Lets see how smart you guys are...

[Heavy Z]

There is one more possiblity. Pick A, shown C and swap to B losing the prize.

 

Wouldn't that make it 2 for 2?

 

I noticed that as well, omitting this possibility will skew your conclusions for sure.

 

No matter how smart Ms VosSavant is, it doesn't change the fact that there is a donkey behind one door, and a car behind another. You have a 50/50 chance, regardless of whether you switch or not.

 

No matter how I think it through from the other side, this is where I also wind up.

 

Good luck sorting things out.

[Pop N Wood]

Look at it this way. If I showed you two doors, told you there was a gift behind one door and nothing behind the other, then told you which door had nothing behind it, what do you think your odds of picking the right door are?

 

With only one choice left you will pick the door with the prize every time.

 

Now start with three doors. Put one door in one group and two doors in the other group. The "group" with 1 door has only a 1 in 3 chance of winning. The group with two doors has a 2 in 3 chance of winning.

 

But picking the group with two doors won't do you any good if you have to randomly pick one of those two doors. But if you pick the group with two doors, then have someone throw out the bad door, the odds of getting the prize are the odds of the prize being in the second group. 2 out of 3.

 

Think of the initial pick as elimnating one of the doors with a 1 in 3 chance of throwing out the good one.

 

The whole trick is that the person throwing out the bad door is not doing so randomly. They are picking based upon your initial pick and their perfect knowledge of the remaining two doors.

[Kevin Shasteen]

.., If no interaction is made with the experiment nothing can be said about where the car is, so it's behind all the doors and so is the donkey..,

 

Ahh, I was wondering when this issue would arise. All the conspiracy nuts salute you-myself included.

 

The ideas on the stats given, prior to your statement, never assumed the game was rigged. So if the game is rigged then the stats are non-statistical and all odds are a wash.

 

Sure the idea of the game being rigged is crazy-but it would not be the first time a tv game show was rigged: therefore the idea that the game is rigged has to be a legitimate possibility.

 

There is only so far logistics can take you-at some point in time the human factor of emotion has to "Make a Decision"...conspiracy or no conspiracy eventually you will have to make a choice.

 

I agree in that if the game is not rigged then anytime you can remove one of the odds your chances of being right will improve.

 

I never took stats-perhaps I should've becasue I always enjoy watching a thought process being worked out.

 

If I had taken statistics the odds are high in that I never would have used them in my choice of occupation : so no harm no foul.

 

2bornot2b? I think that is the real question.

 

Kevin,

(Yea,Still an Inliner)

[RPMS]

Now start with three doors. Put one door in one group and two doors in the other group. The "group" with 1 door has only a 1 in 3 chance of winning. The group with two doors has a 2 in 3 chance of winning.

 

This is where your argument goes awry. They only show you the empty door if you DO NOT CHOOSE IT in the first round. If you do choose it, you're out and have no further chance of participation. Look at it this way:

 

Round One:

Assume: Door A=empty, Door B=Donkey, Door C=Prize

 

If you pick door A, you lose.

If you pick door B, you lose (assuming you don't have need of a donkey)

If you pick door C, you win.

 

One in three chance.

 

Round Two:

Same assumption, nothing moves. Door A=empty, Door B= Donkey, Door C=Prize.

 

You know A is empty, so you won't choose that one EVER. It's removed from the equation (which is what Ms Savant neglected to do, by the way).

You're left with two doors, two possible choices.

If you pick door B, you lose.

If you pick door C, you win.

 

2 possible choices, 1 possible win. 50/50 chance.

 

Okay, let's assume you switch, just to see if your odds improve:

If you originally picked B and switch to C, you win.

If you originally picked C and switched to B, you lose.

 

2 possible choices, 1 possible win. 50/50 chance.

 

Sorry, guys. You've been duped by bad statistics. In the words of someone who's name I can't recall, "Statistics is like a woman in a short skirt. What is exposed is interesting, but what is concealed is critical."

[Guest bastaad525]

I never understand the way any of this stuff works... Odds?? what odds???

