Game One: You are given a choice of three doors. You pick number one. The host opens one of the other two doors, having been given instructions that, if you pick the car, the host is to open one of the other doors, and if you pick a goat, the host opens the other door with a goat. Stalemate. It is a predetermined outcome.
Game Two: The prior game’s outcome stands. The new choice you have is do you keep door number one, or do you switch?
How do you have a 2/3 chance of winning if you switch?
In: 107
Think of it this way: imagine it’s *ten* doors.
You pick one.
What are the chances you picked correctly? 10%
That means there’s a 90% chance the prize is behind one of the doors you did not pick.
Now Monty reveals what’s behind 8 of the other doors (it’s all goats and not random, he knows where the prize is hidden). However there is *still* a 90% chance you picked wrong at the very beginning! The fact that you have been shown what was behind 8 of the other doors doesn’t change that.
By revealing what’s behind the 8 other doors, Monty has given you new information. You already knew there was a 90% chance you were wrong, but you now know that in the event you were wrong (which is 90% likely, remember) that there is only *one* possible place the prize could be.
Or, alternately, imagine you picked a door, then were told “Okay, you can keep what’s behind that door, or you can open up all the other doors and keep whatever you want that’s behind them.
Let’s make a similar example but broaden the math so it’s easier to see. Instead of 3 doors with one prize, let’s make it 100 doors with one prize. You pick door 13, and the host decides to open all the doors except 13 and 46. Doesn’t it make more sense to make the switch?
This is a tricky problem and actually a great example of why statistics Suck^TM
Imagine you’re watching a TV show where they show you three doors, two of which hide a goat and one of which hides a car. Then, the host opens one of the doors that contain a goat. Now, you’re smart and remember the Mont Hall problem, so you pick, from the comfort of your couch, a door in advance so you can switch afterwards and get that sweet sweet 2/3 chance of being right. Even though there’s only two doors left over you have a more than 50% chance of guessing correctly, right?
But how can this be? How can you have a more than 50% chance to guess correctly between two completely random choices?
Easy. You don’t.
Let’s run this through. You have the choice of 3 different doors, and the host can open two possible doors for a total of six possible routes.
If you first choose the car, the host is definitely going to pick one of the other doors to open and if you switch (which you do since you’re so smart) you’re guaranteed to encounter the sweet sweet feeling of shame after your spouse laughs at you for choosing the goat. 2 out of the 6 routes result in goat.
If you first choose a goat (which happens in the other four routes) the host could open another door with a goat and by then switching you’d get the smug satisfaction of having chosen the car, which happen in 2 different routes, so 2 in 6 routes result in a car.
In the remaining 2 routes however, the host opens the door you’ve chosen.
At which point we run into a problem: we didn’t account for this to happen so we’re forced to choose a new door. Since we have no clue about the other doors we have to pick one at random and get it right 50% of the time.
So switching results in 2/6 times 100% car, 2/6 times 0% car, and 2/6 times 50% car, for a total of 50% chance you get the car by switching.
Consider it like this: there’s a 33% chance your door indeed contained the car once you picked it, but your door survived the first trial (which happens to 2/3 doors so 66% of the time) so now (due to Bayes) the odds of your door having the car is 33/66 = 1/2. Fifty-fifty.
However, Monty Hall does not work like that. Not when you’re a contestant.
In the Monty Hall problem, the host specifically states they will never open the door you picked. Which means that in two out of three times (aka when you picked the goat) the host actually doesn’t have a choice which door to open, which gives you more information than usual.
While the other doors are freaking out and panicking about the trial, the door you picked on the other hand never broke a sweat because it never was going to lose the trial anyway. As a result, since the door never went through the trial in the first place, the odds of it having a car never increased.
Compare it with a contest: while there are a thousand participants, you specifically get special treatment for no reason whatsoever to go straight to the finals without ever having to even face a jury, and in the finals you’re facing off against the best of the best who beat all other contestants and survived all rounds. You’re going to have a hard time (and so is your door).
Another way to word the problem:
You choose one of 3 doors. You are asked if you want to “stay” with your choice or “switch” to the other 2 doors.
* If you “stay”, you get what is behind that door. (33% chance of being right)
* If you “switch”, you also get a “free goat pass” in that the first switch door has a goat, you ALSO get to check the next one too. You only loose if there were two goats. (33% chance of being wrong.)
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