Problem: Help!

OneWingedBird

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#61
Assuming that the car is behind door one, then:


a) You pick door one and the host (who has a choice of two doors) opens door two, switching will give a lose.

b) You pick door one and the host (who has a choice of two doors) opens door three, switching will give a lose.

c) You pick door two, and the host has no choice but to open door three, switching will give a win.

d) You pick door three, and the host has no choice but to open door two, switching will give a win.


These are the only four possible scenarios, and there will still be two win situations and two lose situations whichever door the car is behind.

I think where people are going wrong is in forgetting that if you choose the car door outright, then the host has two choices of action both of which will result in switching being a lose - missing one of those out suggests that switching will give a win in 2/3 switches, when it is actually 2/4.
 

ginoide

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#62
there's only one chance out of three that you choose door one, though. we're talking about your chances, not about monty hall's.
(ginoide, the born-agian twothirdian)
 
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Anonymous

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#63
There are three equally likely scenarios. As scotmedia pointed out, one of them subdivides, but this doesn't change your chances of picking the car in the first place.

a) You pick door one and the host (who has a choice of two doors) opens door two or door three, switching will give a lose.

b) You pick door two, and the host has no choice but to open door three, switching will give a win.

c) You pick door three, and the host has no choice but to open door two, switching will give a win.

There's nothing particularly wrong with writing that out as four scenarios, but you must take into account that you are twice as likely to be in one of scenarios c) and d) than one of scenarios a) and b)

We can turn it around the other way and show by reductio ad absurdum that you're wrong:

a) You pick door one...

b) You pick door one...

c) You pick door two...

d) You pick door three...
I've truncated your scenarios. If these scenarios are equally likely, there must be some bias toward choosing door one - your list of scenarios show it being picked twice as much as either door two or three. But there is no such bias in the problem.

David
 

OneWingedBird

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#64
Ok, I played the goat applet and got 7/20 when I stick every time and 9/20 when I switch every time.

I think we're arguing over an ambiguity in the rules here:

If the host always opens the same goat door in responce to you choosing the car door, then there are two switching outcomes which will give a win against one which will give a lose.

If the host chooses which goat door to open in responce to you choosing the car door, then there are two switching outcomes that will give a win and two that will give a lose.
 

OneWingedBird

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#65
Probably not the best 10am explanation, I'd have done better to list a & b as a(i) and a(ii), but I still think the whole disagreement is hindging on wether the host has a choice there.

True there is only one outcome of the three which involves you choosing the car door, but if the host does have a choice, then there are two possible outcomes depending on that where switching will give a lose, as opposed to one each for the other two initial choices.
 
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Anonymous

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#66
True there is only one outcome of the three which involves you choosing the car door, but if the host does have a choice, then there are two possible outcomes depending on that where switching will give a lose, as opposed to one each for the other two initial choices.
If you pick the car, he can choose the same goat door, or he can choose a random goat door. His strategy for picking a goat makes no difference to your optimum strategy, which is to switch.

Your a)i)/ii) scenario has a probability of 1/3 of coming about. The "subscenarios", which are both equally likely, therefore both have a probability of 1/6.

David
 
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Anonymous

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#67
Ok, I played the goat applet and got 7/20 when I stick every time and 9/20 when I switch every time.
For a less ambiguous set of figures, just play the stick each time. Just keep clicking (click-ing, click-ing) on door 1 until you've racked up a 100 trials.

I won 30 out of 100 by sticking - you don't get much closer to 1/3 than that. If I was right to stick 1/3 of the time, then I must have been wrong to stick (and right to switch) 2/3 of the time.

David
 

Breakfastologist

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#68
I sat down last night with some cards and my girlfriend and ran it through manually until I understood it.

The sleight of hand is in the impression you get that your odds change when you have seen an incorrect result. The 50/50 idea is natural and obvious if you have two separate events, but because the odds don't change from start to finish, the 66/33 odds apply the whole way through. Its a case of thinking of the game as a single action, rather than a series of them.

It still does my head in that if I had chosen one, had an incorrect alternative revealed then wandered off before completing it and came back having forgotten my previous choice, the odds would have changed.
 
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Anonymous

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#69
Well, I am very thick. But this is how it seems to me: in the first instance, you're choosing which one of the goats you're going to get rid of. The 50/50 probability applies the whole way through - I see it as 2 consecutive choices to make. The first choice is: which one of the non-car doors are you getting rid of? The second is: which one of these 2 remaining doors has the car behind it?
 

Melf

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#70
i tried the 3 door game and managed to get 80% or 8 out 10 tries
correct

and there is a patten tho
 

Min Bannister

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#71
No thats not true (Wolfie) because you don't know which of the doors has a goat behind it to start with. You would only know that if the host opened a doot before you make your choice, which as I say, WOULD make your odds of getting the car 50/50.

It is the order of events which is important.
 
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Anonymous

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#73
Thank God for that, I do FINALLY get that its always better to swap.

I briefly considered trying to putting down here all the alternate explanations that ran through my brain that came to 2/3 conclusion (such as the odd 'the Monty card turn is completely irrelevant' idea and I had a stab at a bizzare one which I called 'Shrodinger's Goat' where the card that Monty will turn over is both a car and a goat at the same time) but, quite frankly, they're the ravings of a crazy man so I didn't.
 

OneWingedBird

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#74
Ok, it is better to swap. Damn obvious why when you think about it, the problem is a lot less complex than I thought, it just took a music teacher to explain it to me:rolleyes:

The original door you choose still only has a 1/3 chance of being the car, but when you switch, the othe door has a 1/2 chance. So it's better to switch...
 
