This is a great clarification of how this problem works. I STILL can't grasp why you'd switch. Since you have no idea which door it is, couldn't the 2/3 probability be applied to either his door or your door? For all you know, the one that he removed was just one random one of the goat doors. Your chance of picking the car was 1/3 before, if you could have the car already, why would it be better to switch now that he removed which ever would definitely be a goat? It seems like you still have a 50/50 chance of getting the car and your chance of the one that remains (unpicked/unremoved) is just as likely to be the other goat as it is to be the car..

That said, I made a simulation of it where it does the following: 1. Assign random locations behind doors, 2. Player randomly chooses door, 3. Host excludes door (random if both are goats, goat if one is a car), 4. Player either switches, or doesn't switch based on what's chosen for that simulation (the full set gets one or the other)

After running 50,000,000 then dividing success/tries, my number is nearly exactly the same between the two. Is my simulation working incorrectly?

I find it clearer if you have 100 doors. Pick one. The host eliminates 98 doors [with no prize behind them].

Do you switch? Or do you stick with your original guess? As the parent mentioned, switching gives you a 99/100 chance of being correct, staying gives you a mere 1/100 chance.

Thanks for clarifying a very counter intuitive problem. I always struggled with this one, not anymore :) Sometimes scaling things up is the best way to understand things. Who knew!

I don't see why the other door has 99x more probability of being correct. It's still 50/50, original door or remaining door, the other 98 don't change the odds.

Your initial pick has a 1/100 chance of being right. If you switch, and you were right, you lose. On the other hand, your initial pick has a 99/100 chance of being wrong, and if you were wrong, and you switch, you win. So switch.

Let's play a game to make this more clear. I'm thinking of a number between 1 and 1000. First, guess what it is.

2

Well, before you finalize your guess, I'll give you a hint: it's either 2, or 437. It's not any other number.

Keep in mind I made my choice before we started, and it hasn't changed.

Do you want to change your guess to 437, or stick with 2?

I've always flip-flopped on grasping the Monty Hall thing (sometimes I'd think I understand, other times not so much)... but this little guessing game really clarified it for me, once and for all. Thanks!

This is a nice reinterpretation.

the behavior of the game host as he will never reveal the door with the car this is the key bit. If the door you picked is the correct one (1% chance) the host can open any door. If you picked incorrectly (99% chance) he can open any door except the winning door so the remaining door will be the correct one 99% of the time.

From the initial setup you have a 1/3 chance of choosing the correct door and a 2/3 chance of choosing the wrong door.

The key to this problem is the behavior of the game host as he will never reveal the door with the car. So, after he reveals the contents of one of the doors, if you choose not to switch, you still only have a 1/3 chance of having chosen the door with the car.

On the other hand, if you choose to switch, 2/3 of the time you will have chosen incorrectly initially, and under the switching strategy, those 2/3ds of the time you are guaranteed to win the car. Hence, you have a 2/3ds chance of winning under the switching strategy, vs a 1/3d chance of winning under the "staying" strategy.

Hope this clarifies things.

Your simulation is not correct.

Look at it this way: 1/3 of the time, you pick the right door up front. If you switch, you lose.

2/3 of the time, you didn't pick the right door. If you switch, you win.

So switch.

> For all you know, the one that he removed was just one random one of the goat doors.

The key observation is that the problem states that Monty Hall opens an empty door. It's not clear whether Monty knows which doors are empty, but it doesn't actually matter, since the problem states that he will always open an empty door.

As others have said, your choice as a contestant is whether to take one door (the one you initially pointed at), or to take the other two doors (at least one of which is obviously empty). Stated that way, it's pretty obvious that taking 2 doors is better than taking 1 door.

The important bit is that Monty will never choose to show you the winning door, and he will always show you a door after you pick a door and before asking you if you want to switch. He always opens a losing door.

The location of the prize and the player choice of door are independent events. Monty's choice is dependent on whether the player choice and the prize location are the same. Monty will never open the door the player picked. Monty will never reveal the prize. Those rules generate exploitable information for the player.

