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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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Behind door #1 there is a goat named Betsy. Behind door #2 there is a goat named Fred. Behind door #3 there is a car.

Three possible scenarios if you choose to switch doors when asked:

1. You pick door #1 (Betsy). The host will show you Fred behind door #2 and then asks if you'd like to switch your door for door #3 (which has a car behind it). You choose to switch. You win a car.

2. You pick door #2 (Fred). The host will show you Betsy behind door #1 and then asks if you'd like to switch your door for door #3 (which has a car behind it). You choose to switch. You win a car.

3. You pick door #3 (the car). The host will show you a goat behind one of the remaining doors and then asks if you'd like to switch your door for the other one (which also has a goat behind it). You choose to switch. You win a goat.

Out of the three possible scenarios in which you choose to switch doors, you will win the car in two of them.

Thank you Martini, this is a *fantastic* explanation. It's still wrong as the question is posed, though. The question does not state that the host will open a door regardless of whether you are right or wrong. It's possible he only does it when you are right, to try to trick you into switching. If that's the case, obviously you don't switch. That's not how let's make a deal worked, though, so it's pretty clear that OP wanted to assume a let's make a deal situation. but the question should have included this assumption.

Edited by sunshipballoons
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The question does not state that the host will open a door regardless of whether you are right or wrong.

That's true. In post #4 I included "All that matters is that the host's policy is always to reveal a goat behind one of two doors that weren't picked and give the contestant the opportunity to switch. If the host always behaves in this manner, should you switch, not switch, or does it not make a difference?"

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this puzzle is found in the book "The curious incident of the dog in the night-time" by Mark Haddon

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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!)

iv seen this problem before

its a good question !!

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My math teacher actually asked me something similar to this.

She should definetly switch because there is a 2/3 chance that door #3 has it and only 1/3 chance that door #2 has it.

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I was on the 50/50 side until I thought about it more, and read some of the other posts. Here's how it works in my head, and is the most simple way that I can think of it.

First choice: There is a 1/3 chance of picking the door with the car, and a 2/3 chance that the car is behind one of the other doors

Monty reveals which of the other doors has a goat. You knew all along there was a goat behind one of those doors. There is still a 2/3 chance that the car is NOT behind your door. That means there is a 1/3 chance of it being behind your door. Still. Martini's example of getting to choose between your door or the other two is very similar to this explanation, and was what brought me to it. Thanks, Martini, I like learning. :)

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Sorry, but your logic is incorrect. You have a 50% chance of winning the car, right from the start.

Regardless of which door you choose, one of the incorrect doors is removed. The fact that there were 3 doors to start, is irrelevant.

What matters is that when it comes time to make the final decision, there are only 2 doors, and a 50% chance of guessing correctly, regardless of what your first choice was.

If there were 100 boxes (99 tissue and 1 diamond). Regardless of what your choice was, the host then removes 98 tissue boxes, you are still left with 2 boxes. One with tissue, and one with a diamond. You have a 50% chance of guessing correctly.

If 3 doors (2 goats + 1 car) then the possibilities are:

- You pick the car (in which case one goat is revealed, then you're given the choice of changing your mind, if you stay with your choice you win, if you change your choice you lose).

- You pick a goat (in which case one goat is revealed, then you're given the choice of changing your mind, if you stay with your choice you lose, if you change your choice you win).

Those are the only two possibilities. 2 equally likely possibilities = 50%

The first choice (when there were 3 options) is mutually exclusive from the second choice. In other words, it doesn't matter what your first choice was, it doesn't affect the outcome of the second choice.

Edited by Assassin
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I stand corrected.

(Tried to edit my post, and it won't let me)

She should always switch. There's a 2/3 chance that she would have picked the wrong door the first time, which improves her odds during the second choice. The two are not mutually exclusive.

She should always switch.

Edited by Assassin
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Door #2 has goat A (probability 1:3) - MB shows goat B behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances for this scenario (1:3 x 1:1 x 1:1 = 1:3)

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)

Door #2 has the car (probablility 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)

Door #2 has the car (probablility 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)

My Question is then : Why are these number different? shouldnt they be the same?

David

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My Question is then : Why are these number different? shouldnt they be the same?

I'm not sure what you're asking or why you have two scenarios above where door #2 has a car but only one scenario each for goat A and B being behind door #2.

Look at it this way;

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.

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Here's an easy explanation for anyone who says it's 50/50.

