Problem with the Montey Hall puzzle

[Wonko the Sane]

Okay, I'm very very tired of coming to this website, looking through the archives once in awhile, and seeing the very ignorant solution posted to the Montey Hall problem. It says that you have a 1 in 3 chance of picking the right door and that when one wrong door is removed, it increases the chance of the one you didn't pick. That's wrong, you have to look at this from a different perspective to see why.

This puzzle has two distinct parts.

1) There are three doors, two wrong, one right. You pick one. Then the host opens one wrong door and you have the choice two switch.

2) Here the system has been reduced. The denominator has changed. There are not 3 doors anymore, but two. You aren't allowed to pick the third door, thus there is a 50-50 chance that you are right. Here is what is happeneing, in effect. It's almost like the host revokes your guess, opens a wrong door, and tells you to pick again. It isn't one system being reduced. It's one system being destroyed and an entirely new one being formed. Thanks for bearing with me.

[Quailman]

This problem was dealt with in depth in the old forum. There were a lot of explanations, but the one that seemed to me easiest to understand was to alter the situation and consider the extreme example of 100 doors. You know after you choose one:

1. That there is a 99% chance that the prize is behind a remaining door.

2. That there is a 100% chance that there are 98 loser doors remaining. (Maybe 99).

Now if Monte opens the 98 loser doors that you already knew were there, nothing has happened to change the first fact. There is still a 99% chance that the prize is behind one of the 99 unchosen doors.

I forget who posted this, but increasing the odds differential seemed to make it really clear. Anyone who didn't understand it by this point - and there had already been much more discussion - was just pigheaded.

[This message has been edited by Quailman (edited 12-10-1999).]

[Murray]

Note to forum regulars:

Unless there is some honestly new development to discuss in regard to the Montey Hall problem, let's have a moratorium on any topic relating to it.

There is plenty of discussion and insight into the problem in the old forum and there's a fair amount in this one. Anyone who is curious about the problem has plenty of access to information about the problem on this and other websites.

Let them use those resources and let us not drive ourselves mad trying to convince the unconvincible.

END THREAD.

[Ghost Post]

There are three equally likely possibilities:
A - door 1 wins
B - door 2 wins
C - door 3 wins

Suupose I geuss door number 1.
1/3 of the time (A) I am right to begin with, and I will lose if I switch.
The other 2/3 of the time (B & C) I am wrong to begin with, and will win if I switch.

The same argument applies if I geuss door number 2 or door number 3.

So 2/3 of the time I win if I switch.

Of course, you can always convince yourself, beyond all reasonable doubt, experimentally.

Do the following 100 times:
Have someone place three playing cards, 1 red and 2 black, face down, in a random order. Pick which one you think is red. Have the person turn over one of the black cards from among the two you didn't pick. Then, switch. That is, turn over the remaining card you didn't originally pick (not the one that you now know is black, of course). Keep a tally of how many times you win (pick red).

Now repeat again 100 times, but this time don't switch. Stick with your original pick each time.

You'll see that in the first experiment, you'll win about 2/3 of the time, while in the second, you'll win about 1/3 of the time.

Of course, you can do the experiment by using the java applet on the solution page, at the bottom of:
http://www.greylabyrinth.com/Puzzles/answer020.htm

I'd love to know how anyone can argue with their own experimental observations.

And, in case you haven't seen it, the java applet is written in such a way that you can be sure it isn't cheating (for instance, by switching around what's behind the doors after you've made your pick).

[Ghost Post]

Wow, people who don't understand this puzzle sure do get worked up over it. I haven't seen this level of zealotry since Martin Luther King nailed up his 7 deadly commandments or whatever...