Here is the complete probability tree, and two possible strategies of play:

  Prize behind door A;  P = 1/3                           ALWAYS STAY strategy:
    Player picks door A;  P = 1/3 * 1/3 = 1/9               P(WIN) = 1/18 + 1/18 + 1/18 + 1/18 + 1/18 + 1/18
      Monty shows Door B;  P = 1/3 * 1/3 * 1/2 = 1/18              = 6/18
        Player stays with A;  WIN                                  = 1/3
        Player switches to C;  LOSS                         P(LOSS) = 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9
      Monty shows Door C;  P = 1/3 * 1/3 * 1/2 = 1/18               = 6/9
        Player stays with A;  WIN                                   = 2/3
        Player switches to B;  LOSS
    Player picks door B;  P = 1/3 * 1/3 = 1/9             ALWAYS SWITCH strategy:
      Monty shows Door C;  P = 1/3 * 1/3 * 1/1 = 1/9        P(WIN) = 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9
        Player stays with B;  LOSS                                 = 6/9
        Player switches to A;  WIN                                 = 2/3
    Player picks door C;  P = 1/3 * 1/3 = 1/9               P(LOSS) = 1/18 + 1/18 + 1/18 + 1/18 + 1/18 + 1/18
      Monty shows Door B;  P = 1/3 * 1/3 * 1/1 = 1/9                = 6/18
        Player stays with C;  LOSS                                  = 1/3
        Player switches to A;  WIN
  Prize behind door B;  P = 1/3
    Player picks door A;  P = 1/3 * 1/3 = 1/9
      Monty shows Door C;  P = 1/3 * 1/3 * 1/1 = 1/9
        Player stays with A;  LOSS
        Player switches to B;  WIN
    Player picks door B;  P = 1/3 * 1/3 = 1/9
      Monty shows Door A;  P = 1/3 * 1/3 * 1/2 = 1/18
        Player stays with B;  WIN
        Player switches to C;  LOSS
      Monty shows Door C;  P = 1/3 * 1/3 * 1/2 = 1/18
        Player stays with B;  WIN
        Player switches to A;  LOSS
    Player picks door C;  P = 1/3 * 1/3 = 1/9
      Monty shows Door A;  P = 1/3 * 1/3 * 1/1 = 1/9
        Player stays with C;  LOSS
        Player switches to B;  WIN
  Prize behind door C;  P = 1/3
    Player picks door A;  P = 1/3 * 1/3 = 1/9
      Monty shows Door B;  P = 1/3 * 1/3 * 1/1 = 1/9
        Player stays with A;  LOSS
        Player switches to C;  WIN
    Player picks door B;  P = 1/3 * 1/3 = 1/9
      Monty shows Door A;  P = 1/3 * 1/3 * 1/1 = 1/9
        Player stays with B;  LOSS
        Player switches to C;  WIN
    Player picks door C;  P = 1/3 * 1/3 = 1/9
      Monty shows Door A;  P = 1/3 * 1/3 * 1/2 = 1/18
        Player stays with C;  WIN
        Player switches to B;  LOSS
      Monty shows Door B;  P = 1/3 * 1/3 * 1/2 = 1/18
        Player stays with C;  WIN
        Player switches to A;  LOSS

If your simulation does not conform with the mathematically derived result, your simulation is incorrect.

The most clear way that I've heard it described is by scaling it up to 100 doors or so, and then removing 98 doors instead of removing 1. (This is mentioned elsewhere in the thread.) It would be interesting to see the results of your test using this method. It seems intuitively clear with 100 doors that switching is the better method.

This is the one article that made me fully grasp it forever. I've known about this for ages, but it never really clicked until I read it put this way:

http://waitbutwhy.com/2016/03/the-jellybean-problem.html

The table here [1] helped me understand it when I first heard of this problem.

[1] https://en.wikipedia.org/wiki/Monty_Hall_problem#Simple_solu...

I only started to understand how to look at this problem once I realized the host was going to eliminate a bad door _no matter what was chosen_. Of course if you have already selected a bad door he won't eliminate it.

Another useful exercise is to stretch the number of doors to some large N and suppose the choice was posed repeatedly with a decreasing number of doors. If you can have anywhere from 2 to N guesses and as N decreases the number of bad doors decreases, it seems more obvious that you should always switch.

Not switching until the end should work out better than switching multiple times.

4 doors, in 8ths to simplify fractions.

