A fascinating mathematical problem

Matthew

Out of interest, what was your answer to the Mulder/Scully?

Er, I actually gave my answer in replying to Minky.

Mulder is right because they only agreed 50% of the time when opening doors randomly. If it were pre-programmed one would expect to find more than 50% agreement as explained previously.

Matthew

Matthew

The mathematics on paper makes sense, but real-life experiment is quite different. In the case of choosing random colours, left door or right door, you can assume its 50/50 (50%), but in fact, the percentages would vary over TIME. Indeed, TIME is a component of everything you calculate, and your friend would be rightfully impressed if you talked about the time value within a derivative contract. You can also talk about the importance of using 24 iterations or more for Cox Ross binomial fair value models.

PR,

I have no idea what you are on about but my gut instincts are definiitely saying bollocks. FWIW the Mulder/Scully thing has been proved in "real life experiements" although I have no idea if there was some kind of "Er, like, have you considered TIME" footnote somewhere along the way.

If I mentioned "Cox Ross Binomial fair value models" (whatever that might be) my friend would probably laugh and accuse me of trying to impress girls by using complicated sounding phrases I didn't understand.

Matthew

...Returning to the original problem:

The argument that the contestant should switch seems to be based on the assumption that the host's opening a door does not change the probability of of the car being behind the contestant's door. The easiest way of seeing that this is not the case, IMHO, is by listing the contingencies:

(c = car, g = goat)

contestant's door.. c... g... g
host's door........... g... c... g
remaining door...... g... g... c

Beforehand, each of the outcomes represented by a column is equally likely (with probability 1/3).

However, we know the host does not choose the car, so the middle column is excluded. This leaves us with two equally likely outcomes (with probability 1/2).

Hence, the contestant switching doors makes no difference to their likelihood of being right.

At least, as far as I can see...

[This message was edited by JeremyD on Mon 19 April 2004 at 19:17.]

Matthew

Because you don't understand what I'm saying doesn't make it bollocks. 50/50 or 50% can never exist for any lenghtly period of TIME. Several examples are the below -

Pi = 3.14 blah blah blah blah

As Pi is considered totally random, and has a probably infinite number of decimal points, one would would expect (based on your 50% chance), that 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 would exist in equal numbers when calculating a million decimal places, but in fact they are: -

0 = 99,959
1 = 99,758
2 = 100,026
3 = 100,229
4 = 100,230
5 = 100,359
6 = 99,548
7 = 99,800
8 = 99,985
9 = 100,106

It is more than probably that Minky's real-life answer in the Mulder and Scully case would have been correct if the number of matching colours exceeded 50%. Which is more than possible through random selection? As you mentioned 50% was their finding (unlikely in real-life) you are faced with a language test rather than a mathematical one.

PR

"Because you don't understand what I'm saying doesn't make it bollocks"

No, but the poorly written and confusingly expressed nature of your posts made it seem like it might be bollocks.

"one would would expect (based on your 50% chance), that 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 would exist in equal numbers when calculating a million decimal places"

I don't think anyone would expect them to be exactly equal -- rather one would expect them to be something close to equal depending on exactly how random Pi actually is. Whether this is random enough would depend on the tolerances required for a particular example.

"It is more than probably that Minky's real-life answer in the Mulder and Scully case would have been correct if the number of matching colours exceeded 50%. Which is more than possible through random selection?"

I am a little confused again so I shall re-iterate what I believe to be the correct answer:

-- If Scully is correct and the boxes are pre-programmed you would expect a result of around 5555 matches out of 10,000 iterations.

-- If Mulder is correct and the boxes are truly random then you should expect around 5000 iterations.

The difference between 5555 and 5000 over 10,000 runs I take to be statistically significant. If it isn't then replace 10,000 with 100,000.

Does that answer you question?

"As you mentioned 50% was their finding (unlikely in real-life) you are faced with a language test rather than a mathematical one"

This actual experiment does give an answer of 50% (within experimental and statistical error -- so 50.01%, 90.98%, etc). It looks like Maths to me.

Matthew

Ok, I'll play.

Seems to me the arguments favoring the switch assume something - namely, that "goat = goat", i.e., that goat is one choice. Suppose it's brown goat/grey goat, or billy goat/nanny goat:

Intitial choice | Revealed
Billy | Nanny ->>> Switch wins
Nanny | Billy ->>> Switch wins
Car | Nanny ->>> Stick wins
Car | Billy ->>> Stick wins

And, we're back to 50/50.

Saying that the like items are "degenerate" (in the strictly mathematical sense) works for quantum mechanics; we can't say "this electron" or "that electron", just "electron". Not sure it applies to goats, though... I mean, I'm sure they know the difference.