For you to switch and lose, you must have chosen the car at the start. That's 1/3 chance.

For you to switch and win, you must have chosen a goat at the start. That's 2/3 chance.

So switching would always give you a 2/3 chance of winning the car.

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I registered just to post on this thread.

I still do not understand the logic behind this puzzle, none of the examples proved anything to me; I *knew* it was still 50/50 at the end. To prove to myself that it was, I wrote a little Java program:

import java.util.Random;

public class Test {

	public static void main(String[] args) {

		Random rand = new Random();

		int guess;

		int correct = 0;

		int[] doors;

		for (int i = 0; i < 1000; i++) {

			doors = new int[3];

			doors[rand.nextInt(3)] = 1;

			guess = 0;

			if (doors[1] != 1) {

				guess = 2;

			} else {

				guess = 1;

			}

			if (doors[guess] == 1) {

				correct++;

			}

		}

		System.out.println(correct);

	}

}
Much to my chagrin, it seems to come out around 666 every time (between 600 and 750, anyway). If you then take out the switch:
import java.util.Random;

public class Test {

	public static void main(String[] args) {

		Random rand = new Random();

		int guess;

		int correct = 0;

		int[] doors;

		for (int i = 0; i < 1000; i++) {

			doors = new int[3];

			doors[rand.nextInt(3)] = 1;

			guess = 0;

			if (doors[guess] == 1) {

				correct++;

			}

		}

		System.out.println(correct);

	}

}

it comes out to be 250 - 400 (~333) every time.

If you do not believe me, download the Eclipse Java IDE (www.eclipse.org) and try it yourself.

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Posted · Report post

Go Java!

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Obviously you switch, since there is now a 75% of winning.

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Obviously you switch, since there is now a 75% of winning.

That is incorrect and isn't even one of the "logical" incorrect answers.

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QP obviously states that because someone knows what is behind each door, the outcome's probability is something different than if no one knew.

However, I think both 2/3 and 1/2 prob. are correct.

(Regarding the conversation between Babysoots and FSK, Babysoots is Jenny and FSK is Monty. I agree with Babysoots because ultimately it doesn't matter what Monty's perspective is if the goal is to cheer on the contestant or to be the contestant. To view it from Monty's perspective is moot because you ALWAYS know. I think I am just rewording Babysoots post and pardon me if I am butchering it.)

Jenny's probability of winning the car from Monty's perspective is 2/3.

Jenny's probability of winning the car from Jenny's perspective is 1/2.

Because (explanation of Jenny's perspective) she never knows what door the car is behind. In the examples given, people are saying, "Let’s say the car is behind door #3..." That is Monty's perspective and is non-submissable in the circumstance of Jenny's perspective.

From Jenny's perspective there are a 100% probability there is a goat & a car behind all three doors, then the last 2 doors. Once Jenny has decided on her final door, ALL doors become 50% goat/car ratio, and once Monty opens the door the two remaining doors become 100%/0% goat if it is a goat, car if it is a car.

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go for the second door

A similiar riddle: 10 doors, nine goats, one car. U choose a door - you chance is 1/10 of being right. The host reveals 8 doors with goats. What's the probability that the goat is on the door that you haven't chosen????

Do you feel it?

90% !!!!

Go for it!

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Hey,

just discovered this site thanks to google adds. Love it. I spent a few hours going through this topic and I finally agreed with most of you: the player will be more likely to win if they pick the other door, going from 1/3 to 2/3 chance. Here's my explanation, hope it can help those who haven't figured out:

Let's start simple: there are 100 doors, you're the host, and you're playing 100 games in a row. Most of the time, meaning during 99 games (since the player will pick the wrong door most of the time), you will think: "I have to open all the doors, except the one the player chose and the one hiding the car". That seems obvious, doesn't it? It means that the player should think "that guy opened all the doors except mine and the one hiding the car. Let's switch!" The player will be wrong only one game in 100 (unlucky to be lucky, their first pick being right...)

Crystal clear to me, hope it will help.

Hope to be proved wrong too, I like to think about it :-)

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The problem I am having with the whole 100 door example is that I don't think Monty opened all the doors except her first door and 1 more, he actually only opened 1/3 of the doors. Therefore in the 100 (99) door example she would get to pick 33 doors and he would open 33 doors, leaving 33 doors for her to switch to.

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The problem I am having with the whole 100 door example is that I don't think Monty opened all the doors except her first door and 1 more, he actually only opened 1/3 of the doors. Therefore in the 100 (99) door example she would get to pick 33 doors and he would open 33 doors, leaving 33 doors for her to switch to.