 

 

It's this simple.... no matter what you do, the car is gonna be behind whatever door it's behind, which is predertimined well before you ever get the chance to make your guess. The human factor is the ONLY factor... which door are you going to choose? You guys make it sound like the car mysteriously moves around between guesses... well if I pick this one first, and then change my first guess w/o even knowing if my first guess was right, the ODDS go up... like picking them in some arcane mystical order is going to make whatever you finally end up picking the right one. It's not odds... it's just luck, or random, or chaos, or whatever you want to call it. It is the human factor... you have an equal chance of winning or losing really... you either pick right or pick wrong. Now of course... the very choice in itself might be determined by looking at your entire life and every experience you've ever had affecting that moment, as it effects every moment... or you could just be picking at complete random.

 

I know to the educated person I may sound like a 'tard but really... all those 'odds' dont mean anything if you end up picking the wrong door anyways.

[Pop N Wood]

Now start with three doors. Put one door in one group and two doors in the other group. The "group" with 1 door has only a 1 in 3 chance of winning. The group with two doors has a 2 in 3 chance of winning.

 

This is where your argument goes awry. They only show you the empty door if you DO NOT CHOOSE IT in the first round. If you do choose it' date=' you're out and have no further chance of participation. [/b']

 

Your not reading the problem statement. If you are trying to say they don't always show you what is behind one of the doors, then you are asking a different question. If you add the restriction that they will only show you an empty door if you pick the right door to begin with, then the best you should never switch and your odds are back to 1 out of 3. But if you start with the idea that they always show you what is behind one of the doors, then the answer is without question to switch choices.

 

This is why I disliked statistics and enumeration. Sometimes the difficulty is in approaching the problem from the right direction. And when you get done, you are always left wondering if you did it right.

 

But like I said, this problem is a bit of a classic in graduate level engineering courses. I heard the answer years ago, but to be honest I didn't really understand it until I started playing with it on this thread. In fact, I wrote a computer simulation to do a series of random draws (called a Monte Carlo simulation) to try and verify the answer. (I know, get a life) It was only through working out the code logic that it became obvious to me there are only two possible outcomes.

 

BTW, never heard your short skirt analogy in any of my random processes courses.

[RPMS]

Nope - you're still looking at two doors, behind one of which is a prize. No matter how you slice it, you have a 50/50.

 

I think what's happening in your simulations is that you're counting the empty door as a possible choice, the way Eric did in page 2 of this thread.

 

(A' date='B,C: you pick A, are shown B, swap to C to lose the prize.

A,B,C you pick B, are shown C, swap to A to win the prize

A,B,C: you pick C, are shown B, swap to A to win the prize

[/quote']

 

You can't pick C to be shown B, because C was empty in the first place.

 

Even if what you're saying is true and they manipulate the scenario so you can't choose an empty door in the first round, you're still left with two doors, behind one of which is a prize. Ignore all the fluff, and concentrate on this fact. No matter which you choose or whether you switch, you have 2 doors and one hidden prize.

 

If you're going to tell me why I'm wrong in this, I would greatly appreciate it if you'd tell me WHY I'm wrong. I'm not saying this to be pigheaded, I would simply like to know where I'm going wrong.

 

Thanks,

[Pop N Wood]

Well like I said, I don't like the enumeration approach. It is confusing. But I thought the idea of separating the 3 doors into two groups explained it about as well as I know how. Definitely better than the explanation I got when I first heard about this years ago.

 

For what it is worth, I asked this question of a PhD friend of mine, and he practically called me a moron for saying it wasn’t 50-50. After I worked on him and he thought about it awhile he agreed it is 2/3 –1/3. He was still pissed that I annoyed him with it in the first place.

 

This is all fun and games, but try doing something like this for real. I always try to get as many coworkers as possible to agree with me. That way if I am wrong I at least have allies.

[utvolman99]

I can't belive that some of you guys are still arguning that it's a 50 50 chance. If an empty door is opened for you, you would ALWAYS switch. When you switch you have a 2/3 chance.

 

I really dont know how else to explane it to make it any more clear. I can tell you thought that I am certified to teach Six Sigma statstical analysis tools and am also a mechanical engineer working as a Six Sigma Black Belt. My company is considered to be in the top three Six Sigma programs nationally.

