Come forth and be confused!

[aperson]

Hey dumbass, are you saying that if I have a lottery ticket, I have a 50% chance of winning? Because that's what your argument would state in that manner.

The removal of a box MODIFIES THE CHANCES OF THE ORIGINAL BOX BEING CORRECT.

[Lupin_the_3rd]

yes you would have a 50% of winning with the lottery ticket if all 32532343224 others didn't win
there is you and there is another guy
both have like a 99.9999% chance of winning
however there are two, each has an equal chance of winning

[aperson]

Look over the proof, learn math, look at more of my statements.

You have a 33% chance initially.

That chance doesn't just magically turn into 50% chance when a door disappears.


Edit: I'm writing a program to prove you wrong right now.

[Lupin_the_3rd]

if you are talking about you door compared to all the others, your math is correct. If you are talking about your current chance of winning a prize, it is 50%...both doors have a 66% chance of winning, regardless if its "your" door or not.

[aperson]

Sorry, the current chance is 66%, try again.

[Moogy]

Hey people, aperson obviously knows what he's talking about here. You guys do not. Stop trying to prove him wrong, because he IS correct, I will vouch for that.

[Lupin_the_3rd]

so you're telling me that because it's "your" lottery ticket, you have a better chance than another guy?
if 10000000 people have NOT won, there are two people left. Do you say, "i have a better chance than you because its my ticket"? No
you BOTH have an equal chance. there is no way that can be wrong....

[aperson]

No, I'm assuming the condition that YOUR ticket will NOT be removed.

[Lupin_the_3rd]

if thats the case, we are thinking of two separate problems

[jewpinthethird]

What is the prize? Because this is too much work.

[aperson]

Here's the proof.

#include <iostream.h>
#include <stdlib.h>
#include <time.h>

/*This program is to prove the infamous door riddle,
Yes, the chances are 66% that switching will make you win.

Written by Tyler Mitchell &#40;aperson / maphive&#41; Jan 10, 2003*/


//For simplicity, door 1 will always be the original choice.
//Every game will be switched, the projected outcome should be 66% win, 33% loss &#40;approx&#41;

int main&#40;&#41;
&#123;
	int seed;														//Number to be rolled as a seed for choosing box
	int runs;														//How many times you want to be proven wrong
	srand&#40;time&#40;NULL&#41;&#41;;												//Seeds the rand to time to randomly choose a door.
	bool door1, door2, door3;										//Flags for the three doors, bool is a yes or no operand
	int choice;
	int removed;
	int wins = 0, losses = 0;
	
	cout << "Tyler Mitchell &#40;aperson / Maphive's&#41; 'The Door' program &#40;c&#41;2003\n\n";
	cout << "Every run switches doors\n ...A projected 66% wins to 33% losses should be observed\n";
	cout << "For best results try numbers over 1,000\n\n\n";
	cout << "How many runs would you like to play &#40;larger number = more accuracy&#41;&#58; ";
	cin >> runs;

	for&#40;int i = 0; i < runs; i++&#41;									//Loop until 'runs' runs have been done
	&#123;
		door1 = false, door2 = false, door3 = false;				//Reset values each run
		seed = rand&#40;&#41; % 3;											//Roll a random seed to choose which door is a winner
		switch&#40;seed&#41;												//Assign value from seed to the lucky door.
		&#123;															//This does NOT make all doors true, only one.
		case 0&#58;														//Number discrepancies come from the modulus &#40;%&#41; command
			door1 = true;
			break;
		case 1&#58;
			door2 = true;
			break;
		case 2&#58;
			door3 = true;
			break;
		default&#58;
			cout << "Sanity check failed, door seed error!\n";
			return 0;
			break;
		&#125;

		if&#40;rand&#40;&#41;%2 == 0&#41;											//More randomization for further proof
		&#123;
			if&#40;door2 == false&#41;
				removed = 2;
			else
				removed = 3;
		&#125;
		else
		&#123;
			if&#40;door3 == false&#41;
				removed = 3;
			else
				removed = 2;
		&#125;