  2/8,2/8,2/8,2/8 => initial conditions, choose leftmost
  2/8,3/8,3/8 => random bad door removed, choose a random 3/8
  3/8,5/8 => remove another door.

Next choice is only 5/8, whereas if we had waited it would have been 6/8.

>> The probabilities change

> That's the common misunderstanding of the problem...

You've got a choice here... you can either interpret "The probabilities change" as meaning the probabilities of a given door being correct at the start (which would be a misunderstanding) or you could interpret it from the pragmatic angle; the probability of success if you switch doors after one is taken away.

Given the context I'm inclined to believe that wrsh07 intended the latter.

This is good reasoning. Unlike other probability puzzles, the new information isn't useful in this one.

P(my first choice was right) = P(my first choice was right | given that I've been shown a goat in another door)

It's my second-favorite probability puzzle; my favorite is the one in which the new information has no relevance at all but still changes the probabilities, the "not both daughters" problem.

I should have just given the other problem, sorry for the laziness.

Your friend tells you that she has two children. They're not both boys, she adds. What are the odds that they're both girls? (Obviously 1/3).

When she tells you that one child is named Mary, the odds that they're both girls change (to 1/2)! Even though you knew that at least one of the children was a girl, and the name might as well be arbitrary, and the new information seems useless... the odds have changed.

It's nontrivial to work this one out; exercise for the reader.

After first datum (there are two children):

  Presuming A and B to be independent events
  (first child is a girl, second child is a girl)
  P(A) = 1 - P(!A) = 1/2
  P(B) = 1 - P(!B) = 1/2

  P(A and B) = P(B|A) * P(A) = 1/2 * 1/2 = 1/4
  P(A and !B) = P(!B|A) * P(A) = 1/2 * 1/2 = 1/4
  P(!A and B) = P(B|!A) * P(!A) = 1/2 * 1/2 = 1/4
  P(!A and !B) = P(!B|!A) * P(!A) = 1/2 * 1/2 = 1/4

Good so far.

But the probability later on depends on how you find out that at least one of your friend's children is a girl. If your friend tells you a random fact about her whole family, the probability is 1/3. If she tells you a fact about a randomly chosen child, the probability is 1/2.

You can presume from "they're not both boys" that the statement is equivalent to "there is at least one girl". When she tells you the name, you now know "a specific child of mine is a girl".

"Not both boys" gives you the precondition "!(!A & !B)" which equates to "A or B".

  P(A or B) = P(A and B) + P(A and !B) + P(!A and B) = 3/4
  P(A and B|A or B) = P(A or B|A and B) * P(A and B) / P(A or B) = 1 * 1/4 / 3/4 = 1/3

"Mary is a girl" gives you either the precondition "A" or the precondition "B"

  P(A and B|A) = P(A|A and B) * P(A and B) / P(A) = 1 * 1/4 / 1/2 = 1/2
  P(A and B|B) = P(B|A and B) * P(A and B) / P(B) = 1 * 1/4 / 1/2 = 1/2

These are both the same number, so that works out fine.

Now see if you can work it out if your friend told you this beforehand:

Between pregnancies, I had an accident that resulted in an odd medical condition. The doctors said that after it happened, 50% of boy fetuses that otherwise would have resulted in pregnancy would simply fail to implant, and I wouldn't even know about it.

Now there's an all new ambiguity. How do you decide whether your friend was planning on having two children all along, or if she has two children because that's how many times she got pregnant?

Here's an epistemological head-scratcher: do the odds still change if you don't hear the daughter's name accurately? Why?

Remember, those odds are the level of certainty you have that a given statement may be true based only on what you already know. You can also assign a level of certainty to the facts you think you know.

Given the number of ways to mishear "Mary" and the dialectical variant in pronunciation characterized by the Mary-marry-merry merger, I'd tentatively assign the following values:

  25% One child is a girl.  (Heard correctly)
  25% One child is a boy.   (Heard incorrectly)
  50% No new information.   (Incomprehensible or ambiguous)

  In the first case P(A and B|A) = P(A and B|B) = 1/2
  In the second, P(A and B|!A) = P(A and B|!B) = 0
  In the third, P(A and B|A or B) = 1/3

In any case, you're just guessing at the probability you might mishear something, so there's no definitive answer. The mere possibility that I might have heard a boy's name means that the overall probability could be anywhere between 0 and 1/2. By the above, my new odds are 5/12.

