Weird Math Puzzle

[AdD 1]

I read this in a book yesterday, still can't quite wrap my mind around it.

There are 3 doors available to choose from. 2 contain nothing, one contains a prize. You choose one of the three doors. Then one of the doors that you didn't choose but also contained nothing is opened and you are given the option to switch you choice of the remaining two doors or to stay with your original guess. What should you do?

(Hint- it's not a 50/50 chance)

Life is ez
On the dz
Every jumper's dream
3 rigs and an airstream

[quade 4]

You have -slightly- better odds by switching doors.

http://math.ucsd.edu/~anistat/chi-an/MonteHallParadox.html

quade -
The World's Most Boring Skydiver

[Lizard 0]

Kick all three doors in and you win!

[CrazyIvan 0]

I don't like doors...I prefer curtains

__________________________________________
Blue Skies and May the Force be with you.

[rehmwa 2]

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You have -slightly- better odds by switching doors.

It's not nearly 'slightly', it's twice as good a chance by switching. 67% vs 33% (remember, your original pick was only a 1 in 3 chance of getting it right - and the door removed is 'always' one of the two losers (ie it's not removed randomly, but by using special knowledge). Since you were likely wrong in the original choice, it's better to switch once one of the 'wrong answers' is removed.

...
Driving is a one dimensional activity - a monkey can do it - being proud of your driving abilities is like being proud of being able to put on pants

[happythoughts 0]

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I don't like doors...I prefer curtains

Pay no attention to the a man behind the curtain. .

[quade 4]

The Monte Hall Paradox

Run some simulations 10,000 times without the "host" knowing the answer and I think you'll find that the answer really is only -slightly- higher.

Now I know what you're going to say, "but doesn't the host know?" and the answer really should be, "no, he can't, because that would be agaisnt FCC rules".

quade -
The World's Most Boring Skydiver

[Miami 0]

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You have -slightly- better odds by switching doors.

It's not nearly 'slightly', it's twice as good a chance by switching. 67% vs 33% (remember, your original pick was only a 1 in 3 chance of getting it right - and the door removed is 'always' one of the two losers (ie it's not removed randomly, but by using special knowledge). Since you were likely wrong in the original choice, it's better to switch once one of the 'wrong answers' is removed.

There would have to be 4 doors and 2 of the wrong ones would have to be removed to make the odds twice as good to switch. With three doors the odds do become better once one wrong door is removed but only by 16.66%. The more doors there are the better the odds are once all but one of the wrong doors and the winning door are removed.

Miami

[fudd 0]

This is just a faulty back refrence.

in 1/3 of the cases you will pick the price door.
In 2/3 of the cases you will pick a losing door.

The theory may seem to be right, but you can't include the 1/3 of the times you will pick the price door in the calculation for the overall probability for picking the price door when choosing between 2 doors instead of 3.

Doesn't matter if you switch or not.

There are only 10 types of people in the world. Those who understand binary, and those who don't.

[labrys 0]

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The theory may seem to be right, but you can't include the 1/3 of the times you will pick the price door in the calculation for the overall probability for picking the price door when choosing between 2 doors instead of 3.

Huh? In the original question, if you assume that the first door was opened randomly and happened to not have a prize behind it, the odds are 50/50 that the prize is behind one of the remaining doors. The odds are not even *slightly better if you switch, they are exactly even. You would, however, have to run the simulator an infinite number of times before you would get that numerical result.

If you assume that the first door was opened with the knowledge that nothing was behind it, and realize that you have a 1 in 3 chance of your original choice being correct, then it is clear that the remaining door has a 2 in 3 chance of being the right choice. Better to switch.

Owned by Remi #?

[labrys 0]

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There would have to be 4 doors and 2 of the wrong ones would have to be removed to make the odds twice as good to switch. With three doors the odds do become better once one wrong door is removed but only by 16.66%. The more doors there are the better the odds are once all but one of the wrong doors and the winning door are removed.

Nope. With 3 doors you do double the odds. In your 4 door example you triple the odds, not double them. You started out with a 25% chance that the door you chose was correct, but by eliminating 2 remaining doors you increased the odds that the 1 remaining door was the "prize" door to 75%

Owned by Remi #?

[AdD 1]

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Huh? In the original question, if you assume that the first door was opened randomly and happened to not have a prize behind it, the odds are 50/50 that the prize is behind one of the remaining doors. The odds are not even *slightly better if you switch, they are exactly even.

Actually this is what I and apparently a lot of other Ph D's in math thought too, but if you construct a flow chart you will see that switching does give you a better chance of winning. The flow chart is the only way I could understand that it isn't 50/50 but it's counter-intuitive.

Life is ez
On the dz
Every jumper's dream
3 rigs and an airstream

[labrys 0]

Can you describe your flow chart?

In my post I agreed 100% that the odds are better if you switch but ONLY if the first door is opened with the KNOWLEDGE that there is nothing behind it.

If you can show me how the odds change when everything is completely random I'd appreciate it. I don't have a PhD in math, only an MS in applied statistics, sorry.

Owned by Remi #?

[SkydivingNurse 0]

You might like a book titled "The Paradox of God and the Science of Omniscience" by Clifford Pickover, lots of thought experiments like that in it.

[AdD 1]

Sure, here's what I remember about it, don't have the book handy. The "host" never opens a door with the prize behind it, regardless of your choice. You'll see this proves that switching yields a 2/3 probability of a win, not 50/50.

Life is ez
On the dz
Every jumper's dream
3 rigs and an airstream

[labrys 0]

As I have said several times now, I agree that if the "host" knows which door to open the odds are better to switch. If everything is random the odds are even

Owned by Remi #?