		if&#40;removed == 3&#41;											//Switch to the door that wasn't removed
			choice = 2;
		if&#40;removed == 2&#41;
			choice = 3;

		if&#40;&#40;choice == 2 && door2 == true&#41; || &#40;choice == 3 && door3 == true&#41;&#41; //If the choice made is the correct door
			wins++;													//Increment wins by one
		else														//If the choice made is the incorrect door
			losses++;												//Increment losses by one
	&#125;
	system&#40;"CLS"&#41;;
	cout << "After executing " << runs << " runs the following was observed&#58;\n";
	cout << "Wins&#58; " << wins << " or " << double&#40;wins&#41;/runs*100 << "%\n";
	cout << "Losses&#58; " << losses << " or " <<  double&#40;losses&#41;/runs*100 << "%\n";
	
	return 0;
&#125;

Sample runs:

After executing 1000 runs the following was observed:
Wins: 666 or 66.6%
Losses: 334 or 33.4%

After executing 10000 runs the following was observed:
Wins: 6697 or 66.97%
Losses: 3303 or 33.03%

After executing 100000 runs the following was observed:
Wins: 66685 or 66.685%
Losses: 33315 or 33.315%

After executing 1000000 runs the following was observed:
Wins: 666185 or 66.6185%
Losses: 333815 or 33.3815%

After executing 1337 runs the following was observed:
Wins: 883 or 66.0434%
Losses: 454 or 33.9566%



In conclusion, I win.

[VxDx]

I'm really lazy, but your algorithm must be flawed because you are wrong. That or you explained the problem wrong. Quite probably the latter.

It's this simple. There are 3 door. You pick one, and one incorrect one is eliminated. These events are NOT RELATED IN YOUR PROBLEM. there is no relevance in your problem as to which door is chosen to be eliminated, and it really doesn't matter. All that matters is that after an inherantly incorrect door is removed there are two possibilities. Either you chose correctly or incorrectly. The chances are 50-50 that you will win, so which door you pick doesn't matter.

Your logic is flawed nevertheless. If you choose correctly there are two doors that can be opened, if you choose incorrectly, only one. I'll list all the equally likely possibilities in prize, choice, door opened - win if you stay format
112 y
113 y
123 n
132 n
213 n
221 y
223 y
231 n
312 n
321 n
331 y
332 y

that gives even chances of winning at 6/12 or 50-50

[QreepyBORIS]

I believe it is a 50-50 percent chance.

Like the lottery thing- 10 million tickets are bought. Assume each person gets one. All the sudden, we know that 9,999,998 people dont have the winning ticket. You are not one of these people. You could win.

Its a 50-50 chance.

[VxDx]

the lottery doesn't work like that because there is no assurance that there is any winner at all.

[aperson]

I'm really lazy, but your algorithm must be flawed because you are wrong. That or you explained the problem wrong. Quite probably the latter.

It's this simple. There are 3 door. You pick one, and one incorrect one is eliminated. These events are NOT RELATED IN YOUR PROBLEM. there is no relevance in your problem as to which door is chosen to be eliminated, and it really doesn't matter. All that matters is that after an inherantly incorrect door is removed there are two possibilities. Either you chose correctly or incorrectly. The chances are 50-50 that you will win, so which door you pick doesn't matter.

Your logic is flawed nevertheless. If you choose correctly there are two doors that can be opened, if you choose incorrectly, only one. I'll list all the equally likely possibilities in prize, choice, door opened - win if you stay format
112 y
113 y
123 n
132 n
213 n
221 y
223 y
231 n
312 n
321 n
331 y
332 y

that gives even chances of winning at 6/12 or 50-50

Wrong, look at the post I have with the exhaustive proof. See the split in the third list of code? It's amazing.

And you're a programmer, look at my code and tell me where there is something wrong. What's that, it's all correct.

Nice try.