Suppose it all depends on how ( or how well) the problem is defined.

The goats in my post above are degenerate but still give the same 50/50 outcome because there are two goats in each of the three possible outcomes. Using two distinguishable goats simply gives two versions of each of these outcomes. So the distinguishability of the goats is irrelevant, AFAICT.

quote:

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Originally posted by JeremyD:
The goats in my post above are degenerate but still give the same 50/50 outcome because there are two goats in each of the three possible outcomes. Using two distinguishable goats simply gives two versions of each of these outcomes. So the distinguishability of the goats is irrelevant, AFAICT.

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Yes, I agree; one out of two is as good as two out of four.

The flaw in the switcher gambit (which you and I both address) is that they assume the goats are not identical if I choose goat initially, leading to two "switch wins" scenarios (i.e., this goat or that goat), whereas when I pick car initially, the gambit assumes goat=goat, yielding only a single "stick wins" scenario (see John Channing's explanation below):

quote:

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Stick | Switch
Goat | Goat Car ->>> Switch wins
Car | Goat Goat ->>> Stick wins
Goat | Car Goat ->>> Switch wins

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As outlined above, the middle line should rightly have two possible values, representing "reveal goat 1" and "reveal goat 2". Only one of these is shown, therefore, the switch seems to win.

In other words, if lines 1 and 3 in John's scenario are distinguishable, we must have two distinguishable versions of line #2.

Seems like.

quote:

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As Pi is considered totally random, and has a probably infinite number of decimal points, one would would expect (based on your 50% chance), that 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 would exist in equal numbers when calculating a million decimal places

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No.

You confuse the finite and infinite. If you expect after million "tries" all the digits to be exactly 100,000 , then the 1,000,000th digit wouldn't be random would it ?

In theory (using expectancy etc which involves infinite sums and infinite number of tries) the answer is exactly 0.1 for each digit.

This gives us in theory a 50/50 variable.

In real life, we can get as close as we want to 50/50.

So your claim that no exact 50% generator exists might be interesting but not relevant neither to theory nor real life. In the same vain, I can say you can never measure exactly 1 kilo-gram, so let's not use kilos.

regards,
Omer.

quote:

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Originally posted by Matthew Robinson:
A proof for those who insist on seeing numbers rather than relying on one's intuition:

1. Pick a door (A). Chance of being correct is 1/3

2. Chances of it being one of the other two doors (B, C) is 2/3.

[I assume we all agree to this point]

3. Host eliminates door B which, since it never has the car, has a car probablity of 0.

[This is the bit people get wrong and assume this has a probability of 1/3 -- omitting the fact that the host knows how to pick the one without the car]

4. Probability of it being behind door C is 2/3 - 0 = 2/3. Door A's probability remains unchanged at 1/3.

Matthew

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Wrong. It's not allowed to change the point in time for which you calculate the probability.

1. Pick a door (A). The game show host opens another door to reveal the goat. So, in the light of the information you now have, the real probability of you picking the car with your first pick was 1/2, and not 1/3.

2. Whether you stick or switch, you still have a probability of 1/2.

Markus is right.
You are looking at the probability of a single solution from a variable position. You can only equate the outcome as a probability from a fixed position - so the probability changes as the situation progresses.

To put it another way, when the rules state in advance that one of the three doors will be eliminated by the host, and that this door does not have the car behind it, the real choice only ever is between two doors. The third is just decorum, there to amuse the audience and confuse the candidate and learned heads on internet fora.

It seems to me that all the people who struggle with this simple problem have the benefit of some higher education in Maths or Statistics. People who scraped their A-level like myself don't have nearly the same problem

So, answer me this. If you pick door A and I pick door B *and* door C I win 2/3 of the time yes?

If the host wrote down on a piece of paper which one of my doors that definitely has a goat and I still get to open my two doors I win 2/3 of the time. Yes?

If the host whispers in my ear which one to open first I, with this extra information, will open one door and have a 2/3 success compared to the person who picked door A.

How does the host whispering in my ear affect your chances of winning from a choice you made earlier?

Matthew

PS I think I might write a program to demonstrate the right answer.

Matthew,
having just read your answer more carefully I see you've assumed that the host deliberately open a door that he knows has a goat behind it. This is not stated in the question, so I don't understand why you have assumed it.

Markus,

quote:

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To put it another way, when the rules state in advance that one of the three doors will be eliminated by the host, and that this door does not have the car behind it, the real choice only ever is between two doors. The third is just decorum, there to amuse the audience and confuse the candidate and learned heads on internet fora.

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No, it was correct if you knew in advance what door shall be opened with the goat.