Okay, then lets look at the 100 door example your way (only we'll make it 99 doors for simplicity's sake):

1 car and 98 goats

You pick 33 doors. Monty opens 33 of the remaining doors in which goats are revealed (he NEVER reveals a car as that would mean the game is over).

He offers to let you switch your 33 doors for the other 33 doors. Should you? Of course. You went from a 1/3 chance to a 2/3 chance for the same reason that has been discussed in this thread multiple times. Let's look at your earlier post:

Jenny's probability of winning the car from Monty's perspective is 2/3.

Jenny's probability of winning the car from Jenny's perspective is 1/2.

This is wrong. If Jenny makes the switch, her probability of winning is 2/3. If she doesn't, her probability of winning is 1/3. I'll repeat what I wrote earlier as I believe it's the simplest way of looking at it:

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.

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While typing my rebutle, I thought about it, and I have to say I agree with you. My ultimate simplification of the topic is:

When Jenny is faced with the last two doors she thinks to herself, "Initially I could have chosen, a goat, a goat, or a car. Odds are, I picked a goat."

"I'll switch."

Obviously I was trying to solve the solution, making things more complicated rather than think simplisticly. I find in most mathimatical problems you (I) either over think it, or take it for granted.

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I also saw this from Google, and am trying to figure it out. Seems as though there is a possibility that is being dropped, with no reason so far, at least not one I managed to see. :)

Originally there were 6 possibilities for placements at the doors. There were again 6 different possibilities for the picking of the doors and opening of the doors. (Looking at Jennifer's perspective) (I used Excel)

So prior to picking there was a 2/6 chance (1/3) of picking the door with the car.

Each time there was a door opened two of the possibilities were eliminated. The 2 with the cars for that particular door. This left 4 different possibilities, 2 with the car at her picked door, and 2 with the car at the unpicked door. As there are now 4 different possibilities left, 2/4 leaves 50% chance. The opened door has been eliminated, so using 2/6 for each doesn't work, as the opened door possibilities have changed since there is zero probability that that door will be chosen.

Example of possibilities

Door 1 Door 2 Door 3

PICKED OPENED

1 Goat A Goat B Car

2 Goat A Car Goat B

3 Goat B Goat A Car

4 Goat B Car Goat A

5 Car Goat B Goat A

6 Car Goat A Goat B

Couldn't do all, having horrible formatting issues. This one is looking bad enough :(

Now since each alternative had 1/6 chance of happening, and 2 of those chances are gone, there is the discrepancy of how that 2/6 (1/3) is being applied. The 1/3-2/3 people are applying it to the unpicked door, the 50/50 people are splitting it up evenly, which jives with the above example. 2 out of 4 possibilities would leave 50% chance. Since most understand the intuitive logic (correct or incorrect) of the 50/50 answer, the question is why place the entire 1/3 chance on the unpicked door? By doing this there is the elimination of one of the outcomes. Which one, and why? If they are split, again, why, since it does not seem intuitively reasonable.

Another way to look at it:

Scenario 1: She picked door #2. He opened door #1. Answer says door #2 has 1/3 chance of car, door #3 has 2/3 chance.

Scenario 2: Instead of picking door #2 she picked door #3. He still opens door #1. All other things are exactly the same as in scenario 1, including where the car is. But answer now says door #1 now has 2/3 chance and door #1 only 1/3. Why are the chances different? The car is in the same place.

The host knowing seems to be the key, but I can't fathom what that factor is and how it applies. Does anyone?

Thanks!

A different host perspective I was thinking of, the Deal or No Deal scenario. Why, or how, are the statistics different if a contestant picks the same suitcase as the host would have? If they both picked the $.01 suitcase first out, how is there a difference in the chances of the rest? Trying to narrow down the crux.

Edited by Shelan
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A different host perspective I was thinking of, the Deal or No Deal scenario. Why, or how, are the statistics different if a contestant picks the same suitcase as the host would have? If they both picked the $.01 suitcase first out, how is there a difference in the chances of the rest? Trying to narrow down the crux.

The mechanics of Deal or No Deal are very different, in that the host doesn't know which suitcase contains what amount (or he could know, but since he doesn't open any of them, his knowledge is irrelevant).