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Anonymous

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#75
The original door you choose still only has a 1/3 chance of being the car, but when you switch, the othe door has a 1/2 chance.
Nope, the other door has a 2/3 chance. Otherwise P(win or lose) comes out as less than 1.

David
 

nikoteen1

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#77
right to make it fair we will toss a coin, heads its better to switch, tails it isnt...........
 

nikoteen1

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#78
oh its landed on its edge, which MUST mean it makes no difference at all... just as i suspected.
 

Ringo

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#79
This is one of the best threads I have ever read :D

The 2/3 theory has my vote and when I played the online goat game I got 62% success rate. The whole thing reminds me of another problem I heard when I was young.

3 flatmates decide to buy an old television for their flat. It costs £30. They each put in £10 (3 x £10 = £30), pay for the TV and take it home.

The shop clerk realises that the TV was in fact on sale for £25 but the sticker had fallen off. Taking 5 X £1 coins from the cash drawer he walks round to their flat. On his way, he realises that they don't know how much the TV was on sale for, so when they answer the door he hands them £1 coin each, explains that there was a mix up and that they should have some money back. He walks off with £2 left in his pocket.

So, they each paid £10 and recieved £1 back, so they have each spent £9.
3 x £9 = £27 (plus the £2 in the shop clerks pocket) = £29. Where is the missing £1?
 

feen5

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#80
Oh God no Ringo your stirring up another hornets nest with that one :eek!!!!: , i believe there are several variations of that same teaser and the solution is in another thread some where on the board, i'll have a look for it but if i can't find it someone else may know where it is.

PS I recently got my hands on that book and was finished it in about an hour, a fantastic book and the Monty hall chapter had me scratching my head for a while before i could get my head around it, then again maths has never been my strong point.
 

Yithian

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#81
Ringo_ said:
The whole thing reminds me of another problem I heard when I was young...
Right. Stop there. We've had this before and it was painfull enough then. Just use the search function, find the thread, read it, and whatever you do don't post anything on it or bump it.
 
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Anonymous

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#82
I'm sorry, I just couldn't resist! Its all in the way the adding up is presented - it gives you the maths quickly and then hustles you onto the question without giving you time to check the sums. There is a mistake in there :
Ringo_ said:
So, they each paid £10 and recieved £1 back, so they have each spent £9.
3 x £9 = £27 (plus the £2 in the shop clerks pocket) = £29. Where is the missing £1?
You don't add the £2 in the clerks pocket, you take it away. That gives you the £9 each plus the £3 they have just got back. Or to put it another way, There was £30. £25 went on the tv, £2 went to the clerk and £3 went back to the people.

Please don't let this be the start of a big fight!
 

Rrose_Selavy

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#84
theyithian said:
Ringo_ said:
The whole thing reminds me of another problem I heard when I was young...
Right. Stop there. We've had this before and it was painfull enough then. Just use the search function, find the thread, read it, and whatever you do don't post anything on it or bump it.
I second that! Oh no..........





They had this Doors puzzle on Mind Games on BBC4 a while back.


I think the solution was to change the channel.
 
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Anonymous

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#86
feen5 said:
PS I recently got my hands on that book and was finished it in about an hour, a fantastic book and the Monty hall chapter had me scratching my head for a while before i could get my head around it, then again maths has never been my strong point.
Funny - I've just finished reading that book. Yes, the Money Hall problem had me puzzled too, but the diagram Haddon includes makes it a whole lot clearer.
 

Peripart

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#87
My comment on the Monty Hall problem - I've got a degree in Maths but this still makes my brain hurt. I can convince myself that switching tactics both improves and does not improve your chances, so I can only bow to the judgement of the poster who actually did the experiment for real and got a 62% succes rate. That's close enough to 66% for me. Now, where did I put the paracetamol...?

Oh, before I go: Ringo, stop trying to cause trouble, you little scamp!
Your problem looks tricky, but isn't:
Put simply, of the original £30 (or$), £25 is with the shopkeeper/manager/whatever, £2 is with the waiter and the other £3 is with the 3 customers. The "missing" pound comes from the confusion as to whether the £2 should be added to the £27 or subtracted from the £30. Or possibly, it fell on to the floor and rolled under a table while everyone was arguing over the bill. Either way, it's there somewhere!
 

Peripart

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#88
My comment on the Monty Hall problem - I've got a degree in Maths but this still makes my brain hurt. I can convince myself that switching tactics both improves and does not improve your chances, so I can only bow to the judgement of the poster who actually did the experiment for real and got a 62% succes rate. That's close enough to 66% for me. Now, where did I put the paracetamol...?

Oh, before I go: Ringo, stop trying to cause trouble, you little scamp!
Your problem looks tricky, but isn't:
Put simply, of the original £30 (or$), £25 is with the shopkeeper/manager/whatever, £2 is with the waiter and the other £3 is with the 3 customers. The "missing" pound comes from the confusion as to whether the £2 should be added to the £27 or subtracted from the £30. Or possibly, it fell on to the floor and rolled under a table while everyone was arguing over the bill. Either way, it's there somewhere!
 

Peripart

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#89
A couple of apologies:

First, I got a bit over-zealous in my response to Ringo's problem, and completely failed to see Johnny Molten's much clearer explanation. Sorry, Mr Molten.

Second, the computer biffed me offline just as I was submitting my last post, so I ended up resubmitting when I logged back on. It's not that I was deliberately repeating myself, or that I'm a total blithering idiot or anything. Just wanted to clear that one up. Phew!
 

ogopogo3

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#90
Well, since that's been adequately explained, who wants to discuss that pesky little mathematical proof which states that the number 0.999999 to inifinity is actually equal to 1? :twisted:
 
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