Why is the Monty Hall Problem so controversial?

Xittenn · March 30, 2012

Very interesting observations my good man!

Ophiolite · March 30, 2012

Precisely! I might be mucking around a bit yes! But this was what I was getting at when I said "No probabilities form positive interferences with itself PLACE BIG DOT HERE"

I think I hit on a deeper thought but honestly I couldn't make it any clearer if I tried!

So you have this vague disquiet, which you cannot adequately explain, that the 'standard' explanation is faulty and you hand out negative rep to someone just for presenting the standard explanation. :rolleyes:

So, if you didn't like the idea that SNC meteorites probably come from Mars and I made a post saying they probably did, you would give me negative rep? And if you wouldn't do it in that case , why not? Ridiculous!

Xittenn · March 30, 2012

So you have this vague disquiet, which you cannot adequately explain, that the 'standard' explanation is faulty and you hand out negative rep to someone just for presenting the standard explanation.

I'm sorry but wiki is not a means of backing this statement up, show me a source of reputable mathematics where this conclusion has been made. Taking pot shots at my etiquette doesn't add to the strength of the argument either. Roll your eyes elsewhere!

Also the inability of others to understand me doesn't constitute a vague disquiet, it constitutes a failure to communicate.

Ophiolite · March 30, 2012

I'm sorry but wiki is not a means of backing this statement up, show me a source of reputable mathematics where this conclusion has been made. Taking pot shots at my etiquette doesn't add to the strength of the argument either. Roll your eyes elsewhere!

I'm not seeking to add strength to argument: I am very deliberately taken pot shots at your ettiquette which stinks. Do you routinely neg rep anyone who provides wiki as a source? Either way, as I said, ridiculous. I'm done with you. Welcome to my first use of the Ignore function for probably two years.

Schrödinger's hat · March 30, 2012

Xitten, go back to my 1 million doors example.

Are you really saying that the remaining door has just as much chance of having the car?

Perhaps this scenario might be easier to think about:

Shuffle a deck of cards.

Pick a card at random, put it in your pocket without looking at it -- This is your initial door.

Go through the remaining deck and remove any cards that are not the ace of spades until you have one card left (ie. either the ace of spades if you found it, or the last card you looked at).

Which card is more likely to be the ace of spades? The one in your pocket, or the remaining card?

If this scenario is different, please explain the distinction.

Xittenn · March 30, 2012

Xitten, go back to my 1 million doors example.

Are you really saying that the remaining door has just as much chance of having the car?

If the car is never removed, and there is a goat and a car, then there is a 50/50 chance of the door that you are holding being a goat.

I'm sorry the idea is bunk and I have a real maths exam tomorrow . . . night !

** there's an ignore button?? this is news?? everybody ignore me now, I feel an idea is bs and I argued about it profusely without showing proof--wait I did actually show proof **

** I'm getting an A, I'm winning!

the asinine cretin · March 30, 2012

If anyone doesn't believe it I'd say write a simple computer program and behold the results for yourself. It's a simple algorithm and perhaps the exercise of coding it will make it understood. Here is a random sample of my results, each after 1000 iterations.

[stayWins:329][ChangeWins:671]

[stayWins:318][ChangeWins:682]

[stayWins:340][ChangeWins:660]

[stayWins:306][ChangeWins:694]

[stayWins:329][ChangeWins:671]

[stayWins:334][ChangeWins:666]

[stayWins:351][ChangeWins:649]

[stayWins:334][ChangeWins:666]

[stayWins:325][ChangeWins:675]

Edited March 30, 2012 by the asinine cretin

Schrödinger's hat · March 30, 2012

Here's an interactive javascript app I wrote that does it:

http://jsfiddle.net/uLzmY/

Source code if it gets lost somehow, not a complete webpage as jsfiddle inserts some things for me. If you want to make it into a page but cannot, feel free to ask and I'll complete it:

This goes in the body:

