The Monty Hall probability proof always baffled me. Read more in
Wikipedia for the description.
But for those who don't know, and don't want to know the whole deal I'll give the tl;dr friendly version.
Assume that you're presented with 3 boxes (A, B, C) and are asked to choose one. One of them contains a prize. After you've chosen, I'll open an empty box from the other two. Now, you have the option of sticking together with your choice, or, you may change your choice. According to math, your chances of winning are more if you switch.
Let me explain in an easier way, you choose box B. Then I open box C and show that it is empty. Again, I ask you to choose between A and B. Are you better off by choosing A?
Common sense wrongly suggests that switching between A and B makes no difference. (probability remains 1/3rd - wrong intuition). Psychology makes us stay with B (people are reluctant to change). But, probability shows that switching produces a definite boost to our chance of winning - double!
Though I understood the math, I really wanted to verify.
So here is some crappy C simulation.
Let us not switch for 100 times and see how much we win.
Later, we'll switch for 100 times after choosing, and see how much we win.
Code:
#include
#include
#include
int doMontyNoSwitch();
int doMontySwitch();
int main(void)
{
int choice = 0,i;
int winCountNS=0,winCountS=0;
//Seed the randomizer
srand(time(NULL));
//100 times without switching
for(i=0;i<100;i++)
{
//Increment win count for no switch if we win
if(doMontyNoSwitch() == 1) winCountNS++;
}
for(i=0;i<100;i++)
{
//Increment win count for switch if we win
if(doMontySwitch() == 1) winCountS++;
}
printf("\nNS:%d S:%d",winCountNS,winCountS);
return 0;
}
//Here we'll switch
int doMontySwitch()
{
int r1,correct,choice,reveal,switchedChoice,i;
r1 = rand()%3;
//Let me randomly set the correct box/door
correct = r1;
//Choose randomly
choice = rand()%3;
for(i=0;i<3;i++)
{
//Show the door/box which does not contain the prize and what I didn't choose
if(i!=correct && i!=choice)
{
reveal = i;
break;
}
}
for(i=0;i<3;i++)
{
//Reveal the box which was not chosen and the wrong one
if(i!=choice && i!=reveal)
{
//Change our choice her, that is, "the switch"
choice = i;
break;
}
}
if(choice == correct)
{
//return 1 to indicate we've won here
return 1;
}
else
{
//return 0 to indicate we've lost here
return 0;
}
}
//Without switching
int doMontyNoSwitch()
{
int r1,correct,choice,reveal,switchedChoice,i;
r1 = rand()%3;
correct = r1;
choice = rand()%3;
for(i=0;i<3;i++)
{
if(i!=correct && i!=choice)
{
reveal = i;
break;
}
}
if(choice == correct)
{
return 1;
}
else
{
return 0;
}
}
So, when I compile and run this thrice,
NS:31 S:63
NS:32 S:62
NS:25 S:60
The probability is right. So right! Ideally it should have been 33 and 66, but it is so damn close, wow!
So, next time I'm going to switch no matter what!
By the way, the code above is not well designed or thought out, I'll work on it later....
