The Monty Hall problem
Align
Remain Calm Join Date: 2002-11-02 Member: 5216Forum Moderators, Constellation
Remain Calm Join Date: 2002-11-02 Member: 5216Forum Moderators, Constellation
in Discussions
<div class="IPBDescription">3 doors, 2 goats, 1 car</div><a href="http://www.youtube.com/watch?v=mhlc7peGlGg" target="_blank">http://www.youtube.com/watch?v=mhlc7peGlGg</a>
Basically:
<!--quoteo--><div class='quotetop'>QUOTE </div><div class='quotemain'><!--quotec-->You find yourself on a game show, you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?<!--QuoteEnd--></div><!--QuoteEEnd-->
I still think the answer given is stupid as hell, even though I vaguely remember one of my maths teachers proving it to me. There're 2 doors to choose from, not 3, the chance can't be 33% or 66%.
Basically:
<!--quoteo--><div class='quotetop'>QUOTE </div><div class='quotemain'><!--quotec-->You find yourself on a game show, you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?<!--QuoteEnd--></div><!--QuoteEEnd-->
I still think the answer given is stupid as hell, even though I vaguely remember one of my maths teachers proving it to me. There're 2 doors to choose from, not 3, the chance can't be 33% or 66%.
Comments
<span style='color:#000000;background:#000000'>The way that it makes the most sense to me is realizing that there is a *chance* that the "choice" of door the host reveals to have the goat, gives you information about which door has the car. But then the fact that there is a chance it gives you information is information itself.</span>
..
I don't know what else to say. I don't believe math lies?
Short version, personally, I'd go for the middle door. When they open a wrong door (assuming a goat) I'd watch the goat. If it's looking toward my door, change. If not, stay. After all, a goat is more likely to look at another goat, than a car.
Yes, it is a really annoying gotcha, though. It definitely pissed me off until I figured it out. I was way into the presentation of the question at first, thinking like "well, you have to know the motivations of the host...and then whether he thinks you're smart or not, like in a game of rock-paper-scissors..." until I realized he's like an automaton. Then I got to the point of thinking about it being 50:50. Then I finally figured it out.
But I agree with the short version method. Although I do like all things de chevre... mmmmmm goat butter... goat cheese...
I thought the video leaves this perfectly clear. Yes, there are two doors, but the chance of the car being behind either is not the same. The car is twice as likely to be behind the door you DIDN'T choose originally rather than behind the one you did. So the 33% and 66% chances are correct.
To rehash: Assume that you swap after the reveal. If you originally pick a goat, the host will reveal the OTHER goat, so the last door has the car. If you originally pick the car, the host will reveal either goat, and the last door has a goat as well. If you originally pick a goat, you get the car in the end. If you originally pick the car, you get a goat in the end. However, there are initially two goats and only one car. So the chance of picking a goat initially (and therefore winning a car in the end if you swap) is twice as high as picking the car first. In other words, when swapping your chances of winning the car are 66%, and 33% for a goat.
This is like trying to predict the fall of a die. You have 1/6 * 1/6 = 1/36 chance of getting two sixes in a row (or any values where the order is important). Roll the dice once, get a six. What is the chance of another? Still 1/6, because the next fall doesn't care about the previous results. It's the difference between predicting from the start and from when you're actually there.
This is like trying to predict the fall of a die. You have 1/6 * 1/6 = 1/36 chance of getting two sixes in a row (or any values where the order is important). Roll the dice once, get a six. What is the chance of another? Still 1/6, because the next fall doesn't care about the previous results. It's the difference between predicting from the start and from when you're actually there.<!--QuoteEnd--></div><!--QuoteEEnd-->
Fail. But you are in good company, lot of mathematical literate people also thought Marilyn Vos Savant was mistaken back in 1990, even Nobel laureates.
Look at it this way. If you switch, you will only lose if you originally chose a car (1/3). If you originally chose a goat and switch, you will win because the only door left is the one for the car (2/3). If you switch you will win 2/3 of the time, if you don't switch nothing has changed and you still have 1/3 chance of winning.
Just go through it again and again. And your example with dice is irrelevant for the problem, because with the dices the previous outcome really doesn't affect the next throw of a dice, whereas in the Monty Hall problem the host actively changes the chance of picking a car if you swap.
See the <a href="http://Wikipedia%20article"" target="_blank">http://en.wikipedia.org/wiki/Monty_Hall_problem</a> for examples and explanations]Wikipedia article[/url] for examples and explanations.