<form name="form" method="post" action="">
<input name="one" type="button" id="sone" value="1" onclick="selectDoor(1)">
<input name="two" type="button" id="stwo" value="2" onclick="selectDoor(2)">
<input name="three" type="button" id="sthree" value="3" onclick="selectDoor(3)">
<input name="switch" type="button" id="bsw" value="switch" onclick="switchDoor()">
<input name="commit" type="button" id="bcm" value="commit" onclick="commitDoor()">
<input name="reset" type="button" id="bcm" value="reset" onclick="resetDoors()">
<br>
<input name="chsw" type="checkbox" id="chsw"> Auto Switch
<input name="chaut" type="checkbox" id="chaut"> Auto Play
<input name="chrst" type="checkbox" id="chrst"> Auto Reset
<input name="chsh" type="checkbox" id="chshow"> Show
</form>
<div id="win">Pick a door.</div>
<div id="show"></div>
<div id="tally">0/0</div>
<script type="text/javascript">resetDoors();</script>

This goes in a script tag in the head somewhere:

var doors = [], 
   hdoors = [],
   openDoor = undefined,
   winDoor = 0,
   open = false,
   currentDoor,
   score = 0, 
   plays = 0;

var resetDoors = function(){
   winDoor = Math.floor(Math.random() * 3);
   openDoor = undefined;
   for ( var i = 0; i <= 2; i++) {
       doors[i] = ((i === winDoor) ? "Car " : "Goat");
       hdoors[i] = 'Door';
   }
   var sh = document.getElementById('chshow').checked;
   document.getElementById('show').innerHTML = sh ? doors: hdoors;
   document.getElementById('win').innerHTML = 'Pick a door.'
};

var selectDoor = function(sDoor) {
   currentDoor = sDoor - 1;
   open = false;
   for(var i = 0; i <= 2; i++) {
       if ( i === currentDoor ) {
           hdoors[i] = 'Curr';
       } else if ( i !== currentDoor && i !== winDoor && open === false) {
           hdoors[i] = doors[i];
           open = true;
           openDoor = i;
       } else {
           hdoors[i] = 'Door';           		
       }
   }
   document.getElementById('show').innerHTML = hdoors;
   document.getElementById('win').innerHTML = 'Door ' + (currentDoor + 1) + ' chosen.';
   if (document.getElementById('chsw').checked) {
       switchDoor();
   }
   if (document.getElementById('chaut').checked) {
       commitDoor();
   }
};

var commitDoor = function() {
   var win = currentDoor == winDoor;
   score += win;
   plays++;
   document.getElementById('win').innerHTML = (win ? 'Win, car behind ': 'Lose, car behind  ') + (winDoor + 1);
   document.getElementById('show').innerHTML = doors;
   document.getElementById('tally').innerHTML = score + '/' + plays;
   if (document.getElementById('chaut').checked) {
       resetDoors();
   }
};

var switchDoor = function() {
   var newDoor;
   for (var i = 0; i <= 2; i++) {
       if (openDoor !== i && currentDoor !== i) {
           newDoor = i;
       }
   }
   hdoors[currentDoor] = 'Door';
   hdoors[newDoor] = 'Curr';
   currentDoor = newDoor;
   document.getElementById('win').innerHTML = 'Switched to door ' + (currentDoor + 1);
   document.getElementById('show').innerHTML = hdoors;
};

Edit: Don't press the switch button before picking a door, that's the only bug i noticed and I'm too lazy to fix it.

If clicking the buttons is too laborious, I can add a click 1000 times button on request.

Xitten, when you get back, perhaps you could look at it this way:

Go back to the 1 million doors example, 999,999 goats, one car.

You pick a door.

The host offers you something different this time. You can either stick with your door, or switch to the 999,999 remaining doors.

If you switch, the host will then open 999,998 of the doors you picked, leaving one door as part of your selection, and the door you initially chose. You then open the remaining door of the 999,999 and keep a car if it is there.

Is it now sensible to switch?

If so (and if my previous 1 million doors example is a 50/50 chance of getting the car) what is different?

Edited March 30, 2012 by Schrödinger's hat

ewmon · March 30, 2012

Interestingly enough, it all hinges on the fact that the game's host knows where the prize is. I ran a simulation on this years ago, and it turns out to be a 50/50 chance if the host didn't know the location of the prize. If he didn't know and opened the door with the prize, there would be that "waah waah waah" sound, and the host would voice condolences for the contestant picking the wrong door.

Schrödinger's hat · March 30, 2012

Interestingly enough, it all hinges on the fact that the game's host knows where the prize is. I ran a simulation on this years ago, and it turns out to be a 50/50 chance if the host didn't know the location of the prize. If he didn't know and opened the door with the prize, there would be that "waah waah waah" sound, and the host would voice condolences for the contestant picking the wrong door.