[Andy]

Wonko - you are not likely to be convinced unless someone identifies a flaw in your reasoning, so here it is:
You said, "It's almost like the host revokes your guess, opens a wrong door, and tells you to pick again."
If that were really the case, then the odds would be 50/50 as you propose. That would require that the host open a randomly-selected wrong door each time. 1/3 of the time, you would have selected the right door, and you lose by switching. 1/3 of the time, you would have selected a wrong door, the host opens the other wrong door, and you gain by switching. 1/3 of the time, you would have selected a wrong door, the host opens the door you chose; 1/2 of those times (1/6 of all times) you gain by switching, and 1/2 (1/6) you still lose. Total - 1/3 you lose by switching, 1/2 you gain by switching, 1/6 you lose either way; or, 1/3 you win by staying, 1/2 you lose by staying, 1/6 you lose either way. Stay and win 1/3, or switch and win 1/2.
In the actual situation, the last of these three scenarios isn't played out the way I described it. The host never opens the door you chose, but opens the other wrong door instead. The effect of this is that the 1/6 of the time you would lose either way is converted to win by switching. Total - 1/3 you lose by switching, 2/3 (1/2 + 1/6) you gain by switching. You can also work out the odds for other variations where the host might open the winning door but happened to pick another when you were the finalist.

[This message has been edited by Andy (edited 12-13-1999).]

[Wonko the Sane]

Finally gotten back here after awhile. Your logic is missing a very important case. You listed.

1/3 of the time you choose the right door
1/3 of the time you choose the wrong door and win by switching
1/3 of the time you choose the wrong door and lose by not switching

You're missing something. It should be this
1/4 of the time you choose the right door and win by not switching
1/4 of the time you choose the right door and lose by switching
1/4 of the time you choose the wrong door and win by switching
1/4 of the time you choose the wrong door and lose by not switching

This means, 50% of the time you win, 50% of the time you lose.

See? Not matter how you word it, the chance is still 50%.

[Ghost Post]

Wonko writes:
> 1/4 of the time you choose the right door and win by not switching
> 1/4 of the time you choose the right door and lose by switching

How is it, that out of three doors, 1/2 of the time (1/4 + 1/4) you choose the right door?

Also, why is it that if you actually run an experiment simulating the problem, and repeat it many times, and always switch, you win about 2/3 of the time?

[Evil Empire]

Wonko,Wonko,Wonko

You Listed:

You're missing something. It should be this
1/4 of the time you choose the right door and win by not switching
1/4 of the time you choose the right door and lose by switching

So add it up and 1/2 of the time you select the right door. You must be one hell of a guesses to select the right door our of 3 1/2 the time. Hats of to you.

Looks like extro. beat me to the punch.

[This message has been edited by Evil Empire (edited 12-16-1999).]

[Andy]

Wonko: you said that I said
1/3 of the time you choose the right door
1/3 of the time you choose the wrong door and win by switching
1/3 of the time you choose the wrong door and lose by not switching

That's not exactly what I said. Try this:

In the original game:
1/3 of the time you choose the right door. Monty opens one of the wrong doors. You can win by staying or lose by switching.
2/3 of the time you choose the wrong door. Monty opens the other wrong door (it's the only door he can open within the identified practices). You can lose by staying or win by switching.

In the variation where Monty opens either wrong door at random, this becomes:
1/3 of the time you choose the right door. Monty opens one of the other doors. You can win by staying, or lose by switching.
1/3 of the time you choose a wrong door. Monty opens the other wrong door. You can lose by staying, or win by switching.
1/3 of the time you choose a wrong door. Monty opens the door you chose. You can lose by staying (but why would you?), or you can switch. If you switch (the recommended action), you will pick the other wrong door 1/2 of the 1/3 of the time, and you will pick the right door the other 1/2 of 1/3 of the time .
Summary:
1/3 stay and win, switch and lose.
1/3 stay and lose, switch and win.
1/6 stay and lose, switch and win.
1/6 stay and lose, switch and lose.
This looks like 50/50 for win/lose - but it isn't 50/50 for stay or switch; if you always stay, you will win only 1/3. If you always switch, you will win 1/2. If you switch when Monty opens your door, it doesn't matter which you do when he opens another door.