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Originally there were 6 possibilities for placements at the doors. There were again 6 different possibilities for the picking of the doors and opening of the doors. (Looking at Jennifer's perspective) (I used Excel)

You're making a similar mistake others have made in the One girl/One boy riddle.

There aren't six possibilities, there are only three. That the goats could be in each other's spot is irrelevant, as they're equally losing prizes. The goats just represent losing, or winning "nothing". In other words, instead of goats, what if two doors had nothing behind them? You wouldn't think that the "nothing" behind one losing door could be swapped with the "nothing" behind the other losing door, right?

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.

If you insist that the goats' placement is relevant, then we'll look at the six possible scenarios and see the probability remains the same given you make the same choices as above:

Door #1 has a car, Door #2 has goat A. Door #3 has goat B. 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 #1 has a car, Door #2 has goat B. Door #3 has goat A. 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 #1 has goat A, Door #2 has goat B. Door #3 has a car. You choose door #1. The host will show you what's behind door #2. You make the switch. You win.

Door #1 has goat A, Door #2 has a car. Door #3 has goat B. You choose door #1. The host will show you what's behind door #3. You make the switch. You win.

Door #1 has goat B, Door #2 has goat A. Door #3 has a car. You choose door #1. The host will show you what's behind door #2. You make the switch. You win.

Door #1 has goat B, Door #2 has a car. Door #3 has goat A. You choose door #1. The host will show you what's behind door #3. You make the switch. You win.

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

Another way to look at it:

Scenario 1: She picked door #2. He opened door #1. Answer says door #2 has 1/3 chance of car, door #3 has 2/3 chance.

Scenario 2: Instead of picking door #2 she picked door #3. He still opens door #1. All other things are exactly the same as in scenario 1, including where the car is. But answer now says door #1 now has 2/3 chance and door #1 only 1/3. Why are the chances different? The car is in the same place.

The chances are different because the host doesn't randomly choose a door to open; he only opens a door with a goat behind it. Look at it this way: You agree that when you pick a door, you have a 1/3 probability of picking the door with the car behind it, right? If the host asked you if you'd like to switch your door for both the other doors, wouldn't you do it, as doing so would give you a 2/3 chance of winning? That's essentially what the host is doing, only he's revealing what's behind one of the doors before giving you the choice to switch.

In scenario 1, there is a 1/3 probability that door #2 because it's one of three doors that she picked randomly. In scenario 2, door #2 was given special protection by the host; he wouldn't have revealed what was behind it unless it was a goat.

A different host perspective I was thinking of, the Deal or No Deal scenario. Why, or how, are the statistics different if a contestant picks the same suitcase as the host would have? If they both picked the $.01 suitcase first out, how is there a difference in the chances of the rest? Trying to narrow down the crux.

On Deal or No Deal all of the boxes are chosen randomly, so switching doesn't matter. The probability of winning a higher amount doesn't change if you switch. But, if Howie Mandel were the one to eliminate boxes and he didn't do it randomly, he eliminated lower amount boxes first- definitely switch.

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Well, I figured it out. I just couldn't let it go without figuring out why, the answer just didn't really matter too much without why.

And thank you Martini for the time and energy you took to try to help me.

I am posting the answer that made sense to me, to try to help out those who were stuck looking at it in the same perspective as I needed to. I removed the goats at first, as Jennifer wants the car and only the car. The short answer is that the Host's knowledge hides 2/6 of the options from our intuitive viewing, skewing the chances from 2/6 to 2/4.

This is the (very) long answer

There are actually still 6 different scenarios once Jennifer picks a door. I'm going to use Door 1 as the picked door, and since Door 1 is already picked it can't be opened. This is where Jennifer is at prior to one of the doors being opened, where the Host has figure out which door to open, and the percentages are thrown around.

Two scenarios where the car is behind Door 1

1) One of these is where Door 2 is opened

2) One of these is where Door 3 is opened.

Two scenarios where the car is behind door 2

3) One of these is where Door 2 is opened

4) One of these is where Door 3 is opened.

Two scenarios where the car is behind door 3

5) One of these is where Door 2 is opened

6) One of these is where Door 3 is opened.

Now I'm going to explain it as though this actually happened 6 times. As there are only 3 places the car can be, and we are running it 6 times, each car placement has to happen Twice. (This is important) Those 2 times are split up between each of the other doors being opened, so each random scenario would happen once.

The first 2 examples work fine. (Remember, 2 scenarios with car behind Door 1.) One out of 6 times scenario one will happen, one out of 6 times scenario 2 will happen. Each has a 1/6 th chance of happening.