This is like trying to predict the fall of a die. You have 1/6 * 1/6 = 1/36 chance of getting two sixes in a row (or any values where the order is important). Roll the dice once, get a six. What is the chance of another? Still 1/6, because the next fall doesn't care about the previous results. It's the difference between predicting from the start and from when you're actually there.<!--QuoteEnd--></div><!--QuoteEEnd-->
You're partially correct. If you are given two doors to choose from, and there's a fifty-fifty chance that the car is behind either and you have no idea which door it is, then it doesn't matter which door you choose. But that's not the case. You start out knowing nothing, and having a 66% chance to pick a goat. Then the host removes one goat from the equation. This does not affect your original choice - if you stick with what you picked, you still have a 66% chance to get a goat.
If it still doesn't make sense to you, try imagining it with <u><b>100</b></u> doors. 99 goats, 1 car. You ALWAYS switch. You pick a door, the host removes 98 goats, you switch doors. If you initially picked a goat, the door you switch to will have the car. Imagine it with 1000 doors. With 10000 doors. In each case, as long as you switch doors, you only lose if you manage to initially pick the door with the car behind it. In any other case you pick a goat, the host removes X minus 2 goats (where X = number of doors) and the other door (which you then switch to) contains the car.
This is like trying to predict the fall of a die. You have 1/6 * 1/6 = 1/36 chance of getting two sixes in a row (or any values where the order is important). Roll the dice once, get a six. What is the chance of another? Still 1/6, because the next fall doesn't care about the previous results. It's the difference between predicting from the start and from when you're actually there.<!--QuoteEnd--></div><!--QuoteEEnd-->
Except it is relevant. You know that the eliminated door didn't have the car (because of the rules of the game).
The car is either behind the door you picked, or behind one of the doors you didn't, so there's a 1/3rd chance you picked the right door. However, since one of the goat doors is eliminated, "the other two doors" becomes an option for you in the switch.
Say you pick A. There's a 1/3rd chance the car is behind A. There's a 2/3rd chance it's behind either B or C, and you know it isn't behind C. That means there's a 2/3rd chance it's behind B, which means you are better off switching.
~~Sickle~~
NOW it makes sense, thank you.
But what if I want 99 goats instead?
Absolutely. Let's take this yahoo article as gospel: <a href="http://answers.yahoo.com/question/index?qid=20080918103801AA9QZWj" target="_blank">http://answers.yahoo.com/question/index?qi...18103801AA9QZWj</a>
Goat milk can go for 18 USD / gallon. Each goat produces ~1 gallon a day. All you need to do is sell out your stock every day to make 650,000 USD a year!
Goat milk can go for 18 USD / gallon. Each goat produces ~1 gallon a day. All you need to do is sell out your stock every day to make 650,000 USD a year!<!--QuoteEnd--></div><!--QuoteEEnd-->
Ok, but I'm only interested if I could get cheap/slave labour to do all the milking and selling.
<!--c1--><div class='codetop'>CODE</div><div class='codemain'><!--ec1-->#include "math.h"
#include "stdio.h"
int main(int argc, char* argv)
{
int doors[3];
int swapwins;
int keepwins;
int mypick;
int mynewpick;
int hostpick;
int swaps;
int keeps;
swapwins = 0;
keepwins = 0;
mypick = 0;
mynewpick = 0;
hostpick = 0;
swaps = 0;
keeps = 0;
// 0 = Goat, 1 = Car
while (1)
{
doors[0] = 0;
doors[1] = 0;
doors[2] = 0;
doors[rand_minmax(0,2)] = 1;
mypick = rand_minmax(0,2);
do {
hostpick = rand_minmax(0,2);
} while ((hostpick == mypick) || (doors[hostpick] == 1));
if (rand_minmax(0,1))
{
// Swap
swaps++;
do {
mynewpick = rand_minmax(0,2);
} while ((mynewpick == mypick) || (mynewpick == hostpick));
swapwins += doors[mynewpick];
printf("Swap win ratio: %f\n", (float)swapwins / (float)swaps);
} else {
// Don't swap
keeps++;
keepwins += doors[mypick];
printf("Keep win ratio: %f\n", (float)keepwins / (float)keeps);
}
}
return 0;
}
int rand_minmax(int min, int max)
{
if ((max - min + 1) == 0) return 0;
return rand() % (max - min + 1) + min;
}<!--c2--></div><!--ec2-->
Swap win ratio: 0.665485
Keep win ratio: 0.333169
Numbers seem clear.