Both closed doors are not symetric because the other door has "survived" a picking of a door with a goat, while the door you chose didn't have to go through such a try. So your chances didn't rise, while the other closed door "inherits" the chances of the door it competed with.

Omer.

To understand it, think of the following :

Same game, only 1,000,000 doors. You pick one, and the host opens 999,998 doors with goats.

Will you still say now that you have a 50% chance of discovering the car ? This means that you are extremely lucky when choosing your door, you achieve a 50% instead of 0.0001%.

regards,
Omer.

If you stick you will win in 1/3 of attempts.
If you switch you will lose in 1/3 of attempts.

It doesn't need to be more complicated than that, does it?

Paul

quote:

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Originally posted by Matthew Robinson:
So, answer me this. If you pick door A and I pick door B *and* door C I win 2/3 of the time yes?

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But you can't. One door will be eliminated. So you can pick door B OR C, but never B AND C.

Let's look at it this way. A, B and C each have a probability of 1/3 in the first round. That means A and B have an equal probability of hiding the car.

The host then takes away option C. I can see absolutely no mechanism why option B should now be assigned a greater probability than A, when they had an equal probability in the first round.

quote:

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PS I think I might write a program to demonstrate the right answer.

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Do that, I'm looking forward to it.

Jeremy -- Most people assume from the question that the host knows what is behind the door as it's implicit in the question that the host always opens a door with a goat behind it. Besides it doesn't say "the host opens a door and in this case it happens to reveal a goat" so you can argue that it doesn't.

But I agree if the host doesn't know where the goat is then there is no advantage in changing.

Matthew

Matthew,

I don't think this matters. Suppose that if the host opens a door and it's a car - the round's cancelled and we try another one. So you can assume the host always opens a door with a goat, his knowledge is I think irrelvant.

Cheers, Omer.

quote:

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Originally posted by Minky:In a game show there are three doors. Behind two of them are goats. Behind the third is a car. You are a contestant trying to win the car. You choose a door and the game show host opens one of the two remaining doors to reveal a goat. Which door is most likely to have the car behind it (should you stick or switch) ?

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Just to check if we're all singing from the same sheet: this is the set-up I was commenting on. The rules are indeed that one door with a goat behind it will be eliminated.

Yet another way of looking at the problem: Does it matter at all if one picks a door in the first round? If we do away with the first round and let the host eliminate a door from the outset, we have two choices (door A and B) with a 50/50 probability of being correct.

With the set-up described above, there is really no difference. Instead of denominating the choice of door A as door A, we give it the moniker of sticking with our choice from round 1. Instead of saying we are choosing door B, we say we switch. But in reality, it is just the description of the process that changes, not the process itself.

I really think that door C is just there to confuse us.

quote:

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With the set-up described above, there is really no difference. Instead of denominating the choice of door A as door A, we give it the moniker of sticking with our choice from round 1. Instead of saying we are choosing door B, we say we switch. But in reality, it is just the description of the process that changes, not the process itself.

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No, the process changes. If first the host eliminates a door, then whatever door you choose underwent a dice-throw where it could have been opened and revealed to have a goat, but wasn't chosen eventually, (maybe because it holds a car ?). This changes the game.

Don already wrote a spreadsheet program that indeed verified the 1/3-2/3 solution is correct, look at the brain teaser thread, the problem is stated in page 2.

Omer.

PS I think I might write a program to demonstrate the right answer.

did it a couple of years ago...you can see the results earlier in this post....it confirms that if you stick you win 1/3 of the time....if you change you win 2/3 of the time. (approx for the benefit of PR)

Omer had a very helpful illustration above with his 1,000,000 doors, again posted in the Brain teaser thread a couple of years back. ie if you pick ONE door and stick with it, your chances of picking the car door are a million to one. Not very good. This means the chances of the car being behind one of the other 999,999 doors is .999,999. Almost a dead cert. All YOU have to do is FORCE the Host to open ALL these 999,999 doors. It doesn't matter whether he opens them all at once!! or one at a time!! Surely we can all agree on this !!!!

Assume the host knows where the car is - he is going to open 999,998 "goat" doors and you know this. He is simply going to keep the car-door until last.....it creates a bit of tension (or confusion or boredom!). "Change your mind" on the last door and your chances of winning are .999,999.

Of course you didn't REALLY change your mind, you simply chose 999,999 doors rather than ONE and you forced the Host to open all 999,999.


Same principle applies to 3 doors.

Cheers

Don

PS now assume the Host didn't know where the car was. Chances of him opening 999,998 doors without revealing a car are pretty slim.........

[This message was edited by Don Atkinson on Mon 19 April 2004 at 21:59.]