Hacker News Reader

Comment by lqet | Hacker News Reader

It does matter. If you are looking at a single instance where Monty got lucky and chose a goat door, then the probability that the door you have chosen has the car is 1/2, and switching doesn't change anything. This can easily be tested: just write a simulation that runs the experiment with Monty choosing a random door, and discard the instances where the game was "ruined" because he picked the car. In the remaining instances, both strategies will perform the same.

View on HN

Sure, here's the simulation. There are only six distinct cases, so we can "simulate" it exhaustively, and a Monte Carlo will approximate this.

Let "A", "B", and "C" represent the prizes; A is valuable, B and C are goats. They're shuffled behind doors, so the player and Monty choose "1" or "2" or "3", and after the fact we'll map ABC to 123. As such, we can assume without loss of generality that the player always chooses "1" and Monty always chooses "2".

What happens?

  123
  ABC - player was right at first, so they lose if they switch
  ACB - player was right at first, so they lose if they switch
  BAC - player was wrong, Monty chooses 2 which is the car, so the game is ruined
  BCA - player was wrong, Monty finds a goat, player wins if they switch.
  CAB - player was wrong, Monty chooses 2 which is the car, so the game is ruined
  CBA - player was wrong, Monty finds a goat, player wins if they switch.

Huh. Of the 4 non-ruined games, half of them are improved by switching, half are worsened.

I believed you were wrong, but now I believe you.

Exactly and if Monty knows of the car location (as in the original formulation), he will basically save BAC and CAB from being ruined and both of these will become "player wins if they switch" outcomes. So instead of 2 out of 4, we now have 4 out of 6 cases where switching is better.

It matters because in the case where Monty does know to avoid the door, those games end up as wins for the person who selected to switch, as opposed to broken games.

You are taking an empty result and replacing it with a strict win.

I think lqet's point is that if it's just recounted as being thrown into a situation -- you've picked a door, and Monty opens a door, and it's a goat -- you don't know if you're in a world where Monty always picks a goat or one where he just happened to be lucky. In the second scenario, it doesn't matter if you switch.

Well in that situation it's still beneficial to switch, unless the probability of being in the first scenario is exactly zero.

You’re forgetting about the possibility of being in a universe where Monty only opens a door if you chose the correct door originally (otherwise he does something else like ending the game). If you know nothing about the rule set of the game show, Monty opening the door tells you nothing and you have no reason to either switch or not switch.

Indeed. It's like the argument that you may as well believe in God, because if he doesn't exist it doesn't hurt you.

That said, though, if you're trying to convince someone of the correctness of the Monty Hall Problem, and they're stuck on that misconception, saying "well you may as well switch in case I'm right" isn't a great argument.

Can’t that be expressed even more simply as:

1/3: you picked car, switch fails

1/3: you picked goat, monty ruins. Not included

1/3: you picked goat, monty open goat. Switch wins.

Of the two non ruined scenarios, half win

We can also demonstrate by using the 100 door example. Monty opens 98 of the 99 doors you didn’t choose. 98 times out of 99, monty ruins the game.

So random choice dramatically changes things.

I think where I differ (also expressed on the parent comment) is that when Monty ruins the game, you are tossing out many cases where the "I'm sticking with my original choice" strategy failed, while I'm counting them as a failure in the strategy.

You are free to count them as a failure of the "I'm sticking with my original choice" strategy, but only if you also count them as a failure of the "I'm switching" strategy, because when Monty ruins the game there is no way to win the car by switching to the remaining closed door.

If Monty chooses randomly both strategies fail 2/3 of the time.

Precisely. Either way switch or no switch have the same success rate if Monty ruins the game. You can count it as 1/2 or 2/3 but either way both choices are the same. So it’s a very different scenario from when Monty knows the door is a goat.

Why are we discarding the games that are ruined?

Whether Monty knows for sure or is just picking at random only affects whether games are ruined, not the probability that the door the contestant picked is a winner.

If there are N doors and the player picks door 1, there is a 1/N chance he will be correct. If he sticks to that door, there is nothing that will ever change that 1/N odds that it was the winner.

Downthread someone enumerated the cases, but again, they discard the 1/3 of the cases where Monty ruined it. What that does is throw out half of the cases where contestant picked the wrong door and had the wrong strategy. Whether Monty ruined the surprise by showing the winning door or Monty got lucky and exposed a goat, that original door always has a 1/3 probability of being the right one.

Because the problem asks what to do after a goat is shown.

>Whether Monty ruined the surprise by showing the winning door or Monty got lucky and exposed a goat, that original door always has a 1/3 probability of being the right one.

By that logic, so does the remaining door. 1/3 vs 1/3.

Because that is a fundamental part of the "Monty Hall" problem. You are correct in the case where Monty will randomly pick doors, and sometimes he will pick the winning door and ruin the game, but that is a fundamentally different problem than what is described as the "Monty Hall problem", and therefore irrelevant.

Its like saying soccer players could score more if they picked up the ball and ran with it. While true, you are no longer describing the game of soccer.

Reply on news.ycombinator.com