Indeed, this is where the information comes from.

Another interesting fact: If the host just picks the first (left-most) door he sees that has a goat, this also gives you information.

If you see him open the right-most unopened door, you know the car is in the left one.

If you see him open the left-most door, you have a 50/50 chance (i think? I'm very sleepy -- you can tell by my terrible code)

Also, fixed my applet to reflect this, and added a x1000 and random button

Edit, whoops: http://jsfiddle.net/uLzmY/3/

Edited March 30, 2012 by Schrödinger's hat

Xittenn · March 30, 2012

"Why is the Monty Hall Problem so controversial?" <== title of thread; answer, it is not proofed! I can not handle this mathematically in a sensible fashion and so it remains wrong, even after I wake up .. . . . . otherwise the title of the thread would be "Why the Monty Hall Problem makes life so awesome in that it explains stuff and is so not controversial at all?"

Here's an interactive javascript app I wrote that does it:

http://jsfiddle.net/uLzmY/

Xitten, when you get back, perhaps you could look at it this way:

Go back to the 1 million doors example, 999,999 goats, one car.

You pick a door.

The host offers you something different this time. You can either stick with your door, or switch to the 999,999 remaining doors.

If you switch, the host will then open 999,998 of the doors you picked, leaving one door as part of your selection, and the door you initially chose. You then open the remaining door of the 999,999 and keep a car if it is there.

Is it now sensible to switch?

If so (and if my previous 1 million doors example is a 50/50 chance of getting the car) what is different?

I'm not sure what your word choice is saying in this variation . .

If anyone doesn't believe it I'd say write a simple computer program and behold the results for yourself. It's a simple algorithm and perhaps the exercise of coding it will make it understood. Here is a random sample of my results, each after 1000 iterations.

[stayWins:329][ChangeWins:671]

[stayWins:318][ChangeWins:682]

[stayWins:340][ChangeWins:660]

[stayWins:306][ChangeWins:694]

[stayWins:329][ChangeWins:671]

[stayWins:334][ChangeWins:666]

[stayWins:351][ChangeWins:649]

[stayWins:334][ChangeWins:666]

[stayWins:325][ChangeWins:675]

Can I have your code cretin? I want to watch it but I am too lazy to write the ten lines myself!

So if I do this 1,000,000 times what is my probability is it 1/2, 1/3, 999,999/1,000,000 ??? Why or why not? How can we calculate this? Can someone write a test for this one as well?

Joatmon · March 30, 2012

I know that I will be seen as starting from a tangent, but follow it through and it becomes relevant.

I know this changes the problem slightly(?) but let's say the host doesn't know where the car is but if the contestant sees the car he can keep it.

One time in three the contestant will win the car at the first opening of a door.

If he doesn't see the car he can decide whether to switch his choice or not.

If his original choice was the car and he swaps then he loses. If his original choice was a goat and he swaps then he wins. So he has a 50/50 chance on his second bite of the cherry and it doesn't (it seems to me under these conditions) matter whether he swaps or not.

The net result is that he will win the car two times out of three. (because of his two bites at the cherry).

Now moving back to the original problem:-

The presenter tries to minimise your chances of winning by taking, and apparently wasting, your chance of winning on the first door opening.

If , at the start, you pick a door and do nothing, the two bites at the cherry increase your chances to 50/50.( You either picked the car or you picked a donkey)

What needs explaining is why always changing your mind on the second door opening improves your chances to two in three.

Well, two times out of three you will have picked a goat.

The door you have seen opened hid a goat. So these two times out of three changing your mind gives you the car.

One time out of three you will have picked the car so changing your mind this one time out of three gives you a goat. (The goat you haven't previously seen).

This is the reason that changing your mind improves your chances from 50/50 to two in three(IMO).

Edited March 30, 2012 by Joatmon

D H · March 30, 2012

This is like the infinite improbability drive. And there are no proofs!

No, but apparently convincing you takes infinite patience. I don't have infinite patience, but I'll give it a shot anyhow.

Of course there are proofs. This is a simple problem of conditional probabilities.