 

YOU ALWAYS SWITCH!

[280Zone]

Utvolman shame shame : )

 

belive should be believe

arguning should be arguing

explane should be explain

thought should be though

 

I ain't the best speller either but if you are trying to prove your credibility....

[Guest Thurem]

utvollman

If the rule is, you will be shown an empty door and not a car or donkey door.

You're assuming that you would never choose the empty door first, and be shown that empty door and loose right off the bat. When in fact 1/3 of the time you would start out loosing. In your calculation you're adding this possibility to your odds of winning when you switch doors.

So in reality 1/3 of the time you would pick the car, 1/3 the donkey. Whether you switch or go eni meni mini mo is not gonna affect the odds.

Thure

[wheelman]

RPMS,

The mistake you're making is assuming it's against the rules to choose the empty door. When you make your first choice the odds are 1/3 because you can choose from any of the 3 with no knowledge about any of them. Once the first choice is made and a door has been revealed you now have the oppportunity to choose 1 of the 3 doors again but unless the revealed door was the car you won't choose it so that eliminates one of the 3 leaving 2, the one you chose and the 3rd door. If you don't change the choice your odds remain 1/3 because thats what they were when the door was chosen, if you change to the 3rd you are really choosing 2 doors instead of 3 because you chose not to pick the revealed door and chose to pick the 3rd, thus making a choice which includes 2 doors making the odds 2/3. We all tend to make the assumption that the revealed door can't be picked but the rules didn't say that, you could pick it if you wanted.

 

Kevin,

I wasn't trying to postulate a conspiracy theory but introduce quantum theory into the discussion, but nobody bit on it. I guess it either went right past everyone or it was just to esoteric, it also doesn't help to clear-up the confusion. Anyway no conspiracy theory intended.

 

Pop N Wood,

Your explanation was pretty good, I'm not sure mine was any better although we're effectively saying the same thing. These types of questions generate a lot of confusion because they're counter-intuitive, everyone wants to exclude the revealed door from the second choice because we know what's behind it and it's not what we want even though it's a valid option.

 

Wheelman

[Kevin Shasteen]

.., Kevin, I wasn't trying to postulate a conspiracy theory but introduce quantum theory into the discussion, but nobody bit on it.., Wheelman

 

Sorry, your phrase "If there is no interaction" was ambiguous-besides I never took quantum theory-so even if it was specific it probably would have gone right over my head anyway

 

Just curious-since it has been probably 25 years since I last watched Lets Make A Deal...did anyone ever get the car on the first guess? Because I dont really remember anyone ever getting the "Car" on the first guess.

 

If I am remembering correctly-which I probably am not; would this not suggest some hanky panky going on behind the curtains insuring that the first guess, the majority of the time, would not be the correct guess?

 

Just wondering since its been so long ago that I last watched that show: did anyone ever guess the correct door the first time...and how often was the correct door actually chosen over the span of the lifetime of the show?

 

Kevin,

(Yea,Still an Inliner)

[Guest Thurem]

RPMS,

The mistake you're making is assuming it's against the rules to choose the empty door. When you make your first choice the odds are 1/3 because you can choose from any of the 3 with no knowledge about any of them. Once the first choice is made and a door has been revealed you now have the oppportunity to choose 1 of the 3 doors again but unless the revealed door was the car you won't choose it so that eliminates one of the 3 leaving 2, the one you chose and the 3rd door. If you don't change the choice your odds remain 1/3 because thats what they were when the door was chosen, if you change to the 3rd you are really choosing 2 doors instead of 3 because you chose not to pick the revealed door and chose to pick the 3rd, thus making a choice which includes 2 doors making the odds 2/3. We all tend to make the assumption that the revealed door can't be picked but the rules didn't say that, you could pick it if you wanted.

That's obviously BS. Each of the remaining 2 doors each have 1/3 out of 2/3 of the odds of winning. Whether you change you mind or not. Making it a 50 50 chance...

Thure

[utvolman99]

utvollman

If the rule is' date=' you will be shown an empty door and not a car or donkey door.

You're assuming that you would never choose the empty door first, and be shown that empty door and loose right off the bat. When in fact 1/3 of the time you would start out loosing. In your calculation you're adding this possibility to your odds of winning when you switch doors.

