# Yet another test for your reasoning skills!



## Me Died Blue (Oct 19, 2007)

Brian's good thread made me think of this now-famous statistical question:

You are playing a game show in which there are three closed doors: 1, 2 and 3, exactly _one_ of which has a prize behind it. First you point to one of the doors, but it still remains closed for now, just highlighted. After that selection, the host will open _one_ of the _other_ two doors, which will definitely _not_ have the prize behind it. At that point, you know the prize is either behind the one you first highlighted, or else the one that was neither highlighted by you nor opened by the host. And now you have a choice: You can either stick with the one you initially highlighted, and open only it; or you can change from that one and open only the one that hasn't been highlighted or opened.

The question is, at this point, do you have:

A) A greater statistical chance of winning by staying with your initial unopened door?

B) A greater statistical chance of winning by switching to the other unopened door?

C) An identical, 50/50 statistical chance of winning by opening either one of the two remaining unopened doors?

Enjoy


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## turmeric (Oct 19, 2007)

50/50


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## Barnpreacher (Oct 19, 2007)

The real question is - Are you dressed up like a cowboy or a clown, and is Monty Hall still hosting Let's Make a Deal?


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## Mushroom (Oct 19, 2007)

You started with a 33 1/3 % chance of being right with your first choice, and the other two doors represented a 66 2/3 % chance. When the one door was opened, the 66 2/3 % odds fell to the remaining unselected door, so your original door represents 33 1/3 %, whereas the other 66 2/3 % odds, so your odds are better with that one, right? 

Wrong.

As soon as the other door was eliminated, the 100% of the odds were divided by 2 instead of 3, and now you are left with 50/50 odds for either door.

The answer is C, in my mind, but I'm sure you smarty-pants out there will show me how I'm wrong, so I'm gonna dress as both a cowboy AND a clown and wait for Monty.


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## Puddleglum (Oct 19, 2007)

I'm thinking 50/50 . . . but my gut is telling me that that's too obvious an answer.


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## Jim Johnston (Oct 19, 2007)

I say B, for mostly the same reasons as Brad's initial answer.


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## Richard King (Oct 19, 2007)

fiddy/fiddy as they say in the hood


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## No Longer A Libertine (Oct 19, 2007)

Richard King said:


> fiddy/fiddy as they say in the *hood*


aka Lubbock Texas where gangsta' rap icon Buddy Holly hailed from.


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## Croghanite (Oct 20, 2007)

Originally you had a 33 1/3 % chance of winning your chosen door.
After getting rid of one door, it raises the chance of the two remaining doors being the winning door to 50% each. Add the 33 1/3 % to the 50% to get 83 1/3 % statistical chance of winning by staying with the first choice. 

The answer must be : A) A greater statistical chance of winning by staying with your initial unopened door?


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## Civbert (Oct 20, 2007)

50/50

But I can see how it would be easy to analyze yourself into a complicated but incorrect solution.


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## Jim Johnston (Oct 20, 2007)

Civbert said:


> 50/50
> 
> But I can see how it would be easy to analyze yourself into a complicated but incorrect solution.



Here's an example of how I "analyzed myself into a complicated but incorrect solution."

Three doors = D1, D2, D3, P= prize, NP = no prize.

First "complicated but incorrect solution" considered when you choose door one and stay....

D1 = P; D2, D3, = NP

You chose, D1. You stay, you win.

Next....

D2 = P; D1, D3, = NP

You choose D1, you stay, you lose.

Next....

D3 = P; D1, D2 = NP

You choose D1, you stay, you lose.

Same outcome for any door you start with, you lose 2 out of 3 times.


Second "complicated but incorrect solution" considered when you choose door one and move....

D1 = P; D2, D3, = NP

You chose, D1. You move, you lose.

Next....

D2 = P; D1, D3, = NP

You choose D1, you move, you win.

Next....

D3 = P; D1, D2 = NP

You choose D1, you move, you win.

Same outcome for any door you start with, you win 2 out of 3 times.

Thus I reason B gives the greater statistical chance of winning.


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## panta dokimazete (Oct 20, 2007)

***spoiler alert***

Let's make a deal - Play the game!


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## BrianLanier (Oct 21, 2007)

Paul is correct. 

