Guys I get it now! I convinced myself!

I had to drag out the cards and see that all I was doing was picking 1 out of 3 cards. It was simple from there out.

Once again... Let`s be consistent.

Why the probability of car in the unpicked two doors is 2/3, if we know beforehand, that one door is 100% certain to have a goat?

We have one door picked vs two doors unpicked, ONE of which contains a goat 100%, and WILL be excluded.

So, why two doors get 2/3?

"What is the probability of you winning the car if indeed you change your mind and pick up the other door?"

We know that two doors have goats. One of the goats has been revealed. Thus, we are choosing between two doors, where one has a goat and one has a car. We can only pick one door. Initial choice (1/3 chance to win) has no bearing on the second choice; wether to switch or stay. If you choose to switch, you have a 1 in 2 chance to win. If you stay, you've got a 1 in 3 chance.

My results from the Java simulation linked above gives me these probabilities:
P(win if switch) = 0.56 (28/50)
P(win if stay) = 0.32 (16/50)

sigh, i thought i was pretty clear albeit a little wordy in trying to explain to you why its 2/3. let me say this again:

if you picked a door with a goat in the beginning, switching will get you a car. probability of picking a door with a goat in the beginning, or in other words, the probability of winning by switching: 2/3.

if you picked a door with a car in the beginning, switching will get you a goat. probability of picking a door with a car in the beginning, or in other words, the probability of losing by switching: 1/3.

Thus, 2/3. QED

the probability of winning by switching comes not from the current decision (switch or not to switch), but from your decision in the beginning to pick a door. this is due to the host revealing the goat door after your first decision.

"Just admit that I`m right, and we can save you much embarassment later." - http://forums.massassi.net/vb3/showthread.php?p=796818&#post796818

sorry for being sassy. just pissed off at quantum chemistry.

Aww, I voted from memory and got it wrong. :saddowns:

Sigh. I think we need to cool off a little.

Last argument is good enough, I suppose. If we examine only the probability of winning by SWITCHING and not probability of car being behind either of doors...

yeah sorry.

The host will always reveal a non-prize door, regardless of your initial choice. That's a rule of the game. He will also never reveal the door that you chose.

So if you pick the car initially (1/3), and then switch it's obvious that you'll lose since it's only possible to switch to a non-prize.

If you pick a losing door initially (2/3), the host will ALWAYS reveal what the other losing door is, therefore the other door MUST have the car. So in this situation you should switch.

Since the second round doesn't reveal any new information to the player. The only valid strategy is to go with the above probabilities of winning. 1/3 if you don't switch, 2/3 if you do.

This is one of many mathematical problems formulated to demonstrate how people misunderstand mathematics and statistics. Even very smart mathematicians sometimes get it wrong because even they introduce a few flawed assumptions into their reasoning.

P(a) = 1/3
P[sup]-1[/sup](a) = 2/3

P(b) = 1/2

if you don't switch,
P = P(a) = 1/3


if you do switch,
P[sup]-1[/sup](a) & P[sup]-1[/sup](b) = (2/3) * (1/2) = (1/3)
P(a) & P(b) = 2/3

Oh yeah, i just realized i suck at maths. 0.56 is more likely to be ~0.66 than ~0.5 in this case assuming infi, right?

What happened, I think, is that you just didn't run it enough times. The effect of chance error in terms of percentage decreases with repetition.

Yeah, that's what i meant with "assuming infi"(nite repititions).

Ah, gotcha. Didn't understand the last bit.

Yeah, for my first thirty or so repetitions with the program I was winning almost 80% of the time when I switched.

That's a very good explanation. Too bad I didn't think of it like that before I voted. Meh.

Foolish mortal, you can't change the problem

You must work with the data that is given to you.

It's 2/3 on account of the host KNOWING what's behind the doors and HAVING to show you a goat. If you picked a goat first (2 times out of 3) then the host effectively shows you where the car is with absolute logical certainty. If you picked the car first (1 time out of 3) you're still guessing.

That's a really crap way of explaining it, but it does come out in the maths when done by someone not tired or distracted

Haha, as an engineer I've been taught that if I can't solve a problem I should redefine it

How is it that I got this without any maths? It just makes sense to me. You're likely always going to pick the wrong door, but then you're always going to be shown that the right door isn't in another place. So now you know of two statistically incorrect doors. You pick the other one based on logic alone.

This was MY ARGUMENT on the first page with a picture!