Let's first look at what the probability is without this option to switch doors. There are three doors, one of which hides a car and each of the other two hides a goat. You pick one door, and that's it. The probability that you guessed the car is clearly 1/3.

Now let's modify the game a bit by having Monte open a door that hides a goat. Note that with this mod you are still stuck with your original choice. Also note that Monte can always open a door that hides a goat because Monte knows which doors hide which prizes, and because there will always be at least one door that hides a goat even after you picked a door. Since you can't use this new information, the fact that Monte showed you a goat doesn't change the outcome. The probability that you win the car is still 1/3.

The interesting modification is giving you the option of switching doors after Monte shows a door that hides the goat. Note that Monte is an automaton in this non-realistic version of game. Per the rules of this game, Monte has no choice but to show a door that hides a goat. Monte can be a bit capricious in the real game. An slightly evil Monte will offer you a tantalizing choice, but only if you have already picked a big winner. An even more evil Monte will find even better ways to make you switch if you have picked a winner, or to stick if you have picked a loser.The real Monte isn't quite evil, but he isn't an automaton either. The goal in the real game is to maximize profits. Giving away big prizes reduces profits because those big prizes cost money. On the other hand, giving away only goats also reduces profits because very few would watch the show, meaning little or no money from commercials. (The real goal of any TV show is to sell commercials.)

So how does this new version of the game let you change the odds in you favor? The answer is that switching always loses if your initial choice was the prize door but always wins if your initial choice was one of the two loser doors. Your initial choice is twice as likely to be a loser door, so always switching is advantageous. Switching changes the odds.

hypervalent_iodine · March 30, 2012

This PDF made sense to me, though I'm no mathematician.

Edit: That's not to say that what others have said doesn't make sense, it's simply another source of information that provides both an analytical and numerical solution to the problem.

Joatmon · March 30, 2012

No, but apparently convincing you takes infinite patience. I don't have infinite patience, but I'll give it a shot anyhow.

Of course there are proofs. This is a simple problem of conditional probabilities.

Let's first look at what the probability is without this option to switch doors. There are three doors, one of which hides a car and each of the other two hides a goat. You pick one door, and that's it. The probability that you guessed the car is clearly 1/3.

Now let's modify the game a bit by having Monte open a door that hides a goat. Note that with this mod you are still stuck with your original choice. Also note that Monte can always open a door that hides a goat because Monte knows which doors hide which prizes, and because there will always be at least one door that hides a goat even after you picked a door. Since you can't use this new information, the fact that Monte showed you a goat doesn't change the outcome. The probability that you win the car is still 1/3.

The interesting modification is giving you the option of switching doors after Monte shows a door that hides the goat. Note that Monte is an automaton in this non-realistic version of game. Per the rules of this game, Monte has no choice but to show a door that hides a goat. Monte can be a bit capricious in the real game. An slightly evil Monte will offer you a tantalizing choice, but only if you have already picked a big winner. An even more evil Monte will find even better ways to make you switch if you have picked a winner, or to stick if you have picked a loser.The real Monte isn't quite evil, but he isn't an automaton either. The goal in the real game is to maximize profits. Giving away big prizes reduces profits because those big prizes cost money. On the other hand, giving away only goats also reduces profits because very few would watch the show, meaning little or no money from commercials. (The real goal of any TV show is to sell commercials.)

So how does this new version of the game let you change the odds in you favor? The answer is that switching always loses if your initial choice was the prize door but always wins if your initial choice was one of the two loser doors. Your initial choice is twice as likely to be a loser door, so always switching is advantageous. Switching changes the odds.

I can't be dogmatic about this as I'm not too sure myself but:-

I have spent some time thinking this over and it seems to me that the odds of winning the car start at 1/3 if only one door is opened. If two doors are opened, with the first door "failing" the odds increase in the contestants favour and I think become 50/50. I think this is so because the odds against choosing the unseen goat and choosing the car are both the same at 1/3. To take the extreme example of opening all three doors then the contestant must win.

I note that your conclusion matches mine (except for the 50/50) in#37.

Edited March 30, 2012 by Joatmon

Xittenn · March 30, 2012

This PDF made sense to me, though I'm no mathematician.

Edit: That's not to say that what others have said doesn't make sense, it's simply another source of information that provides both an analytical and numerical solution to the problem.

