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Whatchya Gonna Do (2 goats and a car)


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196 replies to this topic

#61 Martini

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Posted 21 May 2008 - 07:13 PM

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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#62 flowstoneknight

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Posted 28 May 2008 - 07:48 PM

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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#63 tenedor

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Posted 28 May 2008 - 10:20 PM

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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#64 Frost

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Posted 28 May 2008 - 10:58 PM

Go Java!
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#65 Hot Soup

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Posted 29 May 2008 - 06:40 PM

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

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Posted 29 May 2008 - 09:27 PM

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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#67 MaximusD

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Posted 03 June 2008 - 06:46 PM

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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#68 mister_mogoo

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Posted 03 June 2008 - 07:41 PM

Spoiler for answer


Spoiler for explanation

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#69 ragepta

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Posted 04 June 2008 - 04:28 PM

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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#70 MaximusD

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Posted 04 June 2008 - 06:15 PM

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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