--Scythe--
Swap win ratio: 0.665485
Keep win ratio: 0.333169
Numbers seem clear.
--Scythe--<!--QuoteEnd--></div><!--QuoteEEnd-->
Shouldn't that add up to 100%? My C is a little rusty so maybe it is accounting for the 0.15% where the goats run away with the car?
~~Sickle~~
Swap win ratio + swap lose ratio should be 100%. But swap win ratio and keep win ratio are two different experiments, so there is no need for them to become 100% together.
Done many enough times they should each approximate two numbers that together make up 100%, but that is beside the point, just because 1/3 chance of winning + 2/3 chance of winning equals 1.
~~Sickle~~<!--QuoteEnd--></div><!--QuoteEEnd-->
Quite so.
Seriously, lolf wins at explanations. I was kinda with you Align (I men, I knew you were wrong, but I couldn't accept it either). But lolf's explanation is great :D
After she explained the monty hall problem in parade magazine she was inundated with letters from academia explaining how she couldn't possibly be wrong.
There's a great discussion on it at her website <a href="http://www.marilynvossavant.com/articles/gameshow.html" target="_blank">http://www.marilynvossavant.com/articles/gameshow.html</a>
with followup discussion on her forum <a href="http://www.marilynvossavant.com/forum/viewtopic.php?t=64" target="_blank">http://www.marilynvossavant.com/forum/viewtopic.php?t=64</a>
here's the key line for me:
"When you first choose door #1 from three, there's a 1/3 chance that the prize is behind that one and a 2/3 chance that it's behind one of the others. But then the host steps in and gives you a clue. If the prize is behind #2, the host shows you #3, and if the prize is behind #3, the host shows you #2. So when you switch, you win if the prize is behind #2 or #3. You win either way! But if you don't switch, you win only if the prize is behind door #1. "
Or in other words, if you stick you win if you were right the first time ( 1/3 ) but if you switch you win if you were wrong the first time ( 2/3 )
With a door swapping strategy, only picking the door with the car (1/3) will make you lose.
(This forum needs more math.)
Anyway: It is pretty easy to grasp, that the chance of getting the prize changes for the chosen doors differently than for the not choosen doors by making the example extreme.
100 doors
1 prize
You select 1 door
98 doors which do not contain the prize open.
Do you switch to the other closed door?
Now slowly reduce the number of doors till you are at 3 doors .
I hope this helps some of you, to get into the right mindset.
------------------------------------------------
Another approach is to look at it the following way:
You choose a door.
Now the chance that the car is behind your door is 1/3
The chance of the car being behind the other 2 doors is 1/3+1/3 = 2/3
When the gamemaster opens 1 door the door that you hold has nothing changed to it. The chance is still 1/3.
And the chance that the car is behind the other 2 doors is also left unchanged: 2/3
BUT one of the unselected doors has now a chance of 0. So in order for the 2 unselected doors to have a total chance of 2/3 the unselected and unchosen doors needs to have a chance of 2/3.
With a door swapping strategy, only picking the door with the car (1/3) will make you lose.<!--QuoteEnd--></div><!--QuoteEEnd-->
That didn't help me at all back then, since the problem was with the reasoning for there being different chances.
<!--quoteo(post=1719404:date=Jul 26 2009, 07:15 PM:name=Faskalia)--><div class='quotetop'>QUOTE (Faskalia @ Jul 26 2009, 07:15 PM) <a href="index.php?act=findpost&pid=1719404"><{POST_SNAPBACK}></a></div><div class='quotemain'><!--quotec-->Back in school I wrote a small delphi program, which would solve it for m doors, n prizes, o opened doors, p chosen doors.
Anyway: It is pretty easy to grasp, that the chance of getting the prize changes for the chosen doors differently than for the not choosen doors by making the example extreme.
100 doors
1 prize
You select 1 door
98 doors which do not contain the prize open.
Do you switch to the other closed door?<!--QuoteEnd--></div><!--QuoteEEnd-->
lolf already showed this, fask
Of course you should change doors regardless. According to the theory that was disproven it would make no difference, while according to the correct theory you should change doors. You have better odds changing doors regardless of what you originally believe since there is always a chance you are wrong.
Thus the discussion is only of academic interest I could care less! Cow says MOOT.