Here's another variation for you to analyze:
You pick a door. Monty then opens a randomly-selected door. He may open the door you chose or one of the others. He may open the winning door or one of the losers. After he opens a door, you may keep the door you chose originally or switch to either of the other doors. Assuming that you will always pick the winning door if he opens it, and will never pick a losing door that he opens, what's you best strategy and what is your expected long-term winning percentage? Assume that 'random' includes 'evenly distributed.'

[Wonko the Sane]

Alright, I'm going to just run my own experiment on this with a C++ algorithm in a few minutes here. If it comes out at close to 2/3s after say 50000 trials, then I'll trust you guys on this one. It just seems to be a strange perversion of statistics.

[Wonko the Sane]

I'm going to have to do more tests, because based on my program, we're both wrong. It yieled a ratio of 58.5:41.5 win-loss for switching. Then about 29:71 win-loss for not switching. Which means that chances of winning for not switching is significantly lower than expected, but the same is true for switching. I got the same results when I ran 5 trials of 50,000,000 guesses each. Someone might want to check my code and see if there's a problem. I'm pasting it at the bottom. My next project is to set this program up for a system of any number of variables. I'll post the results on that once I finish it.

//Compiled with GCC 2.something
#include <iostream.h>
#include <stdlib.h>
#include <conio.h>
#include <time.h>

main()
{
time_t t;
srand((unsigned) time(&t));
clrscr();
int j,k,e,f,w=0,l=0;
bool a[3];
bool c;
cout << "How many trials? ";
cin >> j;

for (; j > 0; --j)
{
c = false;
for (int i = 0; i < 3; ++i)
{
k = rand() % 2;
if (k == 1 && c == false)
{
c = true;
a[i] = true;
}
else a[i] = false;
}
f = rand() % 3;
do
{
e = rand() % 3;
} while (a[e] == true | | e == f);
for (int i = 0; i < 3; ++i)
{
if (i != e && i != f)
{
f = i;
break;
}
}
if (a[f])
++w;
else
++l;
}
cout << endl << "Wins: " << w;
cout << endl << "Losses: " << l;
return 0;
}

[kitakaze]

Sorry, I can't read the java code. I'm not sure what you are trying to do.

Wonko, read Quailman's post carefully. (That was originally my post on the old forum.)

I didn't understand it either, until I analysed it in detail. To me, the answer should have been the same as if I just walked into a room and saw 2 doors. But it clearly isn't.

When I wrote the test program, I didn't even have to run it to recognise that my logic was false.

Think about this carefully, if your strategy is to switch, you want to be wrong on the first pick. Your chances of being wrong on the first pick are 2/3.


That's all I want to say on this.

Kaze

[Andy]

Wonko - nice little program, which suffers from a couple of flaws. First, it's not as easy to read as it could be; more comments and better visual formatting would help (try using the [ CODE ] and [ /CODE ] tags).
Secondly, the winning percentages you are reporting seem close enough for the game you're playing - but it isn't the Monty Hall game; you're hiding the prize as follows:
Door # 0 - 1/2
Door # 1 - 1/4
Door # 2 - 1/8
no prize - 1/8.
Your code:

code:

---

{


c = false;


for (int i = 0; i < 3; ++i)


{


k = rand() % 2;


if (k == 1 && c == false)


{


c = true;


a[i] = true;


}


else a[i] = false;


}

---

Another version:

code:

---

{


   c = false;


   for (int i = 0; i < 3; ++i)


   {


      k = rand() % 3;


      if (k == i)


         a[i] = true;


      else


         a[i] = false;


   }

---

This hides the prize behind each door 1/3 of the time.

[Wonko the Sane]

I agree, the logic on that code wasn't very wonderful. I usually comment more and trust me, my spacing is much better. But as you can see, cut and paste doesn't preserve the spacing and I didn't have a lot of time to work on it. Nevertheless, it wasn't wonderful code. The one I posted on the other thread is much better. Works for any number of items. I think I used the wrong modulus on it again, I'll look at it in a minute. Thanks for the revision.

