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Comment by dllthomas | Hacker News Reader

Interestingly, if the host picks randomly (and if he reveals the car, you... start over, or you get the car, or you get nothing, or ... it doesn't matter because it happens not to have happened in the time we're considering) then you are faced with a 50/50 chance.

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Yeah - and at least for me this was part of the confusion. The problem is often stated incorrectly.

Critically the host always removes a goat.

If the person presenting the problem just says “the host opens one of the doors”, but doesn’t specify he’s always revealing a bad door then it’s not clear why that matters.

The many door example makes it easy to intuit as well.

I think when stating the problem, some people assume the listener has actually watched Lets Make A Deal. You can under-specify the problem for people who have watched it, as the fact that Monty shows the worthless prize is already known.

That’s obvious. If he reveals the car, he isn’t going to ask you if you want to switch. A goat is a goat. It doesn’t matter to you which goat you get.

It’s not obvious if you think revealing the car ends the game (and you lose).

I think when I was first told it the person said “takes a door away” which is even less clear.

Years ago I challenged you on this and you suggested I write a simulation to check. I did just that and my understanding of the problem improved dramatically. That exchange made a pretty big impact on me and is something I think about fairly often, so thanks. :)

https://news.ycombinator.com/item?id=9079260

This made my week :)

I also wrote a simulation, in PHP, when I first encountered this problem.

  <?php
  $wins = 0;
  for ($i = 0; $i < 1000; $i ++)
   {
   $car = mt_rand(0, 2);
   $choose = mt_rand(0, 2);
   if ($choose == $car) $wins ++;
   }
  echo "stay: " . ($wins / 1000 * 100) . "%<br />";
  // switch
  $wins = 0;
  for ($i = 0; $i < 1000; $i ++)
   {
   $car = mt_rand(0, 2);
   $choose = mt_rand(0, 2);
   if ($choose != $car) $wins ++;
   }
  echo "switch: " . ($wins / 1000 * 100) . "%";

I don't understand how this could be. If you start over when the host picks the car, then isn't that the same as the host picking the goat every time, i.e. the same as the host knowing.

Well, I don't know. Advantage moves to the contestant: they close the doors, but everything's gonna remain in the same place. Steathily repositioning an automobile on a soundstage might be possible (albeit difficult), but just you try moving the damn goat

Fair warning: I haven't rigorously verified the math on this, and tfa is all about the importance of vigorously verifying the math.

In the case where the host revealing the car means you restart, there are three equally likely situations after the host opens a door: - You picked the car, and the host revealed a goat - You picked a goat, and the host revealed the other goat - You picked a goat, and the host reveals the car (restart)

Of the two terminal cases, one gets you the car if you switch, and one gets you the car if you don't switch. In the non-terminal case, it doesn't matter what policy you have because you don't even get a chance to apply it.

One way to think about it is to imagine Monty picking randomly in either case, but then (internally and invisibly) correcting his pick.

Assuming, WLOG, that we choose door 1, that leaves us with 6 equally likely cases just before that final correction (or lack thereof):

    A) Car 1, Monty 2
    B) Car 1, Monty 3
    C) Car 2, Monty 2
    D) Car 2, Monty 3
    E) Car 3, Monty 2
    F) Car 3, Monty 3

If Monty doesn't correct in cases C and F, then when he shows us a goat behind (say) 2 then we learn we are in either A or E - it's 50/50. If Monty does correct himself, then we might have been in A or E or F.

Ah, yep, that’s right. Another way to see it is that we’re interested in the probability that your door has a goat behind it, given that you didn’t need to start over:

P(you chose goat | host didn’t choose car) = P(you chose goat, host didn’t choose car) / P(host didn’t choose car).

The numerator is 2/3 * 1/2, and the denominator is 2/3, so the ratio is indeed 1/2.

(A rejection sampling loop, where you repeatedly simulate a process until a condition holds, has the same distribution over final outcomes as the conditional distribution—so repeatedly restarting the game if the host chooses the car induces the same distribution on final results as simply conditioning on the host not choosing the car.)

This is a small enough situation that we can do complete enumeration of the states.

Let's say you pick door 1. Let's go through all the possibilities: the states of the 3 doors, which one monty reveals, and what do we do, and what is the result?

  1 | 2 | 3 || Monty Reveals | You switch | Result
  ---------------------------------------------------
  G | G | C ||      2        | Yes        |  Win
  G | G | C ||      2        | No         |  Lose
  G | G | C ||      3        | N/A        |  Start Over
  
  G | C | G ||      2        | N/A        |  Start Over
  G | C | G ||      3        | Yes        |  Win
  G | C | G ||      3        | No         |  Lose
  
  C | G | G ||      2        | Yes        |  Lose
  C | G | G ||      2        | No         |  Win
  C | G | G ||      3        | Yes        |  Lose
  C | G | G ||      3        | No         |  Win

Count 'em up: When you switch, 2 wins and 2 losses. When you don't, 2 wins and 2 losses. Of course, the situation is symmetrical for any starting guess you make.

What would this table look like if Monty always revealed a goat? Would any win or lose row be added or removed?

I know the result is right, but I don't think this table illustrates the reason.

You are not enumerating the options correctly. You have 3 rows for the GGC and GCG options but 4 for the CGG option, which is assigning the CGG option higher probability. In reality, they have equal weight.

Agh, you are right. There should be two “start over” rows in each of GGC and GCG. I can’t write “N/A;” we must say “yes/no” in those.

Doesn’t change the result though.

No, because the probability of a do-over depends on whether the contestant initially picked the door with a car.

At first I agreed with you, but then I thought about the variation with 100 doors (this is what originally helped me grasp the correct answer): you choose a door out of 100 (with very slim chance of choosing the correct one), the host opens 98 of the remaining 99 doors and there are goats behind all of them. Now it's obvious that the car is in the remaining door.

If the host is opening random doors, then: * If you chose a door with a goat, in 98/99 cases we would start the game over because the host opened the door with the car. * If you chose the door with the car, the game cannot start over because the host can only open doors with goats.

This means that when you choose a goat, the randomness gives you another chance to choose the correct door.

"Now it's obvious that the car is in the remaining door."

No. It could be behind the one you chose, surely.

I understand the explanations, or at least some of them; but I still don't get it. I thought I was intelligent and numerate, and this is making me sad.

I don't see why, after Monty reveals a goat and invites you to switch, you can't re-assess the probabilities from scratch. There are two closed doors, car behind one, goat behind the other, and you have no evidence which is which; therefore 50:50.

This is exactly why the Monty Hall problem doesn’t apply to Deal or No Deal. Intuitively, it disappears when doors are opened randomly because, by Bayes, the likelihood you picked the good prize goes up every time you reveal a bad prize.

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