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Whatchya Gonna Do (2 goats and a car)

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Jennifer was selected to be on the popular TV game show "Whatchya Gonna Do?". As she jumped up and down in excitement, the host, Monty Barker, showed her three doors.

"Now, Jennifer, behind one of these doors I have personally placed a brand new Jaguar XJS"

"Eeeeeeee!", Jennifer squealed in delight.

"But", continued MB, "behind the other two doors, there are goats! Select a door and the prize behind it is yours. Which shall it be: Door #1?".

"Oh-oh-oh-oh", Jenny jumped.

"Door #2?"

"Ah-ah-ah-ah"

or "Door #3?"

"Um-um-um-um....well...ah...okay - TWO! I choose Door #2". The audience cheers.

"Door #2. Okay Jennifer", says the host, "but before we show you what's behind Door #2, let me tell you that I am glad you did not pick Door #1, because behind Door #1 is..." The first door swings open to reveal a shabby looking goat. Jenny bounces around the stage in joy because she was thinking of picking that door.

"Ah, but the name of the game is 'Watchya Gonna Do?'. You picked Door #2, but there is still Door #3. I'm going to offer you a choice: either you can keep the door you picked, Door #2, and if the car is behind it, I'll throw in a dinner for two at the Hoi Poloi Restaurant..." The audience gives a collective "oooooo" (who wouldn't want to eat at the exclusive Hoi Poloi?)

"... or you can trade it for what's behind Door #3 - AND if it is the car, I'll throw in a year's supply of Platypus Wax! Tell us about the Platypus Wax, Joe"

Joe's voiceover proceeds to tell us all about the wonders of using Platypus Wax on a Jaguar. As he drones on, Jennifer is nervously trying to decide. Until, MB spins around, points to her, and says (with the audience chorusing behind him)

"What - Chya - Gon - Na - Do?"

Should Jennifer stay with her original choice (Door #2), switch to Door #3, or does it make any difference?

Assume the dinner at the High Poloi and the Platypus Wax are worth about the same dollar amount.

Aaaaaaand, before any of you wise-acres start, Jennifer lives in Toronto - the Jaguar is of far more use to her than the goat.

Jennifer should switch. Contrary to what may seem intuitive, switching actually doubles her chances of winning the car.

This problem is just a re-wording of what is known as the Monty Hall Problem. The key to understanding it is that the host knows the locations of the car and goats. His knowledge changes his actions and thus affects the odds.

Here is a breakdown of all the possible scenarios that Jennifer faces and why Jennifer should switch:

  • Door #2 has goat B (probability 1:3) - MB shows goat A behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances (1:3 x 1:1 x 1:1 = 1:3)[/*:m:6a353]
  • Door #2 has the car (probability 1:3) - MB shows goat A behind Door #1 (1:2) - goat B is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)[/*:m:6a353]
  • Door #2 has the car (probability 1:3) - MB shows goat B behind Door #1 (1:2) - goat A is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)[/*:m:6a353]

There are (1:3 + 1:3 = 2:3) chances that switching will get Jennifer the car, and only (1:6 + 1:6 = 1:3) chances she would get the car by not switching. She should switch.

A more general presentation of the reasoning is this:

At the start of the game, there is a 2:3 chance that Jennifer will pick a door with a goat behind it. If she does, the host will reveal the other other goat, and switching doors will get Jennifer the car.

There is a 1:3 chance she will pick the car. The host will then reveal a goat. Switching would win Jennifer a good supply of Ch?vre (and the disdain of her neighbours).

So, 2 out of 3 times switching gets the car. Simple - unintuitive, but simple.

Why does the host's knowledge change the odds. Because he does not randomly select a door to open - he always opens a door with a goat. By doing this he reduced the possible scenarios for Jennifer to the four listed above. If he randomly picked, then Jenny's chances, if the show progressed as presented, would be 50/50. However, there would also be a 1:3 chance that MB would open the wrong door and reveal the car's location (followed by a 1:1 chance that MB would be sacked and re-runs of McGyver would fill out the remainder of the season!)

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If you know the host has followed a certain algorithm to choose a door to open, you should swap doors to gain a 2/3 chance of winning. If you know the host has not followed any algorithm, it doesn't matter if you swap or not; it's just 50/50. Explanation follows.

