Comments on: A javascript simulator for The Monty Hall Problem http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/ The world, according to AlternateLifeform. Tue, 07 Sep 2010 08:52:24 +0000 http://wordpress.org/?v=2.0.2 by: ozkuur http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-26286 Mon, 05 Jan 2009 20:42:54 +0000 http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-26286 I have 2 questions regarding this probability issue. Please answer if you know. 1. I have read a few articles and they are saying that if Monty doesnt know where the car is, switching doesnt matter. My question is : Think of 10 doors and you choose no:1, and Monthy starts to open doors randomly and SOMEHOW he opens 8 doors with goats behind them.....Switching will find the car with %99 or %50? 2. Think of 5 doors, you pick no:1 , and monty opens no:5 (He knows where the car is) , I am asking if the probability of other doors 2 or 3 or 4 have bigger probability than the 1 st door after Monty opens. Waiting for your answers, thanks I have 2 questions regarding this probability issue. Please answer if you know.

1. I have read a few articles and they are saying that if Monty doesnt know where the car is, switching doesnt matter. My question is :
Think of 10 doors and you choose no:1, and Monthy starts to open doors randomly and SOMEHOW he opens 8 doors with goats behind them…..Switching will find the car with %99 or %50?

2. Think of 5 doors, you pick no:1 , and monty opens no:5 (He knows where the car is) , I am asking if the probability of other doors 2 or 3 or 4 have bigger probability than the 1 st door after Monty opens.

Waiting for your answers, thanks

]]> by: Stuart http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-23775 Wed, 12 Nov 2008 07:22:07 +0000 http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-23775 But there are actually FOUR possibilities, not three. If you chose the wrong door initially, then there are then two possibilities, both of which you stand to win if you change door. If you happened to have chosen the right door in the first place, then there are also two possible outcomes, both of which you stand to lose if you change. 2 wins, 2 losses. It's 50:50 But there are actually FOUR possibilities, not three.

If you chose the wrong door initially, then there are then two possibilities, both of which you stand to win if you change door. If you happened to have chosen the right door in the first place, then there are also two possible outcomes, both of which you stand to lose if you change.

2 wins, 2 losses. It’s 50:50

]]> by: Alternate http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-291 Sun, 08 Oct 2006 03:04:43 +0000 http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-291 <p>??? But there is a chance that you would <strong><em>pick</em></strong> the "second" goat in your attempt, then the host would eliminate the "first" goat.</p><br /> <p> Put simply here are the three possibilities: 1. you pick the "first" goat, host eliminates the "second" goat. (win if switch) <br /> </p> <p> 2. you pick the "second" goat, host eliminates the "first" goat. (win if switch) </p><br /> <p> 3. you pick the car, Host eliminates either "first" or "second" goats based on his discretion. (loose if switch) </p><br /><br /> Therefore, there's a 2/3 chance of winning by switching. ??? But there is a chance that you would pick the “second” goat in your attempt, then the host would eliminate the “first” goat.

Put simply here are the three possibilities:
1. you pick the “first” goat, host eliminates the “second” goat. (win if switch)

2. you pick the “second” goat, host eliminates the “first” goat. (win if switch)

3. you pick the car, Host eliminates either
“first” or “second” goats based on his discretion. (loose if switch)

Therefore, there’s a 2/3 chance of winning by switching.

]]> by: Phillip http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-267 Tue, 26 Sep 2006 06:24:33 +0000 http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-267 Since you know in advance that one door, a door with the 2nd goat will always be eliminated, you are actualy picking one of two dors and thus a 50/50 chance. The eliminated 2nd goat is a pantom choice that never realy existed. Since you know in advance that one door, a door with the 2nd goat will always be eliminated, you are actualy picking one of two dors and thus a 50/50 chance. The eliminated 2nd goat is a pantom choice that never realy existed.

]]> by: Alternate http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-237 Sun, 03 Sep 2006 12:30:53 +0000 http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-237 <p>I appreciate your taking the time to post. Let me assure that there's nothing wrong with the simulation. </p> <p>Think of it this way: 3 doors: ==========</p> <p> 33.33 % chance to win by sticking to original choice, 66.66% chance to win by switching. Simulate 1000 attempts, chances are you'll win 333 - 334 or so times for sticking, 666 or so times for switching, </p> <p> 4 doors: ========== </p><p> 25% chance to to win by sticking to the original choice, 75% chance to win by switching. simulate 1000 attempts, chances are you'll win 250 or so times for sticking, 750 or so times for switching. </p><p> 1000 doors: ========== </p><p> .1% chance to win by sticking, 99.9 % chance to win by switching. </p><p> ie, Simulate 1000 attempts, chances are you'll will 1 time for sticking, 999 times for switching. If you keep running the simulator repeatedly, every so often, you'll win 2 times out of thousand, and every so often, you won't even win once. </p><p> 10000 doors: </p><p> .01% chance to win by sticking, 99.99 % chance to win by switching. What that means is that when you run a 10000 door simulation, and you have only 1000 attempts, you only have 10% chance of winning even ONCE in those 1000 iterations. </p><p> IE, if you run the simulator 10 times with those parameters, chances are that only one time out of those 10 attempts, the simulator will say: one 1 time out of 1000! </p><p> What I'm trying to say, is that in order to be assured of even reasonably interpretable results, you will have to run the simulator for atleast as many attempts(iterations) as doors. And the more iterations, the better. </p><p> If you're running the simulator for 1000 doors, bump up the number of iterations to atleast 10000. </p><p> Hmm, I should probably put this into the post. :P </p> I appreciate your taking the time to post. Let me assure that there’s nothing wrong with the simulation.