So the odds do change, based on your confidence in what you heard (and your confidence that your friend isn't the kind of parent to name her son "Meriadoc" or something similar).

You are, of course, correct. Sorry about the careless wording [I think the analogy is still interesting, though!]

This assumes you always open a door, that door never holds the prize, and there is exactly 1 prize.

You could run (ed:a similar game with different rules) such that the odds go 1/3,1/3,1/3 to 1/2, 1/2 if you flipped a coin to chose the second door and sometimes show the prize. Alternatively, if you chose when to open the second door, you could make swapping a very good (100%) or very poor choice (0%).

Worse, you could swap what's behind the doors after they chose.

Which IMO is why people find this so confusing. In the real world the odds presented may or may not line up with the actual odds. aka unknown unknowns.

This is incorrect.

Flipping a coin to "to chose the second door and sometimes show the prize." has no impact on the probabilities. The parent is correct - there is a 2/3 chance the prize exists under one of the other two doors, and if one of them is shown not to have the prize (through random flipping, deliberate selection, whatever) - then the final door now has a 2/3 chance to have the prize.

If one of the doors is shown to not have the prize, the odds are now 1/2 when the choice of reveal door is made randomly. That's because 1/3 of the time, you're not even offered the chance to switch, since it's pointless.

I'll run the trials here. Assume the first two doors have the goat, and the third door has the car.

   Pick  Reveal  Switch?  Outcome
   ----  ------  -------  -------
     1      2      Y 3    Car
     2      1      Y 3    Car
     3      1      N 3    Car
     3      2      N 3    Car

     1      2      N 1    Goat
     2      1      N 2    Goat  
     3      1      Y 2    Goat
     3      2      Y 1    Goat

     1      3       -     No chance to switch      
     1      3       -     No chance to switch
     2      3       -     No chance to switch
     2      3       -     No chance to switch

Meaning, given that you've made it to the point that an offer to switch doors is presented, there is no advantage to do so (or not do so).

But in the actual game, you are always given the chance to switch. Here are the possible outcomes of the game, with the third column being what the contestant should do to win:

   You Pick  Car   Switch?
   -------  -----  ------
       1      1      N    
       2      1      Y   
       3      1      Y
 
       1      2      Y
       2      2      N  
       3      2      Y

       1      3      Y     
       2      3      Y     
       3      3      N   

Without knowing anything else, switching means that you win 2/3 of the time.

I changed to rules so in "3, 1, Y" he might open door 3 and show you a car. At that point switching is a meaningless choice.

Your comment (EDIT: Sorry, not yours, but parent) was talking about a random flip of coin determining which door Monty shows. That's not the actual game. If Monty is randomly showing a door without regard to goat/car, then you have to also consider the scenarios where he reveals a car and you lose early.

If, while randomly flipping a coin, he reveals the car then you have a 0% chance of winning. If he doesn't reveal a car then you have a 2/3 chance of winning if you switch doors.

In this scenario, Monty's door choice and your door choice are completely independent actions. Because of this, you gain no new information from the door he reveals. In the classic version of this puzzle (where Monty always reveals a goat), you do gain information, because most of the time Monty's forced to choose the only other goat, and therefore avoid the car.

http://c2.com/cgi/wiki?NotTheMontyHallProblem

The switching strategy is effective because it relies on Monty's avoidance of the car. When Monty isn't trying to avoid the car, his revelation doesn't help you out at all. It just tells you that you either (A) now have a slightly better chance (50%), or (2) you've already lost. (0%).

EDIT: Also see the table on this wikipedia article. It lists probabilities for all the different variants of the MHP. Your scenario is referred to as the Monty Fall or Ignorant Monty variant.

https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...

    "Monty Fall" or "Ignorant Monty": The host does not know
    what lies behind the doors, and opens one at random that
    happens not to reveal the car
    (Granberg and Brown, 1995:712) (Rosenthal, 2005a) (Rosenthal, 2005b).
    Switching wins the car half of the time.