I know this is an old thread, but I just saw the movie 21 and this problem comes up in it. In the movie, switching doors was said to be the right thing to do (it's supposed to be "simple math"). But I have thought about it and I think that it's not just a matter of simple math - you also need to know that the host employed a certain algorithm for choosing a door to open. The algorithm is obvious: if the contestant picked a goat first off, open the door with the other goat, but if they picked the car first off, pick one of the goat doors at random to open. This algorithm is so obvious that most people assume the host followed it. But it could be that host was just picking a door at random and it happened not to be the door you chose and it happened not to be a goat.

I have come up with a few proofs (for the proposition "you need to know whether the host is following an algorithm"), but I won't put them up. I want to know what other people think first.

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If you know the host has followed a certain algorithm to choose a door to open, you should swap doors to gain a 2/3 chance of winning. If you know the host has not followed any algorithm, it doesn't matter if you swap or not; it's just 50/50. Explanation follows.

I know this is an old thread, but I just saw the movie 21 and this problem comes up in it. In the movie, switching doors was said to be the right thing to do (it's supposed to be "simple math"). But I have thought about it and I think that it's not just a matter of simple math - you also need to know that the host employed a certain algorithm for choosing a door to open. The algorithm is obvious: if the contestant picked a goat first off, open the door with the other goat, but if they picked the car first off, pick one of the goat doors at random to open. This algorithm is so obvious that most people assume the host followed it. But it could be that host was just picking a door at random and it happened not to be the door you chose and it happened not to be a goat.

I have come up with a few proofs (for the proposition "you need to know whether the host is following an algorithm"), but I won't put them up. I want to know what other people think first.

If the host chooses a door to reveal at random, you actually only have a 1/3 chance of switching and winning. This is because the host has a 1/2 chance of opening the door with the car behind it, ending the game right there with you not winning the car. So switching and winning requires that the host reveals the goat (so that you get to switch at all) and that you switch and choose the door with the car behind it.

P(switching and winning) = P(host revealing goat)*P(switching to door with car)

We've established that you have a 2/3 chance of switching and getting the car given that the host reveals a goat. Since the host is choosing randomly between the two doors you didn't choose, the host has 1/2 chance of revealing the car and 1/2 chance of revealing the goat. So P(switching and winning)= 1/2*2/3 = 1/3.

EDIT: spelling

Edited by flowstoneknight
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A correction to my original post: the last sentence of the second paragraph ends "... happened not to be a goat" but I meant "... happened to be a goat".

Flowstoneknight, I'm glad that you agree that it makes a difference whether the host uses an algorithm or not. But I can't agree that if he doesn't follow any algorithm and you swap doors, you only get a 1/3 chance of winning the car. I also don't agree that if the host picks randomly from the two doors you didn't pick, that he has a 1/2 chance of picking the car. He would have a 1/3 since every door has a 1/3 chance of being the car - maybe that's where you went wrong.

But that's getting away from my original point a bit. My point was that the host only changes the problem if he reveals some of his knowledge of what's behind the doors. When he opens that door and reveals a goat, you gain two pieces of information. Firstly, that the car must be behind either the door you picked or the other unopened door - that's obvious. And secondly, you get to see the output of an algorithm that the host has executed, which takes as input the door you chose, and (crucially) the hosts knowledge of what's behind the doors. As output, it produces another door: the door he opened. This output gives you some knowledge of what's behind the two unpoened doors.

But we only get that second piece of information if it is true that the host has actually executed such an algorithm, and has actually used his knowledge of what's behind the doors. If the host is not using his knowledge of what's behind the doors, then you the contestant cannot gain any knowledge of what's behind the doors. Remember that the doors are totally sealed and the only one who knows what's behind them is the host - if he hasn't told anyone what's behind them, or revealed any hints (like the output of an algorithm), then the probability that the car is behind a particular door is just one divided by the number of doors left unopened: that's 1/2.

