1. Naturally I would accept a computer program as evidence. Would be odd to say no. However we will have to agree on the algorithm and this whole thing gave me a terrible headache. I'll check macgyver's later.

over and out.

2. Originally Posted by laserlight
EDIT:
Heh, MacGyver already wrote one, apparently.
Not a very good one, I'm afraid. I suspect there may be issues with it, plus it's not suited for receiving lots of data.

I can try rewriting another one later for that purpose, but someone will probably beat me to it.

3. No matter what happens, the host will remove one of the donkeys. So then it's 50/50.

Does anybody care to try to disprove this?

4. Ok, my last attempt to explain it:

There is a 2/3 probablility of winning by switching from the outset. Your odds don't change once you select the first door. This is because, at the start there will allways be a 2/3 probability that one of the doors you do not chose to begin with will have the money in it.

First off these are all possibilities of what can be behind the doors:
Code:
```       Host  |     You
donkey    donkey |  money
money     donkey | donkey
donkey    money  | donkey```
It needs to be understood that the host has to remove a donkey from the following possibilities:
Code:
```donkey, donkey
money, donkey
donkey, money```
This means that there are three possibilities as to what the host can have behind the final door:
Code:
```donkey
money
money```
The door you picked initially also has three possibilities to the outcome, but they are the same as from the start:
Code:
```money
donkey
donkey```
Now compare the possibilities of what the host can have against what your original door can have:
Code:
```Host  |  You
donkey| money
money | donkey
money | donkey```
This is because the donkey that the host removes only affects the probability that one of his doors have money behind it, it has no effect on the door you picked. Therefore there is a predetermined 2/3 probability that by switching you will win the money.

5. >> Does anybody care to try to disprove this?
I don't know about "proof", but unfortunately it is wrong.

As I mentioned earlier, the host's choice of what to open depends on your original choice of your door, so you cannot start over with a new probability just because a door has been removed.

If the host opened a door at random and it ended up being a donkey, then the chance would be 50-50. But the host doesn't open a door at random. The host opens the door based on your choice and his knowledge of what is behind the doors.

6. You can also think about this alternative scenario. There are 50 doors. You pick 1. The host opens 48 doors that he knows all have donkeys. Should you switch?

7. Think I messed up my original program.

Try this one:

Code:
```#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int doRound(void);
int getRandomDoor(void);

int doRound(void)
{
int iRightDoor, iGuess;

iRightDoor = getRandomDoor();
iGuess = getRandomDoor();

return (iRightDoor!=iGuess);
}

int getRandomDoor(void)
{
return rand() &#37; 3;
}

int main(int argc, char *argv[])
{
int i, times = 100, iSwitch = 0, iStay = 0;

srand(time(NULL));
if(argc > 1)
{
times = (int)strtol(argv[1], NULL, 10);
}
for(i=0;i<times;i++)
{
if(doRound())
{
iSwitch++;
}
else iStay++;
}
printf("Switching was appropriate %d times\n", iSwitch);
printf("Staying   was appropriate %d times\n", iStay);

return 0;
}```
My output for 4 iterations:

Code:
```Switching was appropriate 69 times
Staying   was appropriate 31 times

Switching was appropriate 70 times
Staying   was appropriate 30 times

Switching was appropriate 65 times
Staying   was appropriate 35 times

Switching was appropriate 65 times
Staying   was appropriate 35 times```
Provided there are no bugs, I think this could be submitted as evidence that Daved, CornedBee et al are correct, especially since the difference is about 2:1... ie. 66% chance in getting it right to make a switch.

8. Originally Posted by Daved
If the host opened a door at random and it ended up being a donkey, then the chance would be 50-50. But the host doesn't open a door at random. The host opens the door based on your choice and his knowledge of what is behind the doors.
I need to dissect this because I believe herein lies the crux of the problem.

> If the host opened a door at random and it ended up being a donkey, then the chance would be 50-50.

Take notice of this statement: If it happened to be a donkey, the chance would be 50-50.

>
The host opens the door based on your choice and his knowledge of what is behind the doors.

So he knows what door he can open to show the contestant a donkey. He does this all the time, no matter my choice. He can always open a door to a donkey.

So, what you think will be the odds?... 50:50, because that's exactly what the host does: If it happened to be a donkey, the chance would be 50-50.

9. Here's another simulation that runs when your browser loads. It also has an informal explanation of why 2/3 for a switch is correct.

http://edp.org/monty.htm

10. Originally Posted by robwhit
No matter what happens, the host will remove one of the donkeys. So then it's 50/50.

Does anybody care to try to disprove this?
I gave a proof sketch in my last post. The intuition is this:

P(chosen-door) = 1/3
P(host-door) = 0

Thus,
P(remaining-door) = 2/3.

11. dammit.

12. Perhaps you would like to read:
Ask Dr. Math: FAQ: The Monty Hall Problem
Wolfram MathWorld: Monty Hall Problem

For a walkthrough simulation:
Let's Make a Deal

13. If the donkeys and the money were placed behind the doors after the host made his/her choice, I would have been right.

14. >> because that's exactly what the host does: If it happened to be a donkey, the chance would be 50-50.

There's a difference. If the host chooses a door at random, then there is a 1/3 chance he opens the door with the money. But if the host chooses a door on purpose, then that 1/3 chance disappears.

Let's say this happens 30 times. We'll label the doors as such:

Door A: money
Door B: donkey
Door C: donkey

So out of 30 times, 10 times the contestant chooses Door A, 10 times the contestant chooses Door B and 10 times the contestant chooses Door C.

Contestant chooses Door A, host opens Door B five times and Door C five times, all ten times a switch is bad.
Contestant chooses Door B, host opens Door C all ten times (because he can't open door A), switch is good all ten times.
Contestant chooses Door C, host opens Door B all ten times (because he can't open door A), switch is good all ten times.

So the switch is good 20 out of 30 times.

Let's do the same thing when the host doesn't know where the money is.

Contestant chooses Door A, host opens Door B five times and Door C five times, all ten times a switch is bad.
Contestant chooses Door B, host opens Door A five times and Door C five times, switch is good five times, and five times we ignore (because the original problem assumes the host opened a door with a donkey).
Contestant chooses Door C, host opens Door A five times and Door B five times, switch is good five times, and five times we ignore (because the original problem assumes the host opened a door with a donkey).

That's 10 times switch is bad, 10 times switch is good, and 10 times that are ignored because the problem assumes the host opens a door with a donkey.

See the difference?

15. Yup. I can see it now. It's an amazing paradox. Very invigorating. Also took a look at the links and on wikipedia.

It's a pleasure to see how wrong I am. This one is definitely going to my notebook.

Thanks a lot for the patience through the ordeal of trying to make see the light