The Monty Hall Problem: Help!

Ringo

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#91
Is it? I can't see how it would, as it never achieves 1. Would somebody explain it to me as I have a full cup of tea and a packet of chocolate digestives waiting to be nibbled, dribbled and crumbled all over the keyboard. :D
 

Peripart

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#92
I think that the gist of this proof is that, if you add enough 9s to 0.9999..., then you can make the difference between this number and 1 smaller than any other number that you care to name, so small in fact that there is no difference at all.

I could alternatively blather on about recurring decimals, but that would be a bit tedious, I'm sure, so look at it this way:
We're all (I hope) happy to accept that 0.33333... = one third and that 0.66666...= two thirds, so it follows that if you add the two together, you find that 0.99999...= 1.
And if that's a rigorous proof, I'm a teapot!
 

ogopogo3

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#93
Ringo_ said:
Is it? I can't see how it would, as it never achieves 1. Would somebody explain it to me as I have a full cup of tea and a packet of chocolate digestives waiting to be nibbled, dribbled and crumbled all over the keyboard. :D
True, it can never achieve 1.

However, if you try this little math problem:

1.00000000000
-0.9999999999 (to infinity)
_____________=
0.00000000000 (to infinity)


This is indeed a disturbing universe.
 

dan_uid0

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#95
richajc~ said:
You have 3 doors, A B C there is a 1/3 chance the car is behind any one of the doors,and let's say you choose A.

So we have the car has a 1/3 chance of being behind A and there is a 2/3 chance it is behind B or C.

1/3 A or 2/3(B or C)

Now the host comes and opens one of the doors which is never the door you have chosen or the one with the car. Let's say its C.
Now we get:

1/3 A or 2/3 B

I hope that helps understand how the 2/3 odds carry over to the one door. I think one of the confusing things about this problem is the 2nd choice isn't really 50/50 since the odds were 1/3 when you initially chose.
No, I still reckon you're confused here. What everyone on the thirds thing is doing is treating the choice thing as one event, whereas it is in fact two events.

First event: you have three doors, 1/3 chance of getting the car. At this point the host shows you where one goat is.

Second event: you have two doors to choose from, having been shown where the goat is in the aforementioned third door.

This time round you have 1/2 chances of finding the car, and as long as you're not choosing the door you know has a goat behind it, these odds will remain the same no matter if you switch or not.
 

_Lizard23_

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#96
No.
Because you chose before there were 2 doors.
There were 3 doors when you chose. When you chose you had a 1 in 3 chance of getting the right door. That door you chose still has a 1 in 3 chance of being right, nothing's changed except a door got opened. Someone coming into the room for the first time now and choosing one of the two remaining doors would have a 1 in 2 chance of choosing the right door. But you have more information than they do.

Please consider my 100 doors, 99 goats, 1 car example.
You choose a door.
Monty opens 98 doors, all of which have goats behind them.

There was a 1 in 100 chance you got the right door the first time.
There is still a 1 in 100 chance you have the right door now.

Unless you swap.

Make it a million doors. A billion. Make it so many doors the chances of you having picked the right one at the beginning are less than of you being hit by lightning the same day you win the lottery, then open all but 2 doors, the one you originally chose, and one other. The car's behind one of them.

Someone coming into the room now has a 1 in 2 chance of getting the right door, but because you know the door you chose when there were 872364783463786348756424567687653674265234523523153435324523453987563487957347 doors is so utterly and completely unlikely to have been the right one, you have more information than them.

You can stick with your 1 in 872364783463786348756424567687653674265234523523153435324523453987563487957347 chance, of course :)


/me still lurves this one :D
 

rynner2

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#99
Seven pages on the Monty Hall problem!
And it continues to tease and intrigue:


Monty Hall problem: The probability puzzle that makes your head melt

A reference in a recent Magazine article to the Monty Hall problem - where a contestant has to pick one of three boxes - left readers scratching their heads. Why does this probability scenario hurt everyone's brain so much, asks maths lecturer Dr John Moriarty.

Imagine Deal or No Deal with only three sealed red boxes.
The three cash prizes, one randomly inserted into each box, are 50p, £1 and £10,000. You pick a box, let's say box two, and the dreaded telephone rings.
The Banker tempts you with an offer but this one is unusual. Box three is opened in front of you revealing the £1 prize, and he offers you the chance to change your mind and choose box one. Does switching improve your chances of winning the £10,000?

Each year at my university we hold open days for hordes of keen A-level students. We want to sell them a place on our mathematics degree, and I unashamedly have an ulterior motive - to excite the best students about probability using this problem, usually referred to as the Monty Hall Problem.