This analogous problem (with one crucial difference) should make it clear: Three friends are walking through a tunnel when they come to a junction. There are three doors to choose from and a plaque that reads "One of these doors leads to safety and the others to death. You may open only one door.". (They can't just go back the way they came because of a rock slide has blocked the tunnel.) They start trying to figure out which door to open. The first friend has a hunch that door #1 leads to safety and tells the others he thinks they should open door #1. The second friend thinks door #2 leads to safety, but he keeps it to himself. Meanwhile, the third friend notices a crack in the third door, and peering through sees certain death on the other side - he tells the others what he has seen. They sit and think for a while, when the first friend jumps up and says "when i picked door #1, I only had a 1/3 chance of being right; there was a 2/3 chance that safety was actually behind door #2 or #3. Now it has been shown that safety is not behind door #3, so the 2/3 chance of safety being behind #2 or #3 has turned into a 2/3 chance that safety is behind door #2. Since I first guessed door #1, I am wise to swap to door #2". The second friend interjects: "But I first guessed door #2, so surely I am wise to swap to door #1!". The third friend says "We can only open one door, what shall we do?". The friends argue for ages but eventually decide that they really have no way to decide which door to open, door #1 or door #2; their initial guesses seem to count for nothing. Finally, they realise that all they can do is to pick a door at random.

In this problem, there is no host with knowledge of what's behind the doors, so no information can be passed on about any door which has not actually been revealed. So the chance of safety is spread evenly over the unopened/unrevealed doors.

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im not sure mathmaticlly but after seeing the the summer blockbuster movie 21 this question was in the movie so from that im gonna stick with my door thing.

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you schould make a new choice first you has a 1/3 chance of getting it right. if you eliminate one that your chances increase it is the other door because the old choice was a chance of 1/3 and the other door has a chance of 1/2. so this means the chance that you door is wrong is 5/6

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Posted · Report post

You're not even in the ballpark, beckey. It might help if you read through this thread.

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Well the correct answer is

when the host offers you 2 change the doors you should change them

suppose that there are 3 doors A,B,C

when u had initial chance u had 33.3% each

now suppose u selected A

so u had 33.3% chance of winning

now when the host opens a door suppose C

and it appears a Goat

so u r left with doors A and B

now u selected A and had a 33.3% chance of winning and 66.6% chance of losing

ie. B and C have a 66.7% chance of winning

now u know that c is out of game

which leaves u with a 66.7% chance of winning if u select B

so itz a better option to go for change when the host provides u with that.

so if u initially selected door A then go for B

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THIS RIDDLE IS STOLEN FROM THE MOVIE "21." WORD FOR WORD.

NICE WAY TO BE ORIGINAL.

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Word for word, eh? 21 came out this year. This thread began in 2007. Maybe the OP should be getting some royalties. Did you bother reading the rules? One of them is "don't be a jerk". You should work on that.

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Ok...your resolution involves 3 doors and knowing what is behind one of them already....however...once you know what's behind one of them...the problem doesn't involve 3 doors...in your case...you are still involving the 3rd door in the ratio factor. However...after the point of knowing what is behind the 1st door...you only have 2 choices...so now there is a NEW equation...one that states...."You have 2 doors...behind one is a car...and behind the other is a goat"...there can be no other influences at this point....whether or not there was a third door and an additional goat has no bearing anymore. Sorry...your resolution is technically incorrect...but theoretically correct in the sense of including all 3 doors as a factor in a ratio sense.

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your resolution is technically incorrect...but theoretically correct.

Hi HB,

Can you share with us the difference between technical and theoretical correctness?

Thanks.

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Posted · Report post

isnt this explained in the movie 21?

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Here is the simplest way to put this.

Jennifer has a 1/3 chance of choosing the car

Monty has a 2/3 chance of being left with the car.

Eliminating one of the doors is just for show and a false sense of confidence.

You always switch because Monty always holds the 2 out of 3 chance versus Jennifer's 1 out of 3

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A lot of people don't believe the solution until they've seen it in action. I've made a spreadsheet to make it easy to test this problem. Unfortunately, you can't upload xls files to this forum, but rest assured it's very convincing once you've gone through it.

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A lot of people don't believe the solution until they've seen it in action. I've made a spreadsheet to make it easy to test this problem. Unfortunately, you can't upload xls files to this forum, but rest assured it's very convincing once you've gone through it.

Check this out:

http://math.ucsd.edu/~crypto/Monty/monty.html

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Nearly everyone who has posted on this problem is incorrect. Most people are making the mistake of thinking that since the probability of a correct choice on the first try is 1/3, that it remains 1/3 when one door is opened. When one door is opened, you have new information. You now know that the opened door was an incorrect choice. That new information changes the probability that your initial choice was correct from 1/3 to 1/2. It also changes the probability that the unchosen unopened door is the correct choice to 1/2. Again, the correct answer is that you cannot improve your chances by switching nor by keeping your original choice.