[Miami 0]

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There would have to be 4 doors and 2 of the wrong ones would have to be removed to make the odds twice as good to switch. With three doors the odds do become better once one wrong door is removed but only by 16.66%. The more doors there are the better the odds are once all but one of the wrong doors and the winning door are removed.

Nope. With 3 doors you do double the odds. In your 4 door example you triple the odds, not double them. You started out with a 25% chance that the door you chose was correct, but by eliminating 2 remaining doors you increased the odds that the 1 remaining door was the "prize" door to 75%

Ahh, I get it now...

Staying with your choice when there were 3 doors makes your chances 33.33% of having the correct choice. Removing one of those doors does not increase the odds of the one you picked, so switching to the remaining door would increase your odds of winning to 66.66%.

Is the simulator in the link quade posted correct then? When going with the not switching option it was still about 50-50.

Miami

[labrys 0]

Yup.... the simulator will be right 50% of the time as long as you chose the option that the host DOES NOT know which door the prize is behind :-)

If the host DOES know which door to open, the odds go up to ~67% that the right door is the one you did not chose

Owned by Remi #?

[Miami 0]

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Yup.... the simulator will be right 50% of the time as long as you chose the option that the host DOES NOT know which door the prize is behind :-)

If the host DOES know which door to open, the odds go up to ~67% that the right door is the one you did not chose

Ok, now I'm beginning to feel like a real retard, cause I'm usually really good at math puzzles like this. How does it affect the odds of what you picked, when you are not going to swap, if the host knows which door is the winner or not. I mean you are picking 1 out of 3 and sticking with it. Regardless of what the host knows you should have a 33.33% chance of being right...right?

Miami

[labrys 0]

Aww.. don't be so hard on yourself. You are 100% right in a way. You are correct in saying that no matter if the host knows which door to open you have a ~33% chance of getting it right if you stick with your original choice. You always have a 1 in 3 chance no matter what.

Look at it this way then: If the host opens a door at random without knowing where the prize is, there is an ~33 percent chance that the host will open the door that hides the prize, right? That makes the whole question pointless because the question stated that the prize was not there. So if the host "randomly" opens the door with nothing behind it ( a 1 in 3 chance) then you have a 50% chance that the door you picked is correct....

But, if you start with the idea that you have a 1 in 3 chance then that means that the other 2 doors combined have a 2 in 3 chance. If you add in the fact that the host KNOWS which door has the prize behind it and open one that is not correct, that means that the single remaining door that you did not chose has a 2 in 3 chance of being the winner.

Make sense?

Owned by Remi #?

[Miami 0]

Ah, I see...wasn't thinking along the lines that winning door was an option to be opened. Thanks for the explanation!

Miami

[rehmwa 2]

Yup, it makes little sense to run the sim when Monte Hall doesn't know. he always knows.

...
Driving is a one dimensional activity - a monkey can do it - being proud of your driving abilities is like being proud of being able to put on pants

[rehmwa 2]

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The Monte Hall Paradox

Run some simulations 10,000 times without the "host" knowing the answer and I think you'll find that the answer really is only -slightly- higher.

Now I know what you're going to say, "but doesn't the host know?" and the answer really should be, "no, he can't, because that would be agaisnt FCC rules".

Just a reply, if it's set up right, as you increase your samples, whatever setting you choose would approach 50/50 or some multiple of a 3rd. there's no slightly here. You can do it directly by running all combos instead of a simulation.

(It's still moot - the host always knows)

But in any case, I ran all 18 variations. So you were subject to natural variation even with the huge sample size. (when toggling the "Monte doesn't know" - I didn't run it, but anyway as below).

1 - if monte shows the winning curtain and you don't get to take it as a choice, then you can consider it a loss. Then switching gives you a 1 in 3 chance of winning, not switching gives a 1 in 3, and 1 in 3 automatically lose, because Monte found it instead - so you have a 1 in 3 chance to win no matter what you do.

2 - if you get to take the curtain that Monte showed, then you get a 2/3 chance of winning by either switching to Monte's curtain or, if it's a dud, taking the third curtain. The only loss is if you chose right the first try.

3 - If you toss out Monte's choices and just not count them as either a win or a loss, then the odds are Exactly 50/50, not slightly higher or lower - per number 1 above. (edit - I'll look closer at this one, something doesn't ring right)

But that's not really the point of the exercise as Monte always knows.....

...
Driving is a one dimensional activity - a monkey can do it - being proud of your driving abilities is like being proud of being able to put on pants

[indyz 1]

This is the problem that my discrete math professor used to point out that "discrete math is tricky, and if you try to use common sense it will bite you in the ass."

[whatever 0]

ha-ha, this is hilarious!
my dynamics lecturer told us this one in my final year at university.... totally unrelated to my engineering studies, but as a fun logic puzzle... he had lots like that

so there was much arguing and trying to come to a consensus among the people at that university about the odds

me and my friend disagreed too, so we ended up resorting to writing a VB program to simulate the problem and ran upwards of 10000 repetitions each time... (quick and dirty solution, just what we mechanical engineers like)

and the answer was, like stated here by several people that by switching you double your odds from 1/3 to 2/3....

knowing that, this seemed to be the best way to convince others about this result:

-your odds of picking the right one first time are 1/3, if you don't change this stays your odds. now, the host will always open a door to an empty room, and you will never choose the empty room, so by switching, you effectively pick both of the doors you did not pick originally, making your odds 2/3.

and if that failed we showed them the VB code and results....


another cool one he had was about the three guys who get 3 hotel rooms but pay together, the hotelier does not have the right change and sends the bellboy up later with change

(it's not really a stats problem, though)

soon to be gone