While making a table of all possibilities is generally a good idea, you have to watch out that all the cases you are mapping out are equally likely. If you look at your table again, you initially pick the right door 50% of the time. That can't be quite right...
1 2 3 K S P N N W L P N N W L
P N N L W
P N N L W
(picking the right door 2 times out of 4?)
You are probably confused because in that case the host has two doors he can open, so he will open each of them 50% of the time. If you compensate for that by weighting these lines with a factor of 0.5, than you'll come out at the right 1/3 against 2/3 probability.
But probably the best way to get a feeling for this problem and how weird probability can be would be to find someone to play the host for you and just try it;)
If it turns out 1/3 to 1/3 no matter if you switch, then your program sometimes opens the door with the big prize in it, I guess.
It really works like this: Assuming you picked door A, then there are three possible situations: -The prize is behind door A -The prize is behind door B -The prize is behind door C
So if you stick to the door you just picked, then you have a 1/3 chance of winning. Now obviously there's a 2/3 chance that the prize is behind one of the two other doors. And the host is basically giving you the information "IF the prize is behind one of the two remaining doors, then it is behind THIS one" by opening an empty door. So switch, it will give you a 2/3 chance of winning.
And yes, I had to write a program, too, before I actually believed it.
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While making a table of all possibilities is generally a good idea, you have to watch out that all the cases you are mapping out are equally likely. If you look at your table again, you initially pick the right door 50% of the time. That can't be quite right...
1 2 3 K S
P N N W L
P N N W L
P N N L W
P N N L W
(picking the right door 2 times out of 4?)
You are probably confused because in that case the host has two doors he can open, so he will open each of them 50% of the time. If you compensate for that by weighting these lines with a factor of 0.5, than you'll come out at the right 1/3 against 2/3 probability.
But probably the best way to get a feeling for this problem and how weird probability can be would be to find someone to play the host for you and just try it ;)
If it turns out 1/3 to 1/3 no matter if you switch, then your program sometimes opens the door with the big prize in it, I guess.
It really works like this:
Assuming you picked door A, then there are three possible situations:
-The prize is behind door A
-The prize is behind door B
-The prize is behind door C
So if you stick to the door you just picked, then you have a 1/3 chance of winning.
Now obviously there's a 2/3 chance that the prize is behind one of the two other doors. And the host is basically giving you the information
"IF the prize is behind one of the two remaining doors, then it is behind THIS one" by opening an empty door.
So switch, it will give you a 2/3 chance of winning.
And yes, I had to write a program, too, before I actually believed it.