Actually, let's enumerate the cases for the three door problem to show that the probability is 1/2.

1) You select door A. Door A is the correct door. The host opens door B. You stick with your initial choice. You win.

2) You select door A. Door A is the correct door. The host opens door B. You switch to door C. You lose.

3) You select door A. Door A is the correct door. The host opens door C. You stick with your initial choice. You win.

4) You select door A. Door A is the correct door. The host opens door C. You switch to door B. You lose.

5) You select door A. Door B is the correct door. The host opens door C. You stick with your initial choice. You lose.

6) You select door A. Door B is the correct door. The host opens door C. You switch to door B. You win.

7) You select door A. Door C is the correct door. The host opens door B. You stick with your initial choice. You lose.

8) You select door A. Door C is the correct door. The host opens door B. You switch to door C. You win.

Since "door A" is arbitrary, the cases are repeated when you choose any other door initially. So there are 8 cases. If you keep your original choice, you win in two cases and you lose in two others. If you switch, you win in two cases and you lose in two others.

I think most people miss the point that when you initially choose the correct door, the host has two other doors that he can open, while if you choose an incorrect door, he only has one other door he can open. This is why there are 8 cases per initial door choice and not 6 or 12 as most people enumerate them.

If you disagree with this, please provide an explanation of what you think is incorrect in my analysis.

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Nearly everyone who has posted on this problem is incorrect. Most people are making the mistake of thinking that since the probability of a correct choice on the first try is 1/3, that it remains 1/3 when one door is opened.

It does remain 1/3.

"Now, Jennifer, behind one of these doors I have personally placed a brand new Jaguar XJS"

Since Monty knows where the Jaguar is, he wouldn't open the door showing the Jaguar, would he? Think about it. He offers a contestant a chance of winning a Jaguar by allowing her to choose a door. He then randomly opens one of the other doors and there is a Jaguar behind it? Now what? Does he offer her a chance to switch? Of course that's not how game shows work. He only opens a door which will reveal a goat.

But you already understand this, because you wrote the following:

"I think most people miss the point that when you initially choose the correct door, the host has two other doors that he can open, while if you choose an incorrect door, he only has one other door he can open. "

No one missed that point, by the way.

Since we already know that there is at least one goat behind one of the remaining doors and a door with the Jaguar behind it would not be revealed, the probability of her choice being the winning selection remains 1/3. We already know in advance that no matter what door she picks, Monty can choose to reveal a goat behind one of the other doors. His doing it does not make her choice any better or worse. Exposing information that is already known does not affect probabilities. However, the probability of the remaining door being the winner is 2/3 and the rationale for this has been explained many times.

Some web pages that discuss the solution:

http://en.wikipedia.org/wiki/Monty_Hall_problem

http://montyhallproblem.com/

http://mathforum.org/dr.math/faq/faq.monty.hall.html

Did you bother clicking on the link Scraff provided in the post immediately before yours? You can experiment with switching every time and see for yourself that it is the correct choice 2/3 of the time.

http://math.ucsd.edu/~crypto/Monty/monty.html

We can simplify the possibilities thusly:

Suppose beforehand you decide you will initially pick door #1 and make the switch when offered. The following are the three possible scenarios:

Door #1 has a car and the others have goats. You choose door #1. The host will show you what's behind either door #2 or door #3. You make the switch. You lose.

Door #2 has a car and the others have goats. You choose door #1. The host will show you a goat behind door #3. You make the switch. You win.

Door #3 has a car and the others have goats. You choose door #1. The host will show you a goat behind door #2. You make the switch. You win.

By switching you have a 2/3 probability of winning.

This is a simple way of looking at the problem which may help a lot.

You pick a door. The probability that it is the correct choice is 1/3 and the probability that the Jaguar is behind one of the other two is 2/3. If given the choice to switch your one door for the other two, would you do it? This is essentially the choice Monty has given the contestant, only he revealed a goat behind one of those doors first (which we already knew he could do).