When you originally pick a door - there are three. (We all agree on this). Because you are picking at random, and the prize is only behind one door, there is a 1/3 chance that you have picked the right door. (likewise, We can all agree on that.) Therefore, I think we can all agree that there is a 2/3 chance that the prize exists behind one of the other two doors.

Let us take your approach, and have Monty flip a coin, revealing ANY door at random.

If he flips your door, and you have the prize - You have a 100% chance of winning. If he flips your door, and you have the goat, 0% chance of winning. Those two scenarios are easy, and we can ignore them.

Likewise, if he flips the coin, and shows one of the other two doors, and they have the prize - then, once again - easy - 0% chance of winning. We can ignore these scenarios.

I think up to this point, we can agree on all of the above scenarios. (They are win/lose)

Where it gets complex, he flips a coin, selects one of the doors you aren't on, and reveals a goat.

Given that you had a 1/3 chance originally of winning, and a 2/3 chance of not winning, and given that Monty has now revealed that one of the other two doors (By chance) does not have the car, your 2/3 chance of not winning is still the same, but now it applies to only a single door - I.E. You have a 66% chance of winning by selecting that other door - and the fact that Monty selected that door by chance, is irrelevant.

Unfortunately, despite my (flawed) apparently logical argument - I'm completely wrong, and you are correct.

   from random import *
   def putPrizeInDoor():
       doors=[0,0,0]
       door_winner=randint(0,2)
       doors[door_winner]=1
       return doors

   trials=100000
   testCount=0
   win=lose=0
   for x in range(trials):
       doors=putPrizeInDoor()
       userSelects=randint(0,2)
       montyReveals=randint(0,2)
       if montyReveals==userSelects: # Monty reveals the door user is on
           pass
       else:
           if doors[montyReveals]==1: # Monty Revealed the prize.
               pass
           else: # this is where it's interesting - Monty reveals, at random, door not winning that user not on.
               revealedDoors=set([montyReveals,userSelects])
               newUserChoice=list(set([0,1,2])-revealedDoors)[0]
               testCount+=1
               if doors[newUserChoice]==1:
                   win+=1
               else:
                   lose+=1
        
            
   print ("Tested: {}  Wins: {}  Lost: {}  Percentage: {:.2f}".format(testCount,win,lose,win/testCount))
 

Tested: 44484 Wins: 22220 Lost: 22264 Percentage: 0.50

It's interesting that the deliberate selection by Monty of the goat, increases my chance from 33% to 66% if I switch, but the random selection by Monty of the goat, only increases my chance from 33% to 50% if I switch. I'm not really sure where the argument I made breaks down, but, empirically, it's clearly broken. I'll have to meditate on it a bit to see where I've gone awry.

I must say this is one of the best comments I have seen on HN. I don't know if this helps you meditate on the game or not, but this is how I walk though the logic.

1/3 of the time you picked correct: 100% of those worlds he shows a goat. 1/3 of the time you picked wrong and he shows a goat. 1/3 of the time you picked wrong and he shows a car.

So your overall odds are 1/3 of winning if you don't change AND 1/3 of the time you don't get the chance to change.

Thus, your overall chances of winning in this case is 1/3. But, if he shows you a goat your odds are bumped to 1/3: 1/3 or 50/50.

Thanks for this comment! I thought this thread was dead.

I was searching for online simulators that let me tweak the rules of the game, but I could only find those that follow the original rules of the puzzle.

In thinking more about the Ignorant Monty scenario, I realized it's equivalent to having 3 contestants pick doors:

    Me
    Montgomery
    Switchy McSwitcherson

First I pick a door. Then Montgomery picks a door. Finally Switchy McSwitcherson gets the remaining door. Once all the picking is done, Montgomery opens his door first, and is saddened to see a goat. At this point, who is more likely to win? Me or Switchy McSwitcherson? Since we both had 1/3 initial odds, there's symmetry there, and one of us can't reap all the odds-boosting.

But the classic puzzle is more like a 2-contestant game, where I pick a door, the host selects a goat to reveal (and reveals it), then Switchy McSwitcherson is assigned the remaining door. In that scenario, I'm mad that they made me pick first before the host eliminated a door for Switchy. Switchy enjoys all the odds boosting.