It was the only one that I could find of its kind, and I think you know what motivates me in my terms which is why you posted it. When the Monty Hall Problem was first postulated for me in the movie 21 back a few years ago I thought it was the greatest thing since American cheddar whipped cheese. And then I forgot about it because I am not really into gambling.

So I guess what everyone is saying is the answer to the question is it isn't? I'm saying it needs a more introspective look. I don't think the question I posed to the Hat can be answered logically by any of the posters and I am not comfortable with my own conclusions so I remain in the area of I don't know what the hell this is saying exactly. The word problems posted have clearly stated their point from the beginning, the fact that anyone would think I don't have the common sense to grasp the concept, from my POV doesn't fully understand the concept themselves. When I have a firm grounding on the problem I will change my conclusion, and until then there is controversy for me. Until I can clearly trace the pathways and interactions I stand where I stand. There are insufficient resources explaining in mathematical terms how this works out numerical on all levels. This suggests to me that from a mathematical perspective there is something that doesn't sit well. Reiterating the problem in a variety of wordings without being able to manipulate the numbers beyond the obvious is just thinking games.

My personal opinion on the wikipedia-MHP-wars is that they are fights about the

wrong question. Craig Whitaker, through the voice of Marilyn vos Savant, asks for

an action, not a probability. I think that game theory gives a more suitable framework

in which to represent our ignorance of the mechanics of the set-up (where the

car is hidden) and of the mechanics of the host’s choice, than subjectivist probability.

Gill, Richard D. "The Monty Hall problem is not a probability puzzle* (It's a challenge in mathematical modeling)." Statistica Neerlandica, 65/1. (2011)

Edited March 30, 2012 by Xittenn

D H · March 30, 2012

The probability is 50/50 if you flip a coin to decide whether or not you will switch. If your strategy is always stick with your original choice the probability is 1/3 (two to one against). This reverses to two to one in favor of winning the car (probability of winning the car is 2/3) if your strategy is to always switch doors.

John · March 30, 2012

Here's a more simply worded proof:

http://www.remote.org/frederik/projects/ziege/beweis.html

and a Java applet to simulate as many games as you like:

http://www.remote.org/frederik/projects/ziege/empirie.html

Xittenn, this isn't some competition. I, and everyone else, would be just as happy if the probability did turn out to be 1/2. However, you have been shown in many different ways (including intuitive explanations, mathematical proofs, and simulations) that the probability of winning is 2/3 if the contestant switches, and 1/3 if he sticks with his original door. You've always seemed fairly intelligent, so I'm sincerely not sure if you're just messing with us now or if you've fallen into the classic trap of being too attached to what you think the answer should be. If the former, then good job. If the latter, then perhaps we've all been posting in vain.

As for the whole "it's not a probability puzzle" bit, that doesn't really matter. If we're seeking to decide what the contestant should do (i.e. what the correct action is), then, assuming the contestant wishes to win a car, he should switch. This is based on the probabilities involved.

Joatmon · March 30, 2012

The probability is 50/50 if you flip a coin to decide whether or not you will switch. If your strategy is always stick with your original choice the probability is 1/3 (two to one against). This reverses to two to one in favor of winning the car (probability of winning the car is 2/3) if your strategy is to always switch doors.

I know "in my heart" that what you say is true concerning my 50/50 assessment because if you take a bet with fair odds of 1/3 and let it run its course without interference the intermediate steps are immaterial and you can expect to win one time in three.

I was unsure about that myself, but am entirely satisfied that I have shown logically (using the same argument as you) that always swapping doors improves the odds to 2/3.

Thanks for your time.

Edited March 30, 2012 by Joatmon

Xittenn · March 30, 2012

I don't see where I am profusely contesting anything other than the fact that the idea is not well founded in a mathematical sense. Empirically correct or otherwise the content of your posts have little meaning to me and I don't feel compelled to accept the facts presented as such. I've tried at numerous points to point at ideas that I feel require further clarification under the conditions presented. I really don't know why you guys are so concerned with my standpoint if we ignore me at this point I believe the definitive answer to the question posted by the OP is that there is in fact no controversy. I vote this down, in a democracy my single vote would be meaningless. Is there a reason why others feel that I need to come along before the conclusion can be made otherwise? I simply gave my opinions, as it stands I feel that the conclusion and the empirical evidence are rooted in statements that are not quite as presented and that require deeper reflection. Statistics takes on a very formal structure and if there wasn't something questionable about the statement no one would have cared about the preposition otherwise. The preposition for me is very intuitive, questioning it is what is counter-intuitive.