This mind-melter was alluded to in an AL Kennedy piece on change this week and dates back to Steve Selvin in 1975 when it was published in the academic journal American Statistician.
It imagines a TV game show not unlike Deal or No Deal in which you choose one of three closed doors and win whatever is behind it.

One door conceals a Cadillac - behind the other two doors are goats. The game show host, Monty Hall (of Let's Make a Deal fame), knows where the Cadillac is and opens one of the doors that you did not choose. You are duly greeted by a goat, and then offered the chance to switch your choice to the other remaining door.

Most people will think that with two choices remaining and one Cadillac, the chances are 50-50.

The most eloquent reasoning I could find is from Emerson Kamarose of San Jose, California (from the Chicago Reader's Straight Dope column in 1991): "As any fool can plainly see, when the game-show host opens a door you did not pick and then gives you a chance to change your pick, he is starting a new game. It makes no difference whether you stay or switch, the odds are 50-50."

But the inconvenient truth here is that it's not 50-50 - in fact, switching doubles your chances of winning. Why?

Let's not get confused by the assumptions. To be clear, Monty Hall knows the location of the prize, he always opens a different door from the one you chose, and he will only open a door that does not conceal the prize.

For the purists, we also assume that you prefer Cadillacs to goats. There is a beautiful logical point here and, as the peddler of probability, I really don't want you to miss it.

In the game you will either stick or switch. If you stick with your first choice, you will end up with the Caddy if and only if you initially picked the door concealing the car. If you switch, you will win that beautiful automobile if and only if you initially picked one of the two doors with goats behind them.

If you can accept this logic then you're home and dry, because working out the odds is now as easy as pie - sticking succeeds 1/3 of the time, while switching works 2/3 of the time
.

Kamarose was wrong because he fell for the deception - after opening the door, the host is not starting a new 50-50 game. The actions of the host have already stacked the odds in favour of switching.

The mistake is to think that two choices always means a 50-50 chance. If Manchester United play Accrington Stanley in the Cup then, with the greatest respect to proud Stanley, it's more likely that United will progress to the next round.

Still not convinced? You are in good company. The paradox of the Monty Hall Problem has been incredibly powerful, busting the brains of scientists since 1975.

In 1990 the problem and a solution were published in Parade magazine in the US, generating thousands of furious responses from readers, many with distinguished scientific credentials.

Part of the difficulty was that, as usual, there was fault on both sides as the published solution was arguably unclear in stating its assumptions. Subtly changing the assumptions can change the conclusion, and as a result this topic has attracted sustained interest from mathematicians and riddlers alike.
Even Paul Erdos, an eccentric and brilliant Hungarian mathematician and one-time guest lecturer at Manchester, was taken in.

So what happens on our university's open days? We do a Monty Hall flash mob. The students split into hosts and contestants and pair up. While the hosts set up the game, half the contestants are asked to stick and the other half to switch.

The switchers are normally roughly twice as successful. Last time we had 60 pairs in 30 of which the contestants were always stickers and in the other 30 pairs always switchers:

Among the 30 switcher contestants, the Cadillac was won 18 times out of 30 - a strike rate of 60%
Among the 30 sticker contestants, there were 11 successes out of 30, a strike rate of about 36%
So switching proved to be nearly twice as successful in our rough and ready experiment and I breathed a sigh of relief.

I had calculated beforehand the chances of ending up with egg on my face and the team of 30 stickers beating the 30 switchers. It was a risk worth taking, but one shouldn't play Russian Roulette too often.

etc...

http://www.bbc.co.uk/news/magazine-24045598
 

Zilch5

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The Mythbusters tested this one not too long ago - and conclusively proved that your odds are better if you change your mind.

It's on YouTube somewhere, I'm sure.
 

Anome

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In a way, it really wasn't necessary for them to do the experiment, since the logic and statistics are sound. Switching doubles your chances from 1 in 3 to 2 in 3.

However, by doing the experiment, they illustrated it to people who have a hard time getting round the numbers, some of whom I've had involved arguments with on the subject.
 

Fluttermoth

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I've had loads of arguments with people about it too!

I think humans are particularly bad at probabilities.
 

Ringo

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Fluttermoth said:
I've had loads of arguments with people about it too!