[Wonko the Sane]

Now that I had a chance to look at it again, I was right. The way it works is like this.

c = false; //initialize c (tells if a number has been chosen or not)
for (i = 0; i < 3; ++i) //i is a door
{
k = rand() % 2; //get 1 or 0
if (k == 1 && c == false) //if k is "true" and flag is false
{
c = true; //number chosen
a[i] = true; //make a[i] true
}
else a[i] = false; //set a[i] to false
}

Then it randomizes a number, call it j, between 0 and 2 (references in C++, for those who don't know, start at 0). If j is true, then you selected the right door and lose by the switch, otherwise you win. So it just returns not j.

[Andy]

Better comments - and you're right, cut-and-paste and haste are hard on spacing.
Still, you're not always hiding a prize (you can get 0 three times). Is this intended?

[Andy]

Wonko - P.S. I just read your code in the other thread (I missed it before) and it has the same problem - You're hiding the prize behind door # 0 half the time, etc. - and have no prize once in 2^q times (rarely, when q = 52).

[Wonko the Sane]

I noticed that. I wrote a new verion that just gets two random numbers and returns the inverse of the comparison. Runs a load faster. I was posting it and locked up, but I'll post it again this afternoon. Thanks for the debug.

[HyToFry]

I dont know if anyone has ever pointed this out, so forgive me if i "beat a dead horse" .

The quickest way to explain this is to say "Would you rather choose two doors, or just one?". You will most likely want to choose two doors (unless your goofy) and you are essecially choosing two doors if you deciede that you will ALWAYS switch.

Lemme give you my explanation of this.

Door "A" contains a chocolate bunny, "B" a chocolate bunny and "c" the golden bunny.

You can choose two doors IF AND ONLY IF you are 100% sure that the host will open one of the incorrect doors after you choose.

if you think the bunny is either in "A" or "C", YOU'LL WIN by choosing "B" having the host open door "A" and switching to "C"

if you think the bunny is in either "B" or "C", YOU'LL WIN by choosing "A", having the host open door "B" and switching to "C"

if you think the bunny is in "A" or "B" however YOU WILL LOSE by picking "C", having the host open "A" or "B" and switching to the other.

Now i have given you three possible sinerios two of which turn out to win and 1 that losses.

this means 2 out of 3 times, YOU'LL WIN or 66% you get the golden bunny
and 1 out of 3 times you'll lose or 33% you'll lose and get chocolate (which sounds good to me anywayz).

FIND A FAULT IN MY LOGIC, I DARE YA

I found out that blazer has already touched base on this subject, sorry bout that blazer

[This message has been edited by HyToFry (edited 02-04-2000).]

[Wonko the Sane]

Here's an easier way to think it out. Much, much easier that I had to think of when I wrote the piece of code that I never posted.

Imagine taking 2 random numbers from 1 to 3. The first is what door the prize is in. Once that's set, it's static. The second one is the one that matters. That one has a 33% chance of being the same as the first number. If it is, then you lose by switching. The other 2 times, however, the numbers are different, in that case, the wrong choice is removed and you win when you switch, 66% of them time.

Switching, in essence, removes options. In a game with 3 doors, it removed 2 options. The two that it removes are the one you have, and one of the wrong doors. if 66% of the time, the one you have it wrong, then switching eliminates it (You have no risk of picking the other wrong door since it has been removed) and you win. If you had the right door, then 33% of the time you eliminate the correct door. Simple as that.

[Ghost Post]

1/3 Pick right door - switch you lose
2/3 Pick wrong door - switch you win

2/3 chance of winning by switching

[Ghost Post]

It doesn't get any simpler than that. Yet still complete.

[Wonko the Sane]

It might not get simpler than that, but saying 1/3 switch you lose, 2/3 switch you win is exactly what the solution says. That wont convince anyone that doesn't believe it already.