D H · March 30, 2012

I don't see where I am profusely contesting anything other than the fact that the idea is not well founded in a mathematical sense.

The idea is very well founded in a mathematical sense. This is an extremely simple probability problem.

What exactly is your objection here? Are you truly contesting that switching doubles your chances of winning?

Xittenn · March 30, 2012

The idea is very well founded in a mathematical sense. This is an extremely simple probability problem.

Sounds good to me, thanks for clarifying that!

Schrödinger's hat · March 30, 2012

I don't think the question I posed to the Hat can be answered logically by any of the posters and I am not comfortable with my own conclusions so I remain in the area of I don't know what the hell this is saying exactly.

I'm not 100% sure on what you asked me. Perhaps you could re-state the question, maybe adding a little redundancy just in case I still don't follow. I would be happy to attempt a formal answer if I knew exactly what was being asked.

I think part of the issue here, is that the simplest way to proove this is by simple exhaustion. There are numerous tables and tree diagrams on the wiki page that constitute a proof from a frequentist standpoint.

If you prefer symbol pushing, then you can throw Bayes theorem at it repeatedly, and there are a number of ways to approach the answer.

I've tried at numerous points to point at ideas that I feel require further clarification under the conditions presented. I really don't know why you guys are so concerned with my standpoint if we ignore me at this point I believe the definitive answer to the question posted by the OP is that there is in fact no controversy.

The amount of exposition present is because the answer depends so much on the host's behavior. It's very important that he makes the same decisions every time, otherwise you get different priors, or a different distribution of outcomes. Also, many of us consider the proof to be rather trivial (as I stated, exhaustion is sufficient, or even pointing out the number of distinct states and outcomes) and that all of the difficulty to be had is in adjusting one's intuition.

I think part of the reason you haven't gotten satisfactory clarification is that we (or I at least) are(am) having a great deal of difficulty understanding exactly what it is you are pointing out or what you're asking. This is probably due to an unstated assumption on someone's part, or an unstated knowledge differential. Perhaps you could break things down into much simpler statements (and over-explain slightly) until said confusion clears up.

The reason we're concerned with your standpoint is that this community is here to help people learn and understand things. If we perceive you do not understand something we will endeavour to help you (even if we sometimes do it rather inefficiently).

This is also an excellent example of how intelligent and mathematically experienced people can have difficulty understanding the Monty Hall problem or accepting the conventional answer. Ie. a good example of 'why is it so controvercial?'

Edited March 30, 2012 by Schrödinger's hat

the asinine cretin · March 30, 2012

Can I have your code cretin? I want to watch it but I am too lazy to write the ten lines myself!

So if I do this 1,000,000 times what is my probability is it 1/2, 1/3, 999,999/1,000,000 ??? Why or why not? How can we calculate this? Can someone write a test for this one as well?

Sure. The little test program I wrote yesterday was in C# (using Microsoft's free C# "express edition" IDE - available here). It didn't take long to write and would be easy to rewrite using Python, JavaScript, or any other language. I wasn't intending to share this so don't diss it. hehe. I've commented it and simplified it a bit.


       const int Goat = 0;
       const int Car = 1;

       private void RunSimulation(int iterations)
       {
           // Win counters
           int StayWins = 0;
           int ChangeWins = 0;

           // Doors array
           int[] doors = new int[3];
           Random rand = new Random();

           // Selected index variable
           int index = 0;

           // Perform iterations
           for (int i = 0; i < iterations; i++)
           {
               // Reset default doors value
               for (int x = 0; x < doors.Length; x++)
                   doors[x] = Goat;

               // Set random index to "Car"
               int car = rand.Next(0, 3);
               doors[car] = Car;

               // Randomize selected index
               /* Uncomment following line to randomize selected index on each iteration */
               //index = rand.Next(0, 3);

               // Set "revealed" door where != index and contents != Car
               int revealed = -1;
               for (int x = 0; x < 3; x++)
                   if (x != index && doors[x] != Car)
                       revealed = x;

               // Test success if stay and success if change
               if (doors[index] == Car)
                   StayWins++;
               else
                   ChangeWins++;
           }

           // Output results
           Console.Write("[stayWins:{0}][ChangeWins:{1}]", StayWins, ChangeWins);
       }

P.S. If you think the last part is dubious consider that there are only three options and that after the reveal there are only two options and just one goat. Thus, every time the selected door doesn't contain the car, changing to the only other available door would win.