I think humans are particularly bad at probabilities.
And what are tha chances of that!. :D
 

Bad Bungle

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It has only just occurred to me to look on the forum for help with the 'Monty Hall' (3 doors, 2 goats, 1 car) problem I have been mulling over since 1991. Of course back then I had no internet nor did I know what the probability puzzle was called. An article in the Daily Telegraph mentioned Marilyn vos Savant, an American columnist who postulated an answer to the Monty Hall problem in 1990 and was savaged by other Mathematician for being guilty of three unforgiveable sins: she was a woman, she was attractive, she was popular.

"the host (who knows what's behind the doors) always opens a losing door on purpose". Makes sense now.

https://en.wikipedia.org/wiki/Marilyn_vos_Savant
 

Mythopoeika

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It has only just occurred to me to look on the forum for help with the 'Monty Hall' (3 doors, 2 goats, 1 car) problem I have been mulling over since 1991. Of course back then I had no internet nor did I know what the probability puzzle was called. An article in the Daily Telegraph mentioned Marilyn vos Savant, an American columnist who postulated an answer to the Monty Hall problem in 1990 and was savaged by other Mathematician for being guilty of three unforgiveable sins: she was a woman, she was attractive, she was popular.

"the host (who knows what's behind the doors) always opens a losing door on purpose". Makes sense now.

https://en.wikipedia.org/wiki/Marilyn_vos_Savant
Yes, she has the highest IQ on the planet.
 

blessmycottonsocks

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That's a great puzzle! Whilst I can understand the hard logic, it still somehow feels wrong.

Reminds me of another conundrum which was (to me anyway) hard to get your head around.

Three friends go to a café, order a set meal at £5 each and hand over a total of £15.
The meal is poor and they complain to the waiter.
He notifies the manager who says OK knock £5 off their total bill.
The waiter though is dishonest. He pockets £2 and refunds each customer only £1 each.
So, instead of paying £5 each, they only paid £4.
But 4 x 3 = 12. Add to that the £2 the waiter kept and you get £14.
What happened to the other £1?
 

INT21

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Got to admit I can't see how switching improves your chances.

Agreed, the adds are now 1 in two, and no longer 1 in three. But the chances are still 50/50. Either you have picked the right box or you have not. Moving to the only other available box does not change that.
 

Bad Bungle

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That's a great puzzle! Whilst I can understand the hard logic, it still somehow feels wrong.

Reminds me of another conundrum which was (to me anyway) hard to get your head around.

Three friends go to a café, order a set meal at £5 each and hand over a total of £15.
The meal is poor and they complain to the waiter.
He notifies the manager who says OK knock £5 off their total bill.
The waiter though is dishonest. He pockets £2 and refunds each customer only £1 each.
So, instead of paying £5 each, they only paid £4.
But 4 x 3 = 12. Add to that the £2 the waiter kept and you get £14.
What happened to the other £1?
See #79 & #82 in this thread
 

Bad Bungle

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So, does anyone still think the Monty Hall Problem sidesteps the rules of chance? :confused:
Yes, there are specific circumstances to the problem :the Host never opens a door until one has been picked by the contestant, the Host only opens a door not picked by the contestant, the Host only opens a door that has a goat behind it.
 

JamesWhitehead

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What happened to the other £1?
£13 divided by three = £4.33.33 etc. each

The statement that they only paid £4 each is a lie, taking the problem out of pure maths anyway. :jugg:

Change of mind . . .

Edit: Well it is a trick question based on a deception but my version is maybe too sceptical about the voice of the questioner!

See below for the standard solution, copied from the interweb!
 
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INT21

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That's still wrong.

Knock 5 of the total of 15 and you get 10

10 divided by 3 means each person should end up paying 3.333 each., not 4.

They should get back 5/3 = 1.66 each
 

JamesWhitehead

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should end up paying 3.333 each
Edit, scrub my own answer: here is one I copied.

"The problem is that the question is cleverly phrased to conceal what is really going on.
Let’s locate all that money. There are two ways to think about how much money is out there to be found. The way that this question is tricky is that it combines that two ways.
The first way is this. How much money did the three men pay originally?
The second way is this. How much money did they end up paying?
So if it is the first way, then clearly the total we need to account for is the £15.
So let’s see what happens to that £15. The chef gets £10, the waiter gets £2 and the guys get £3 back. That adds up fine.
Now let’s look at the second way. How much money did they end up paying? Well £12, and £10 of it went to the chef and £2 to the waiter. Now, that adds up too.
The problem with the question is that the the £2 that the waiter took is contained in the £12 that they end up paying so we shouldn’t expect them to add to anything meaningful."
 
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INT21

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So it pivots on the customers not knowing that the manager had said charge them £10 ?

Also there is no missing £1
 
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