At first reading (or glance), the correct answer seems to be 50/50, however, once you run some tests, you find out that you win twice as much if you switch than you do if you stay. My buddy and I ran about 50 tests yesterday. (I haven't quite mastered probability formulas yet so I wasn't sure if I was analyising the question correctly or not.) So if you stay your probability of winning is 1/3 and if you switch it is 2/3.


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## Mushroom (Oct 21, 2007)

> Paul is correct.
> 
> At first reading (or glance), the correct answer seems to be 50/50, however, once you run some tests, you find out that you win twice as much if you switch than you do if you stay. My buddy and I ran about 50 tests yesterday. (I haven't quite mastered probability formulas yet so I wasn't sure if I was analyising the question correctly or not.) So if you stay your probability of winning is 1/3 and if you switch it is 2/3.
> __________________


I thought this as well, but did not run any probability tests. (wouldn't know how, sheesh Brian, you are such an egghead ) I changed my position when I realized that the first "choice" made is no choice at all, since that door has not been opened. Regardless of past occurances, at the actual time of decision there are two doors, and neither are open. One has the prize and the other does not. A completely new and unconnected decision must be taken, and either choice has a 50% chance of being right. C - 50/50


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## BrianLanier (Oct 21, 2007)

Brad said:


> > Paul is correct.
> >
> > At first reading (or glance), the correct answer seems to be 50/50, however, once you run some tests, you find out that you win twice as much if you switch than you do if you stay. My buddy and I ran about 50 tests yesterday. (I haven't quite mastered probability formulas yet so I wasn't sure if I was analyising the question correctly or not.) So if you stay your probability of winning is 1/3 and if you switch it is 2/3.
> > __________________
> ...



I'm ok with being an 'egghead', just as long as I don't have egg *on* my head 

So even after Paul's post (and mine), you're still sticking to option C? Perhaps, you should watch out for those eggs!


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## Mushroom (Oct 22, 2007)

> I'm ok with being an 'egghead', just as long as I don't have egg *on* my head
> 
> So even after Paul's post (and mine), you're still sticking to option C? Perhaps, you should watch out for those eggs!


S'OK, Brian, around here we like eggheads (here meaning my house), because we're always wishing we were one. ...There's a grammatical error in there somewhere, I know.

Egg on my face is OK, too. I'm used to it being there, and I hear it's good for the complexion. I guess I'm gonna hafta go read one of those books you recommended on the other thread (where's my ibuprofen?) to understand it, but it just seems that after all is said and done, you still just have 2 doors - a 50/50 chance. So I'll stick with C.

After the elimination of the third door, doesn't the equation change? Isn't it:

D1 = P or NP vs. D2 = P or NP?

In Paul's equations a third door is factored, but at the actual time of decision, there is no third door. If I have a three sided coin, then remove one of the sides, don't I return to the standard 50/50 odds?


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## Mushroom (Oct 22, 2007)

OK, so I read JD's spoiler, and I now see where the odds are not 50/50. Please pass the egg.


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## Me Died Blue (Oct 22, 2007)

The correct answer has been stated by more than one person...but since that doesn't really give much away, I'll wait some more time before "officially" revealing it.

If nothing else, there's more "to" the answer (the _why_ behind it, if you will) than might initially appear.


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## Civbert (Oct 22, 2007)

I was wrong. The odds of winning are 2/3 for the second door. 

Think of it with 10 doors. You pick 1 door. You have a 1/10 chances of winning. There's a 9/10 chance one of the remaining 9 doors win. Now say the host eliminates 8 of the 9 other doors that are losers. Now you have your original choice which still has 1/10 chance, or you can switch to the other door which has a 9/10 chance of winning. Your odds on the first door are the same because you made it when there were 10 doors to pick from. But the host just improved the odds one of the other doors wins by removing 8 losing doors. 

Ask yourself this - did the host increase the odds that you choose the winning door right off just by eliminating 8 losing doors _after _you picked? He can not change the odds that your first choice is right unless he reduces the choices _before _you pick. And the odds of the remaining doors must still be 9/10 no matter how many doors remain. Theres a 10/10 chance that _one _door is a winner no matter how you slice it.


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