This could be written quite redundantly in this way.


	// Test success if stay and success if change
               if (doors[index] == Car)
                   StayWins++;
               else
               {
    		// The following will always increment ChangeWins
                   for (int x = 0; x < 3; x++)
                       if (x != index && x != revealed && doors[x] == Car)
                           ChangeWins++;
               }

I think if you understand what's happening here (that is, the "reveal" rule and the then perfectly straightforward logic of winning) you will understand why the often counter-intuitive Monty Hall Problem odds are correct.

So if I do this 1,000,000 times what is my probability is it 1/2, 1/3, 999,999/1,000,000 ??? Why or why not? How can we calculate this? Can someone write a test for this one as well?

I'm not 100% sure of what you're asking. You can change the iterations parameter in the code I posted to 10^6 if you want. Here is a sample of the results after doing so. Statistically it isn't going to be different than the 10^3 iterations.

[stayWins:332556][ChangeWins:667444]

[stayWins:333503][ChangeWins:666497]

[stayWins:333336][ChangeWins:666664]

[stayWins:333693][ChangeWins:666307]

[stayWins:333350][ChangeWins:666650]

[stayWins:333231][ChangeWins:666769]

[stayWins:332359][ChangeWins:667641]

[stayWins:333530][ChangeWins:666470]

[stayWins:333827][ChangeWins:666173]

[stayWins:333831][ChangeWins:666169]

[stayWins:332490][ChangeWins:667510]

[stayWins:333110][ChangeWins:666890]

[stayWins:333443][ChangeWins:666557]

This is effectively: StayWins=1/3; ChangeWins=2/3. Same as before. If you sum the StayWins and divide by the total number of samples you get the mean.

mu = 4332259 / 13 = 333250.69230769230769230769230769.

The tendency as you increase the sample size is to approach 333333.33...

Since a failed StayWin implies a ChangeWin we can infer that its tendency is to approach 666666.66...

And of course their sum, namely 999999.99..., is equal to 1000000. The code is there for you to play with if you don't believe me.

The interesting question to me at this moment is psychological. Why is this utterly counter-intuitive for so many people? I fell for it initially. Is this an example of a logical defect in human cognition? What other defects of this kind exist? How does this bias our ability to make rational decisions? And so on.

Edited March 31, 2012 by the asinine cretin

Appolinaria · March 30, 2012

I'm not 100% sure on what you asked me. Perhaps you could re-state the question, maybe adding a little redundancy just in case I still don't follow. I would be happy to attempt a formal answer if I knew exactly what was being asked.

I think part of the issue here, is that the simplest way to proove this is by simple exhaustion. There are numerous tables and tree diagrams on the wiki page that constitute a proof from a frequentist standpoint.

If you prefer symbol pushing, then you can throw Bayes theorem at it repeatedly, and there are a number of ways to approach the answer.

The amount of exposition present is because the answer depends so much on the host's behavior. It's very important that he makes the same decisions every time, otherwise you get different priors, or a different distribution of outcomes. Also, many of us consider the proof to be rather trivial (as I stated, exhaustion is sufficient, or even pointing out the number of distinct states and outcomes) and that all of the difficulty to be had is in adjusting one's intuition.

I think part of the reason you haven't gotten satisfactory clarification is that we (or I at least) are(am) having a great deal of difficulty understanding exactly what it is you are pointing out or what you're asking. This is probably due to an unstated assumption on someone's part, or an unstated knowledge differential. Perhaps you could break things down into much simpler statements (and over-explain slightly) until said confusion clears up.

The reason we're concerned with your standpoint is that this community is here to help people learn and understand things. If we perceive you do not understand something we will endeavour to help you (even if we sometimes do it rather inefficiently).

This is also an excellent example of how intelligent and mathematically experienced people can have difficulty understanding the Monty Hall problem or accepting the conventional answer. Ie. a good example of 'why is it so controvercial?'

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