3 Doors...1 car = 1/3 ??? I thought.

[aommaster]

Hehe. I like to think of it like:

 

You are more likely to choose the wrong door from the start, than after one of the doors have been shown to you.

 

I don't like to go into the maths. Its complicated, even though I understand it!

[Guest Anenkefali]

Hehe. I like to think of it like:

 

You are more likely to choose the wrong door from the start' date=' than after one of the doors have been shown to you.

 

I don't like to go into the maths. Its complicated, even though I understand it! [/quote']

 

Yes, whatever suits you! The explanation I gave was the one that helped me understand the problem, together with the model where you have 1 car, and an infinite amount of goats (just as Hailstorm said).

 

It is an interesting problem, and the answer is far from obvious the first time you encounter it.

 

When speaking of probabilities, I also like the fact that the probability of at least 2 persons in a soccer-game having the same birthday is ~50%

[BrainMan]

The math really doesn't have to be that complicated. It is enough just to say that the probability of choosing right on the first pick is 1/3, and therefore the probability of it being one of the other two doors P(C1 U C2) is 2/3. We are shown a door that is not a winner, so P(C1 U C2) = P(C1) = 2/3.

[aommaster]

hehe. Yup!

[Guest slacker]

I think the confusing bit is that there are FOUR possible event paths before the contestant gets to switch or not, switching wins twice and sticking wins twice so it looks like 50:50...

 

c = car, g0, g1 = goat box

 

pick c, shown g0, stick = win OR change to g1 = lose

pick c, shown g1, stick = win OR change to g0 = lose

pick g0, shown g1, stick = lose OR change to c = win

pick g1, shown g0, stick = lose OR change to c = win

 

But the event paths do not all have an equal probability, a random sequence shouldn't go picking "c" two times out of four on average! If the contestant chooses a car, Monty can show either goat box and it doesn't matter which so rewrite the possible events thus:

 

pick c, shown a goat box, stick = win OR change to a goat box = lose

pick g0, shown a goat box, stick = lose OR change to c = win

pick g1, shown a goat box, stick = lose OR change to c = win

 

which are equally likely and gives a "change = win" two times out of three.

 

Here is a C program (linux)

 

/* mh.c - monty hall problem demo */

#include <stdio.h>
#include <stdlib.h>

div_t work;

int rnd() {				/* function returns a random 0, 1 or 2 */

work = div(random(), 3);
return work.rem;
}

int main(int argc, char *argv[]) {

int winsFirst = 0;
int winsSecond = 0;
int numTries = 10000;
int tries = -1;
int car, first, monty, second;

printf ("\nMonty Hall demo\nUsage: \"mh [number of tries (default %d)] [verbose (default FALSE)]\"", numTries); 
if (argc > 1 && atoi (argv[1]) > 0) numTries = atoi (argv[1]);

srandom(11);

while (++tries < numTries) {
	car = rnd();		/* choose a box for the car */
	first = rnd();		/* contestant chooses a box, first choice */

	monty = 0;		/* find a goat box for monty to show */
	while (monty == car || monty == first) monty++;

	second = 0;		/* find box for second choice : NOT contestant's first choice or Monty's */ 
	while (second == first || second == monty) second++;

	if (argc > 2 && strcmp (argv[2], "FALSE")) {
		printf ("\nCar: %d, First: %d, Monty: %d, Second: %d", car, first, monty, second);
		}

/* since Monty's box is a goat, then if first choice isn't a car the second choice must be the car but make it obvious ... */		
	if (car == first) winsFirst++;
	else if (car == second) winsSecond++;
	}
printf ("\nNumber of Tries: %d, Stick with First wins: %d, Second choice wins: %d\n", tries, winsFirst, winsSecond);
return (0);
}

 

$ ./mh

Monty Hall demo
Usage: "mh [number of tries (default 10000)] [verbose (default FALSE)]"
Number of Tries: 10000, Stick with First wins: 3309, Second choice wins: 6691

[Dave]

I don't think anyone can now argue this any other way

 

Thanks slacker.

 

(somehow that thanks seemed quite amusing )

[123rock]

Its the last round of a game show. You have to choose between three doors labeled One' date=' Two, and Three. Behind each door is a prize. Now two of the doors contain goats while one of the doors contains a brand new car.

 

Your asked to choose a door number. Lets say you choose 1. The Host then reveals a door other than the one that you choose, lets say that it is door number 3, and it has a goat behind it. You are then asked if you would like to change your choice to number 2 or stick with number 1.

 

Now the riddle is, does this make a difference? Should you change your answer or stick with it?

 

 

Have you all heard of this wierd puzzle? I have a book about it, but the answer is surely misleading.[/quote']

 

This is the difference between chaos theory and chance.

 

Chance is incomplete because it includes only a few factors, but in most cases that is sufficient.

 

However chaos theory incorporates everything that has happened and is happening, and thus your chances would improve by chaos theory if you switched the door.

 

Take this example, which I was puzzled after my combinatorics' test, but I later realized that that wasn't the question even asked.

 

You have two dice. You are going to roll both of them, but at different times, one after the other. You have to get a six on at least 1 die to win say a million bucks.

 

The question is, does your chance of getting a six improve when you have a second die to roll if you don't get a six on your first one.

 

In chance, it doesn't, but in chaos theory it does, since chance only accounts for the 6 numbers on the die, while chaos theory incorporates not only that, but the way your hand throws the die, the way it hits the table, the position, literally everything, and thus that is the reason why you have a better chance of switching the door than staying.

[matt grime]

Chaos theory has nothing to do with this question; let's not obscure the important probability theory involved. (Conditional probability.)

 

Your grasp of chaos seems informed by popular and misleading sources. Indeed your conclusion that chaos theory tells you to switch is utterly false. I don't know what topological transitivity has to do with this question, do you?

 

The probability of obtaining a 6 in two throws of a die given the first is a six is 1, rather obviously, and has nothing to do with chaos but with the assumptions of probability.

[123rock]

Chaos theory is what actually happens, and chance is what might. That's the connection.

[matt grime]

Erm, no, that's not actually what Chaos theory is. A dynamical system is chaotic if it displays topological transitivity and sensitive dependence on initial conditions (or at least that is the more usual definition). Chaos theory has no need to be used here since we aren't treating this as a dynamical system, we are treating it purely as a probability question. The combinatorics model of dice throwing doesn't ask you to model the throw with a dynamical system, does it?

[123rock]

Chaos theory has nothing to do with this question; let's not obscure the important probability theory involved. (Conditional probability.)

 

Your grasp of chaos seems informed by popular and misleading sources. Indeed your conclusion that chaos theory tells you to switch is utterly false. I don't know what topological transitivity has to do with this question' date=' do you?

 

The probability of obtaining a 6 in two throws of a die given the first is a six is 1, rather obviously, and has nothing to do with chaos but with the assumptions of probability.[/quote']

 

I don't think several books and a "Mathematical Excursions" course in University of California Santa Cruz (UCSC) on chaos theory counts as a popular misleading internet source.

 

I understand that it has no application here, but I just thought another example would explain what I'm trying to say.

[matt grime]

Did I mention the word "internet"? Which books? And what level is this course at UCSC? Graduate?

 

There is a vague and fuzzy notion that chaos theory is that explanation in Jurassic Park. It is non-linear dynamics. The model of throwing a die will be chaotic, it will be non-linear and display TT and SDIC. That isn't the same as a question of probability. Indeed, I do something completely deterministic and write down a number. I tell you it is in the range 1-10. You guess. I do it again, you guess again. We do it 1,000 times, what do you reckon is the approximate number of times you've got it right? Any mention of Chaos?