[contoraria]

I still doubt the Monty Hall problem (mentioned in the wiki-article). I understand the reasoning, summing the probability tree, but there are two ways to build the probability tree, with the step of reveal or without. Since the player doesn't know that step, why would it be part of the tree? Because we know the setup a priori, in a way.

I'm assuming the bigger tree is still incomplete. More levels can be inserted, reflecting the reasoning of the a priori knowledge. I assume the supposed advantage would thus prove to be imaginary. Or in simpler terms:

Normally the tree would have the sequence in the following order: box distribution, choose a box, reveal one empty box, switch choice or don't.

However, if you are predetermined to switch, then the order would be changed to switch before the reveal. Thereby, the reveal is irrelevant to the result of switching and the probabilities are equal again.

[Myrmornis]

You make your initial guess. What's the probability you're right on your initial guess? 1/3. What's the probability you're wrong? 2/3.

So now, you're presented with your second decision. Do you stick or switch? Well, if you stick, that's going to lead to success if and only if you were right on your first guess. And if you switch, that's going to lead to success if and only if you were wrong on your first guess.

We all agree that you were more likely to be wrong on your first guess. So you should bet on the fact that you were initially wrong. I.e. switch.

[taejo]

>However, if you are predetermined to switch, then the order would be changed to switch before the reveal. Thereby, the reveal is irrelevant to the result of switching and the probabilities are equal again.

Before the reveal, you don't know which box to switch to. Monty knows which boxes are empty, and he gives you partial information about that by revealing one of them.

[not_kurt_godel]

I feel similarly about Monty Hall. I literally just made a spreadsheet of all the possible outcomes, and sure enough, switching leads to winning 2/3 of the time. Even though I can see the indisputable data in front of my eyes, my brain just won't fully accept it. Looking at it, I realized an interesting little wrinkle that I think might have to do with the problem's psychological quirkiness in some way I can't fully articulate: When you stay you win 1/3 of the time. When you switch, you win 2/3 of the time. BUT if you randomly pick between staying and switching, you win 50% of the time because (1/2)(1/3) + (1/2)(2/3) = 1/6 + 2/6 = 3/6. There is a correct-ish-ness to the intuitive answer of 50%...it's just not...actually correct.

[contoraria]

PS: of course you can't switch before the game maker reveals a box. But isn't that immaterial? In one of three cases, your first choice is correct. Therefore, in two of three cases, the first choice is incorrect ... so switching should be correct in those two.

[iak8god]

The best way to become convinced of Monty Hall may be to grab a buddy and some playing cards, and run a simulation. 30 rounds or so should be convincing.