Think of it this way:
3 doors:
==========

33.33 % chance to win by sticking to original choice,
66.66% chance to win by switching.
Simulate 1000 attempts, chances are you’ll win 333 - 334 or so times for sticking, 666 or so times for switching,

4 doors:
==========

25% chance to to win by sticking to the original choice,
75% chance to win by switching.
simulate 1000 attempts, chances are you’ll win 250 or so times for sticking, 750 or so times for switching.

1000 doors:
==========

.1% chance to win by sticking,
99.9 % chance to win by switching.

ie, Simulate 1000 attempts, chances are you’ll will 1 time for sticking, 999 times for switching. If you keep running the simulator repeatedly, every so often, you’ll win 2 times out of thousand, and every so often, you won’t even win once.

10000 doors:

.01% chance to win by sticking,
99.99 % chance to win by switching.
What that means is that when you run a 10000 door simulation, and you have only 1000 attempts, you only have 10% chance of winning even ONCE in those 1000 iterations.

IE, if you run the simulator 10 times with those parameters, chances are that only one time out of those 10 attempts, the simulator will say: one 1 time out of 1000!

What I’m trying to say, is that in order to be assured of even reasonably interpretable results, you will have to run the simulator for atleast as many attempts(iterations) as doors. And the more iterations, the better.

If you’re running the simulator for 1000 doors, bump up the number of iterations to atleast 10000.

Hmm, I should probably put this into the post. :P

]]> by: Will http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-236 Sun, 03 Sep 2006 08:57:59 +0000 http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-236 Your simulator grossly inflates the odds of winning after switching doors in the case where the number of doors is greater than 3 because it counts any case in which you lose by not switching as one in which you win by switching, when in fact you still have to choose a door to switch to. This explains the obviously false results for a very large number of doors where you win 100% of the time by switching according to the simulation. Your simulator grossly inflates the odds of winning after switching doors in the case where the number of doors is greater than 3 because it counts any case in which you lose by not switching as one in which you win by switching, when in fact you still have to choose a door to switch to. This explains the obviously false results for a very large number of doors where you win 100% of the time by switching according to the simulation.

]]> by: Alternate http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-209 Sun, 20 Aug 2006 10:48:40 +0000 http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-209 Hi Tar, Thanks for the reply. and I do appreciate your link! You're absolutely right that humans, specifically the game show host, can influence the results of this test - simply by not following the rules. On the actual game show, there's no guarantee that the host <i>will</i> open a door. Which means that the host could choose to open a door <i>only</i> if the player has chosen the winning door - in which case the chances of winning by switching when the game show host opens a door, would be 0% ! ie, for a mathematical simulation to be accurate, the game show host should ALWAYS exhibit consistent behavior – he should ALWAYS be required to open a door. Hi Tar,

Thanks for the reply. and I do appreciate your link!

You’re absolutely right that humans, specifically the game show host, can influence the results of this test - simply by not following the rules.

On the actual game show, there’s no guarantee that the host will open a door. Which means that the host could choose to open a door only if the player has chosen the winning door - in which case the chances of winning by switching when the game show host opens a door, would be 0% !

ie, for a mathematical simulation to be accurate, the game show host should ALWAYS exhibit consistent behavior – he should ALWAYS be required to open a door.

]]> by: Tar Martel http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-208 Sat, 19 Aug 2006 22:13:49 +0000 http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-208 Thanks for the simulation. The first reaction I had when I read about the riddle, was to create a simulation... but then I found your JavaScript one. It is simple and to the point, I like it and it saved me the effort. The solution: The results are not so obvious when using the constraints of the riddle (three doors) but bump it up to a 100 doors and the odds are completely in your favour to change your mind, but then bump it up to a 1000 doors, and I’m afraid your simulation stops working. – More often than not you stand a 100% chance of loosing if you decide not to swap. I don’t believe those odds. Mathematically yes, but in my heart, no. See my expanded comments at http://journal-of-a-novel.blogspot.com/2006/08/monty-hall-problem.html I don’t believe that the solution is completely correct; I believe that humans can influence the results of this simple test in a way that cannot be mathematically calculated. Regards, Tar Thanks for the simulation.
The first reaction I had when I read about the riddle, was to create a simulation… but then I found your JavaScript one. It is simple and to the point, I like it and it saved me the effort.

The solution:
The results are not so obvious when using the constraints of the riddle (three doors) but bump it up to a 100 doors and the odds are completely in your favour to change your mind, but then bump it up to a 1000 doors, and I’m afraid your simulation stops working. – More often than not you stand a 100% chance of loosing if you decide not to swap. I don’t believe those odds. Mathematically yes, but in my heart, no.

See my expanded comments at http://journal-of-a-novel.blogspot.com/2006/08/monty-hall-problem.html

I don’t believe that the solution is completely correct; I believe that humans can influence the results of this simple test in a way that cannot be mathematically calculated.

Regards,
Tar

]]> by: Alternate http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-149 Fri, 09 Jun 2006 13:30:21 +0000 http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-149 "THEY" ??? Sheesh, all this fuss over a typo? <i>OKAY</i>. I get the message. I'll fix it whenever I have time. “THEY” ???

Sheesh, all this fuss over a typo? OKAY. I get the message. I’ll fix it whenever I have time.

]]> by: Steh http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-148 Fri, 09 Jun 2006 11:04:11 +0000 http://alterlife.org/2006/04/16/a-javascript-simulator-for-the-monty-hall-problem/#comment-148 I always wonder what they write when they actually mean "loose". I always wonder what they write when they actually mean “loose”.

]]>