The role of probability in solving the Monty Hall problem

[Aractus]

Here's the easiest way I've found to convince people why switching is better:

Suppose instead of three doors, you have 100. There's a car behind one and goats behind 99 of them. You get to choose a door, and then 98 other incorrect doors will be eliminated.

So, you choose door 34. The host opens eliminates 1, 2, 3 ... 33, 35, 36... 85, and 87 through 100.

So either the prize is behind door 34, which you picked right off the bat, or 86, which is the only one the guy didn't open.

Intuitively, it's clear that they don't have the same probabilities of being correct, right? Essentially, you've always had a 1% chance of it being in 34, and the other 99% collapses into 86.

But see here's the whole problem. If you don't know that Monty Hall intentionally picks a Goat and his selection is truly random then it actually does not matter if you switch or not.

Here are your choices:

Door A | Door B | Door C
(Goat) | (Goat) | (PRIZE)

In two scenarios you initially pick the wrong door, but in the third you pick the Prize. Therefore if Monty intentionally removes a Goat it is better to switch.

BUT

What happens when Monty picks the doors randomly?

Door A | Door B | Door C
(Goat) | (Goat) |(PRIZE)

Non-Switching Scenarios:
Scenario 1 - You pick Door A, Monty Picks Door B, Monty Picks Door C, you Lose
Scenario 2 - You pick Door A, Monty Picks Door C, you Lose
Scenario 3 - You pick Door B, Monty Picks Door A, Monty Picks Door C, you Lose
Scenario 4 - You pick Door B, Monty Picks Door C, you Lose
Scenario 5 - You pick Door C, Monty Picks Door A, Monty Picks Door B, you Win
Scenario 6 - You pick Door C, Monty Picks Door B, Monty Picks Door A, you Win

Now we can see what's going on more clearly - Monty in fact has a 1 in 3 chance that the first door he opens is the Prize, and a 2 in 3 chance that it is a Goat. Assuming no manipulation this is what happens if you switch:

Door A | Door B | Door C
(Goat) | (Goat) |(PRIZE)

Switching Scenarios:
Scenario 1 - You pick Door A, Monty Picks Door B, you switch to Door C, Monty Picks Door A, you Win
Scenario 2 - You pick Door A, Monty Picks Door C, you Lose
Scenario 3 - You pick Door B, Monty Picks Door A, you switch to Door C, Monty Picks Door B, you Win
Scenario 4 - You pick Door B, Monty Picks Door C, you Lose
Scenario 5 - You pick Door C, Monty Picks Door A, you switch to Door B, Monty Picks Door C, you Lose
Scenario 6 - You pick Door C, Monty Picks Door B, you switch to Door A, Monty Picks Door C, you Lose

It makes no difference. And the reason is that you only have a 4/6 chance (2/3) of having the option to switch, and since your original odds of winning were 1 in 3 they remain 1 in 3 if you switch. You don't do your chances any harm, but there's no benefit. You lose 2/3rds of the time in either scenario. But that's not how Monty plays the game. He never picks the door with the prize first. And that changes the probability like so:

Non-switching scenarios:
Scenario 1 - You pick Door A, Monty Picks Door B, Monty Picks Door C, you Lose
Scenario 2 - You pick Door B, Monty Picks Door A, Monty Picks Door C, you Lose
Scenario 3 - You pick Door C, Monty Picks Door A, Monty Picks Door B, you Win
Scenario 4 - You pick Door C, Monty Picks Door B, Monty Picks Door A, you Win

Or

Switching scenarios:
Scenario 1 - You pick Door A, Monty Picks Door B, you switch to Door C, Monty Picks Door A, you Win
Scenario 2 - You pick Door B, Monty Picks Door A, you switch to Door C, Monty Picks Door B, you Win
Scenario 3 - You pick Door C, Monty Picks Door A, you switch to Door B, Monty Picks Door C, you Lose
Scenario 4 - You pick Door C, Monty Picks Door B, you switch to Door A, Monty Picks Door C, you Lose

Although there are only four scenarios, each of the first two is twice as likely to happen as the last two, that's because Monty's former choice of "Door C" has been removed, forcing him to chose the Goat even if he has a choice to open the Prize. So while all the outcomes we were talking about had a probability of 1 in 6, now the top two have a probability each of 2 in 6 (1 in 3), while the last two scenarios (formally scenarios 5 and 6) each have a probability of 1 in 6. Therefore:

Non-switching scenarios:
Scenario 1 (p. = 2/6)- You pick Door A, Monty Picks Door B, Monty Picks Door C, you Lose
Scenario 2 (p. = 2/6) - You pick Door B, Monty Picks Door A, Monty Picks Door C, you Lose
Scenario 3 (p. = 1/6) - You pick Door C, Monty Picks Door A, Monty Picks Door B, you Win
Scenario 4 (p. = 1/6) - You pick Door C, Monty Picks Door B, Monty Picks Door A, you Win

Or

Switching scenarios:
Scenario 1 (p. = 2/6) - You pick Door A, Monty Picks Door B, you switch to Door C, Monty Picks Door A, you Win
Scenario 2 (p. = 2/6) - You pick Door B, Monty Picks Door A, you switch to Door C, Monty Picks Door B, you Win
Scenario 3 (p. = 1/6) - You pick Door C, Monty Picks Door A, you switch to Door B, Monty Picks Door C, you Lose
Scenario 4 (p. = 1/6) - You pick Door C, Monty Picks Door B, you switch to Door A, Monty Picks Door C, you Lose

So I'm not disagreeing about the probabilities, but I'm pointing out that it's only valid to switch if the host is intentionally choosing the goat. If the Host randomly chooses a door it makes no difference.

There is yet one more way for Monty to play the game. This would incredibly cruel, but let's examine it none the less. This time Monty will pick the prize intentionally:

Door A | Door B | Door C
(Goat) | (Goat) | (PRIZE)

Non-Switching Scenarios:
Scenario 1 (p. = 2/6) - You pick Door A, Monty Picks Door C, you Lose
Scenario 2 (p. = 2/6) - You pick Door B, Monty Picks Door C, you Lose
Scenario 3 (p. = 1/6) - You pick Door C, Monty Picks Door A, Monty Picks Door B, you Win
Scenario 4 (p. = 1/6) - You pick Door C, Monty Picks Door B, Monty Picks Door A, you Win

Switching Scenarios:
Scenario 1 (p. = 2/6) - You pick Door A, Monty Picks Door C, you Lose
Scenario 2 (p. = 2/6) - You pick Door B, Monty Picks Door C, you Lose
Scenario 3 (p. = 1/6) - You pick Door C, Monty Picks Door A, you switch to Door B, Monty Picks Door C, you Lose
Scenario 4 (p. = 1/6) - You pick Door C, Monty Picks Door B, you switch to Door A, Monty Picks Door C, you Lose

In this scenario switching reduces your chances of winning to zero!