The second 2 examples have a problem. Again, since we are running 6 scenarios, and there are only 3 places the car can go, each car placement happens twice. We have to play out two times with the car behind Door 2. We can open Door 3 without a problem, and it would normally have a 1/6 chance of happening. But the Host Cannot open Door 2 without being replaced by a younger, more muscular blond boy. So, the second time the car placement happens at Door 2, he has to pick Door 3, AGAIN. This is the Host's Knowledge Factor. It is doubling the chances Door 3 would open. So Door 3 has a 2/6 chance of happening, or 1/3. This means every time the car is at Door 2, Door 3 will be opened. This makes it look like one of the scenarios has been removed, but it hasn't. The chance of scenario 3 was added to what had to be the only outcome when the car is at door 2, scenario 4. Remember there were 6 scenarios, 2 at each car placement. The car will still be at Door 2 2 times out of six (1/3)

The same thing in the last 2 scenarios. 6 scenarios divided between 3 car placements, 2 scenarios occur at each placement. The Host can open Door 2, giving 1/6 chance, but can't open Door 3. His knowledge is forcing him to open Door 2 again, when randomly Door 3 should have been opened. So the Host is giving that other 1/6 chance to open to Door 2, making it also occur 2 out of six times, (2/6) or 1/3 chance of happening. Again this doesn't mean one of the scenarios disappeared. We still did this 6 times.

As you can see, although intuitively there seems to be only 4 scenarios, there are still 6. (Besides, we ran it 6 times.) The Host chose to merge 2 due to his knowledge.

So the six scenarios look like this:

Two scenarios where the car is behind Door 1

1) One of these is where Door 2 is opened

2) One of these is where Door 3 is opened.

Two scenarios where the car is behind door 2

3) One of these is where Door 3 is opened

4) One of these is where Door 3 is opened.

Two scenarios where the car is behind door 3

5) One of these is where Door 2 is opened

6) One of these is where Door 2 is opened.

The reason I was stuck is that I saw it as 4

Two scenarios where the car is behind Door 1

1) One of these is where Door 2 is opened

2) One of these is where Door 3 is opened.

One scenario where the car is behind door 2

4) One of these is where Door 3 is opened.

One scenario where the car is behind door 3

5) One of these is where Door 2 is opened

Now these are the only options you will SEE, but the problem is that it eliminated 2 simulations that did happen. You have to remember that in 6 scenarios the car will be at Door 2 twice, and Door 3 twice, as well as Door 1. So instead of Door 1 having 2 chances out of 4 to have the car, it still has only 2 chances out of 6.

Adding goats, making them different from each other, you'd have to run 12 scenarios. Again there are only 3 places you could place the car, so the car will be placed at each door 4 times. This time I compressed the info, crossing out the random choice and replacing it with the Host's knowledge choice. Here it seems that 4 scenarios have disappeared, leaving the car at Door 1 four times, at Door 2 two times, at Door 3 two times. This is where the visual cues that makes 4/8 chance for the car to be at Door 1, is again just hiding the other 4 chances that are still there, just doubled. We still ran 12 scenarios, and four times the car had to be at Door 1, four times at Door 2, and four times at Door 3. The original placement of these things were still random, it is just the opening of the door that's not. Each Door still only gets a 4/12 (or 1/3) chance of having the car.

Four scenarios where the car is behind Door 1

A) Two of these goat A is behind Door 2 and goat B is behind Door 3

1) One of these is where Door 2 is opened

2) One of these is where Door 3 is opened.

B) Two of these goat B is behind Door 2 and goat A is behind Door 3

3) One of these is where Door 2 is opened

4) One of these is where Door 3 is opened

Four scenarios where the car is behind door 2

A) Two of these goat A is behind Door 1 and goat B is behind Door 3

1) One of these is where Door 2 Door 3 is opened

2) One of these is where Door 3 is opened.

B) Two of these goat B is behind Door 1 and goat A is behind Door 3

3) One of these is where Door 2 Door 3 is opened

4) One of these is where Door 3 is opened

Four scenarios where the car is behind door 3

A) Two of these goat A is behind Door 1 and goat B is behind Door 2

1) One of these is where Door 2 is opened

2) One of these is where Door 3 Door 2 is opened.

B) Two of these goat B is behind Door 1 and goat A is behind Door 2

3) One of these is where Door 2 is opened

4) One of these is where Door 3 Door 2 is opened

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