Monte Hall Hoax

[DBTrek]

Looks like a simple decision tree that covers all combinations of possible outcomes.
What did they get wrong?

"Monty opens 2 or 3"

This collapses two loss conditions over multiple iterations of the game into 1. And it happens three times. There aren't 9 possibilities with 6 victory conditions for switchers and 3 for stickers. There are 12 possibilities evenly split between switchers and stickers.

I don't think so man, take another look at it.
It says Monty opens 2 or 3 because you can't choose the door he opens after its contents are revealed.
Ergo, there is no option for Monty opens 2 you choose 2, or Monty opens 3, you choose 3.

It can only be Monty opens 2 you choose 3, or Monty opens 3 you choose 2 (in rows where a switch is made).
Nothing is being collapsed.
The row represents a single situation where Monte opens one door and you select the other.

[Hanarchy Montanarchy]

Like I said, over multiple iterations of the game that is two discreet loss conditions.

That is the bait and switch. The decision tree applies to one iteration where any time you get a goat it is coded as a loss. But, we are told that over multiple iterations of the game the advantage goes to switching, but they don't introduce the two discreet loss conditions that exist when I initially choose the door with a a car back into the game.

Again, the assumption is that you are still playing an n/3 game on the second choice, but, as you point out, I am not able to choose the door with a revealed goat on the second choice. Thus, I am playing an n/2 game. Different odds, different game, stick or switch - it doesn't matter.

[DBTrek]

It's the XOR operand in play, dude.
Exclusive or.
Monte opens 2, and you're switching, you MUST choose 3.
Monte opens 3, and you're switching, you MUST choose 2.

Monte is opening one door, hence the activity exists in one row.

If you don't switch, you're not playing a different game. You're playing the game where (in my example) out of 100 doors, you picked one with a 99% chance of being wrong. Now Monte has shown you 98 sheep and a closed door. But that doesn't retroactively apply to your choice and mean your initial guess was 50% likely to be correct. It was still 99% likely to be wrong. Just as with three doors your initial decision is still 66% likely to be wrong.

[Martin Hash]

Whatever happened previously is immaterial: what if you just walked in when there were only 2 doors left and you had to choose? Can't you tell that the contestant's odds are the same as yours?

[Hanarchy Montanarchy]

You're right, the odds don't retroactively change for your first decision - the new odds merely render your first decision irrelevant.

Take the 100 door version, but now assume you want a goat. I choose door one, Monty opens door 2 and shows me a goat. Then he says, "okay, you can choose door 1 or another door." I say, "Nah, bruh, gimme that goat, I choose door 2."

Then the switch - Monty calls me a fucking dickhead for wanting a goat and tells me to choose another door. Now my odds are 98/99 for every door including door 1.

Now I get the game - Monty is taking away precious goats each time, reducing my chances of winning.

edit: I put the car odds instead of the goat odds.

[DBTrek]

Whatever happened previously is immaterial

That's only true if the odds of a series of events is the same as the odds of a single event.
But that's not the case.
Combinatorics studies the counting of possibilities, and would not even exist if all probabilities were equal to single events.

Like the coin-flipping example. The single event of you flipping a coin and getting heads is 50/50. The odds of flipping heads twenty times in a row is not 50/50 - even though each individual flip has a 50% chance of being heads.

At least - that's what makes sense to me.
Maybe you both are just more keyed in to the real deal - in which case you should fleece me for 66% of my $1000.

[Hanarchy Montanarchy]

Your odds of guessing a coin flip twice in a row are worse than once.

Monty tells me if I guess a coin flip right twice, I get a car - my odds of getting a car is 1/4.

Then he says, actually, you just have to guess right once. Boom! 50/50 I get the car. The last game I played (where I didn't get a car, regardless of whether or not I guessed right or not because Monty always planned on changing the rules) are totally irrelevant.

[Martin Hash]

I’m just here to be the guy who pulls back the curtain; if you choose the illusion even though I’ve completed refuted it, that’s on you.

p.s. The story of my life.

[Hanarchy Montanarchy]

So far you all still think you’re smarter than me...

It’s just a statistical illusion: the odds are 50/50 once one of the goats is revealed.

Sorry, I didn't see this.

You're right on both counts. I think I am smarter than you and the odds are 50/50 on the second choice.

[Martin Hash]

You guys think: how can this unfamous guy possibly be right when all these famous people think otherwise?