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Nearly everyone who has posted on this problem is incorrect. Most people are making the mistake of thinking that since the probability of a correct choice on the first try is 1/3, that it remains 1/3 when one door is opened. When one door is opened, you have new information. You now know that the opened door was an incorrect choice. That new information changes the probability that your initial choice was correct from 1/3 to 1/2. It also changes the probability that the unchosen unopened door is the correct choice to 1/2. Again, the correct answer is that you cannot improve your chances by switching nor by keeping your original choice.

Actually, let's enumerate the cases for the three door problem to show that the probability is 1/2.

1) You select door A. Door A is the correct door. The host opens door B. You stick with your initial choice. You win.

2) You select door A. Door A is the correct door. The host opens door B. You switch to door C. You lose.

3) You select door A. Door A is the correct door. The host opens door C. You stick with your initial choice. You win.

4) You select door A. Door A is the correct door. The host opens door C. You switch to door B. You lose.

5) You select door A. Door B is the correct door. The host opens door C. You stick with your initial choice. You lose.

6) You select door A. Door B is the correct door. The host opens door C. You switch to door B. You win.

7) You select door A. Door C is the correct door. The host opens door B. You stick with your initial choice. You lose.

8) You select door A. Door C is the correct door. The host opens door B. You switch to door C. You win.

Since "door A" is arbitrary, the cases are repeated when you choose any other door initially. So there are 8 cases. If you keep your original choice, you win in two cases and you lose in two others. If you switch, you win in two cases and you lose in two others.

I think most people miss the point that when you initially choose the correct door, the host has two other doors that he can open, while if you choose an incorrect door, he only has one other door he can open. This is why there are 8 cases per initial door choice and not 6 or 12 as most people enumerate them.

If you disagree with this, please provide an explanation of what you think is incorrect in my analysis.

yet another way to look at it is like this:

after your first pick there are 3 situations that can occur for the remaining 2 doors (each with equal probability)

=>

1:[goat] [goat] (1/3) : you picked right

2:[goat] [car] (1/3)

3:[car] [goat] (1/3)

the guy is so kind of removing a goat from one of these remaining doors, which leaves the following situations

(still 3 situations, but in the 1st situation the guy has a choice of 2 doors to open, which he does at random)

=>

1a: [OPEN] [goat] (1/6)

2b: [goat] [OPEN] (1/6)

2: [OPEN] [car] (1/3)

3: [car] [OPEN] (1/3)

=> it is here you make your mistake with the chances that situation 1a and 1b occur

this means we have 1/6 + 1/6 = 1/3 chance on a goat behind the last door, and a 1/3 + 1/3 = 2/3 chance on a car.

Edited by dfhwze
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Nearly everyone who has posted on this problem is incorrect.

...

So there are 8 cases.

If you keep your original choice, you win in two cases and you lose in two others. If you switch, you win in two cases and you lose in two others.

I think most people miss the point that when you initially choose the correct door, the host has two other doors that he can open, while if you choose an incorrect door, he only has one other door he can open. This is why there are 8 cases per initial door choice and not 6 or 12 as most people enumerate them.

If you disagree with this, please provide an explanation of what you think is incorrect in my analysis.

That's easy to explain.

Your 8 cases do not have equal likelihood.

Specifically, cases 1-4 double count the same event.

post-1048-1223574957_thumbgif

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Can someone explain me the differance of those 2 events.

1) For start you pick a door x, the goat appeart from door y and you are asked to choose x or z.

2) For start you dont pick anything and the goat appears from door y and you are asked to choose x or z.

However I would use another logic.

You have 3 doors I will mark them with X-es

1 2 3

X X X

So your initial pick must be 2.

Now that TV show guy must choose either 1 or 3 showing the goat.

So say he opens door 3. If the goat appears to be somewhat scared and standing away from 2 you can conclude that Jaguar is in door 2 and your inital choice is correct one.

In other case goat might not be that scared but anyway she would try to get close to her fellow goat to comfort her since she is scared (goat behind door 2) of Jaguar on her left side so you know where Jaguar is.

So with a little bit of observation you have 100% of getting the Jaguar if you pick door 2.

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Can someone explain me the differance of those 2 events.

1) For start you pick a door x, the goat appeart from door y and you are asked to choose x or z.

2) For start you dont pick anything and the goat appears from door y and you are asked to choose x or z.

However I would use another logic.

