Statistics: My Brain Hurts

So last night I went to see the other couple. This was a lot of fun in general and probably includes a pornalong post in there, though I'm not sure I'm in the mood to tap it out. Sex, oral sex, spanking, threesome, orgasms... it's funny isn't it? Kinky sex is just like regular sex after a while! So that is not the point of today's post. But it does come about because of events from last night.

Earlier in the evening we watched the movie '21' which was a pretty good movie, though it oversimplified a lot of what actually happened and then also gave it a complete Hollywood twist and ending which also deviated from the original true story. Nevermind though. The point is, at some point early on, they are discussing the 'Monty Hall Problem'. Now, don't strain your head, I've read the entire Wikipedia, and I understand the argument of the complex problem.

Except I don't believe in it.

To recap, imagine you are on the show 'Lets Make a Deal' and you can win a car, or a goat. There are three doors, one car, and two goats. The rules are, you must pick a door to start, the host is going to show you a door with a goat (not the one you picked) and then ask you if you want to switch doors or not with the remaining two. Should you switch? Gut instinct tells you that there is one goat and the car left, so you have a 50/50 chance of winning, so you can stay put. But statistical logic tells us that if you switch doors, your chance of winning just went up to 66%. let me explain.

Option 1- You pick the door with the car. The host shows you a goat, you switch, you lose.
Option 2- You pick a goat(A). The host shows you a goat(B), you switch you win.
Option 3- You pick a goat(B). The host shows you a goat(A), you switch you win.

So from your first pick you see that you set in motion a series of choices in which in two out of three cases you win if you switch. Read it again if you don't get it.

See, I read it, I get it, I just don't believe in it. Let me explain my problem.

Lets say the doors open on two sides and that there is an individual on each side who has no contact or knowledge of the other. One one side, there is a person who gets to make an initial choice which influences how the host chooses, and according to the premise, if they switch doors, their odds of winning go up. Now lets say on the other side of the door, there is another person who is simply shown three doors and told that there are two goats and a car. They do NOT get to make the first choice, instead, the same door with the goat is revealed to them (based on the person choosing from the other side) and they are told to choose for the first time. For that person, there is one goat, and one car, and their odds are actually 50/50.

So here is my problem. For the same event, and the same possibility of choice (in the final round of choosing, you have two doors and you can pick one), the probability for the person on one side of the door is higher than the other. Does that make sense? It does not. It does not make sense because the person who had the opportunity to pick in the first round has no higher knowledge of which door the car may be under. Still, if you go back to the wikipedia article, you can read a million proofs about why it works, I just cannot believe that it works.

Now I'm sure that I'm wrong (because I am not a mathematician) but here is my argument for why it's still 50/50.

Option 1- You pick the door with the car. The host shows you a goat(A), you switch, you lose.
Option 1a- You pick the door with the car. The host shows you a goat(B), you switch, you lose.
Option 2- You pick a goat(A). The host shows you a goat(B), you switch you win.
Option 3- You pick a goat(B). The host shows you a goat(A), you switch you win.

My argument is that the host knows he must always show you a goat. But he can show you either goat. When you pick a door with a goat, the host has no choice but to show you the remaining goat (Options 2 and 3). However, when you pick the door with the car, the host has two choices- to show you either goat(A) or goat(B). Following the possible sequence of events, that makes four possible outcomes where the odds are 50/50 each time.

What I understand is the statistical issue with my solution is that you can only pick once, and the host can only pick once. So Option 1 is irrelevant as to whether or not the host picks goat(A) or goat(B) because the outcome is the same.

So mathematics wants me to understand and believe that in my two sided door scenario, the person who got to choose twice will always have an advantage and should always switch doors to have a higher probability of winning, while the person who had no previous knowledge is simply making a random guess between two doors every time and this lowers their odds of winning.

The problem is I simply can't believe it. I cannot accept that at the moment of the second choosing (do you stick with the door you had, or do you switch) that you are in any way better off than the person who is presented with the same doors at that point and told to pick one. Choosing a door the first time around shouldn't carry over. You shouldn't be able to add probabilities. I understand it on paper, but it just does not make sense in my head that the real world works that way.

Can anyone provide a better exlplanation?? Brain. Hurts.