Monte Hall Hoax

[Martin Hash]

Let’s go back to DB’s 100 doors: by his logic, you should switch when Monte opens a door, but wait, we’re not done; Monte opens another door and asks if you want to switch again; but wait, there’s more: fucking Monte does it 98 times, virtually guaranteeing the very last time you switch, the 100th door is a car. Fucking A! Who would of thought flipping heads 99 times in a row to get the right answer was mathematically proven?!

To reiterate: DB thinks the odds of getting the car on the 99th pick are still 1:99 if he doesn’t switch.

[Hanarchy Montanarchy]

Also always remind yourself that just because you are presented 2 choices, that doesn't imply a 50/50 probability of 2 possible outcomes.

If I provide you sufficient information to know - using only observed facts and logical reasoning - that one choice involves a 2/3 chance of an awesome blow job and the other involves a 1/3 chance of getting your johnson cut off by a nihilist, you ain't going to leave that decision up to a coin flip.

Probabilities update constantly under given constraints.

Nobody beats the house. Best not play for bjs against the nihilists.

[Hanarchy Montanarchy]

Let’s go back to DB’s 100 doors: by his logic, you should switch when Monte opens a door, but wait, we’re not done; Monte opens another door and asks if you want to switch again; but wait, there’s more: fucking Monte does it 98 times, virtually guaranteeing the very last time you switch, the 100th door is a car. Fucking A! Who would of thought flipping heads 99 times in a row to get the right answer was mathematically proven?!

To reiterate: DB thinks the odds of getting the car on the 99th pick are 1:99 if he doesn’t switch.

Which, incidentally, isn't the Monty Hall equation. In the 100 door Monty Hall problem, your chances are supposed to be 99/100 if you switch.

[Fife]

Let’s go back to DB’s 100 doors: by his logic, you should switch when Monte opens a door, but wait, we’re not done; Monte opens another door and asks if you want to switch again; but wait, there’s more: fucking Monte does it 98 times, virtually guaranteeing the very last time you switch, the 100th door is a car. Fucking A! Who would of thought flipping heads 99 times in a row to get the right answer was mathematically proven?!

To reiterate: DB thinks the odds of getting the car on the 99th pick are still 1:99 if he doesn’t switch.

Which, incidentally, isn't the Monty Hall equation. In the 100 door Monty Hall problem, your chances are supposed to be 99/100 if you switch.

Yes!

In the 100 door game if the host let you chose your first door, then decide whether to have that door, or all 99 of the remaining field, what would you do?

That's what is happening when the host shows you 98 losing doors first (all but one door in the remaining field, and when the host knows the location of the winning door at all times and never reveals it). You get the entire field by simply chosing the only door left. The same logic/math/reasoning applies in the 3 door game, or any game of n doors with n>2.

[Martin Hash]

That's the illusion: that you get more doors... You don't. Only 1 at a time, bub. You don't get to pick any head you want out of the next 98 coin flips: they're independent tosses.

[Fife]

The doors aren't reshuffled before the player makes his choice to stick or switch. The array is static.

You could come up with a game where the bag is constantly shaken, so to speak. No problema. But that's not the game being discussed.

[SuburbanFarmer]

Jesus Christ you guys did 8 pages on this?

The odds were always 1 in 3. Those odds didn’t change when Monte decided to show you one of the goats. It makes zero difference that he showed you a ‘wrong’ choice’, you still don’t know if you were right or wrong.

You now have a 50/50 choice between 2 doors, to see if you hit the 1 in 3, but it makes no difference if you switch or stay. It’s not a random re-shuffled choice. You either hit it or didn’t.

[Martin Hash]

I’m abandoning my 100-door argument because it’s using illusory statistics; I want to stay with logic, but here’s my last stab at it:
Each time Monte takes away a door, the odds of your original door having the car are increased by 1, all the way up to 1:2, the same odds as the unopened door.

[Hanarchy Montanarchy]

Jesus Christ you guys did 8 pages on this?

The odds were always 1 in 3. Those odds didn’t change when Monte decided to show you one of the goats. It makes zero difference that he showed you a ‘wrong’ choice’, you still don’t know if you were right or wrong.

You now have a 50/50 choice between 2 doors, to see if you hit the 1 in 3, but it makes no difference if you switch or stay. It’s not a random re-shuffled choice. You either hit it or didn’t.

The interesting question is: if this is true (as most peoples' intuition suggests), then why are the computer models wrong?

I maintain it is because the "Monty opens door 2 or 3" assumption in the decision tree version is a bad assumption that gives switchers better odds in the model than they really have. If that bad assumption is built into the computer models, whelp, bing bang boom - you get 30 years of people spouting claptrap telling you that they are the ones 'following the science.'

I could be wrong, though - math is gay, and statistics are the gayest kind of math.

[Martin Hash]

Extend from a 100 doors to a million doors to a billion doors to a trillion doors to a trillion trillion trillion doors, all opened before you came on the scene but all you see is 2 planets, one has life and the other is barren but you don’t know which one. All you know is that one had a trillion trillion trillion false starts, and the second had none: what are the odds that the planet with all the false starts has life? What about the other one?

I waited awhile to stick this up here.