You have 3 doors I will mark them with X-es

1 2 3

X X X

So your initial pick must be 2.

Now that TV show guy must choose either 1 or 3 showing the goat.

So say he opens door 3. If the goat appears to be somewhat scared and standing away from 2 you can conclude that Jaguar is in door 2 and your inital choice is correct one.

In other case goat might not be that scared but anyway she would try to get close to her fellow goat to comfort her since she is scared (goat behind door 2) of Jaguar on her left side so you know where Jaguar is.

So with a little bit of observation you have 100% of getting the Jaguar if you pick door 2.

Cute idea. ;) But maybe the goats are married and fight with each other. :o

As to your original question, the difference between cases [1] and [2] keep in mind that a goat is certain to appear.

The host will under no circumstances open the door that has the car behind it.

So, don't start thinking about the "chances" that the open door will show a goat - it always will show a goat.

That said, what information do you get when you see a goat? Simply this: your choices have been reduced by one door.

Let's take case [2] first:

There are 3 doors, and if you choose one, your chances of having picked the car are clearly 1/3.

But first, a door opens, showing a goat.

Now you know the car is behind one of two doors, not three, so your chances of picking the car have become 1/2.

How does that differ from case [1]?

There are three doors, and you pick one. You know that your chances of having picked the car are 1/3.

That means the chances that the car is behind one of the other two doors is 2/3.

But you wouldn't switch your pick, because you'd have to guess, and that reduces the 2/3 into two 1/3 choices.

But suppose you didnt have to guess? Suppose the two doors become one door. Then you'd switch and double your chances.

That's what happens when the host shows you a goat.

Your choices have been reduced by one door, leaving you an opportunity to pick the door with the 2/3 chance.

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Obviously apples and oranges are mixed here.

Case 1 and 2 differs only form someone pick of one door.

I will try to simplify the problem.

Same show same same goats same Jaguar.

Instead of being asked to pick a door where someone thinks Jaguar is, one is asked to pick a door where someone is certain goat is, still you aim for Jaguar.

After your marking of one door TV show guy opens one door with goat and says like goat is here and ask you like, are you still sure goat is behind the door you have picked or would you like to swap?

Lets make another differance, you will win the price behing the door that aint opened so you need to guess where goat is.

Would you switch?

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I honestly dont think math is involved here. It is just a matter of luck.

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Obviously apples and oranges are mixed here.

Case 1 and 2 differs only form someone pick of one door.

I will try to simplify the problem.

Same show same same goats same Jaguar.

Instead of being asked to pick a door where someone thinks Jaguar is, one is asked to pick a door where someone is certain goat is, still you aim for Jaguar.

After your marking of one door TV show guy opens one door with goat and says like goat is here and ask you like, are you still sure goat is behind the door you have picked or would you like to swap?

Lets make another differance, you will win the price behing the door that aint opened so you need to guess where goat is.

Would you switch?

Switching gives you a 2/3 chance for the car.

Therefore, sticking gives you a 2/3 chance for the goat.

I'm not sure how you try to pick a goat while aiming for a Jaguar,

but it's your decision: decide what you want, and take the winning strategy.

By the way, there are a number of places on the web where the whole scenario is simulated.

You can play the thing all day long and discover for yourself what happens.

If you're ever in town and have some extra cash, I'll show you myself, even. ;)

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Switching gives you a 2/3 chance for the car.

Therefore, sticking gives you a 2/3 chance for the goat.

I'm not sure how you try to pick a goat while aiming for a Jaguar,

but it's your decision: decide what you want, and take the winning strategy.

By the way, there are a number of places on the web where the whole scenario is simulated.

You can play the thing all day long and discover for yourself what happens.

If you're ever in town and have some extra cash, I'll show you myself, even. ;)

(Door1)(Door2)(Door3)

Goat Goat Jaguar

1) Initial pick Door 1.

You are shown Door 2.

Switching wins Jaguar.

2) Initial pick Door 2.

You are shown Door 1.

Switching wins Jaguar.

3) Initial pick Door 3.

you are shown Door 1.

Switching wins Goat.

4) Initial pick Door 3.

You are shown Door 2.

Switching wins Goat.

I think I would pass on gambling with my extra cash on this one :)

My initial idea was incorect so I switch :D

Edited by OstapBender
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