So in reality 1/3 of the time you would pick the car, 1/3 the donkey. Whether you switch or go eni meni mini mo is not gonna affect the odds.

Thure[/quote']

 

Actually, no... You need to go back and read my original question. I didn't ask, should you switch if you have a chance to pick 1000 times. I said you pick a door and an empty one is opened. We are talking about that one situation. In order to prove the statistics you can play the odds out over multiple iterations but you must keep the same parameters. You pick a door and an empty door is opened.

 

People really get thrown off by the donkey with this one. An easy way to look at it is that the door with the donkey can also be considered to by empty also. The donkey is only there to throw you off. I guess it works?

[utvolman99]

Utvolman shame shame : )

 

belive should be believe

arguning should be arguing

explane should be explain

thought should be though

 

I ain't the best speller either but if you are trying to prove your credibility....

 

Well' date=' I hate to admit it but I was in a hurry and didn't have time for spell check. I am dyslexic and have a real problem with spelling.

 

SORRY

[utvolman99]

That's obviously BS. Each of the remaining 2 doors each have 1/3 out of 2/3 of the odds of winning. Whether you change you mind or not. Making it a 50 50 chance...

Thure

 

Ummm...

 

I suppose thats new math. Like I said above you must take the question as it is written. For the one time you actually have to make the decision you picked a door that is not empty and the empty one was opened. So therefore when calculating the odds you must assume that you never start off with the empty door. Because of the way the question is stated this is a valid assumption. One easier way to look at this is to just throw the donkey out all together. Look at it as two empty doors and one car. This would mean that the only way you would ever lose if you always switched would be if you correctly picked the car from the beginning.

 

Below are two columns of randomly generated numbers ranging between 1 and 3 (Minitab generation). The first column is the door that you initially pick. The second column is the door the car is behind.

 

Pick Car

1 3 -On this pick you would pick door 1 and door 2 would be

opened. If you switched you would win

1 2 -Door 3 would be opened if you switch and win

3 2 -Swithch and win

3 3 -Switch and lose

2 1 -Win

3 2 -Win

2 3 -Win

1 1 -Lose

3 1 -Win

1 1 -Lose

2 1 -W

2 2 -L

3 2 -W

2 1 -W

3 1 -W

3 2 -W

3 3 -L

2 1 -W

2 1 -W

2 2 -L

1 3 -W

2 2 -L

3 1 -W

1 2 -W

3 3 -L

 

As you can see from the numbers above if you switch you will win 17 times out of 25 or aprox 2/3 of the time.

[Guest 502ZX]

Maybe I a little cracked but here goes....

 

Assuming based on the story two things:

 

1. The first door opened is always empty

 

2. One of the ramaining two doors has the "big prize"

 

By the first door always being revealed as being empty that would mean regardless of the original choice you make you are in the group of two doors that have a 2/3's chance of winning. By not changing your selection at this phase of the game you are still making a choice! You will still have a selection within the 2/3's odds grouping, which one it is will have the same odds.

 

Hey, whats that I smell? Does anyone else smell red herring?

[utvolman99]

Maybe I a little cracked but here goes....

 

Assuming based on the story two things:

 

1. The first door opened is always empty

 

2. One of the ramaining two doors has the "big prize"

 

By the first door always being revealed as being empty that would mean regardless of the original choice you make you are in the group of two doors that have a 2/3's chance of winning. By not changing your selection at this phase of the game you are still making a choice! You will still have a selection within the 2/3's odds grouping' date=' which one it is will have the same odds.

 

Hey, whats that I smell? Does anyone else smell red herring?[/quote']

 

Like I said above you guys are making this too hard. Please see my model above. It clearly shows the math behind the always switch theory.

[utvolman99]

I did a search on Yahoo using the words donkey, door and switch. I figured that I would be able find something out there. Sure enough here is a link to the very same problem. Only difference is that the door with the donkey is opened in this one as opposed to the empty door. Here is the link for all you nay sayers.

 

http://www.metamath.com/webstat/b2.rules/dealdiag.html

 

I suppose though that someone will reply and say once again that its a fifty fifty chance!