Monty Hall Paradox: Why Does It Matter What Host Knows?

[Warp]

Suppose you are participating in a game show, and at one point three doors are presented. It's announced that one of them has a car prize behind it and the other two have a goat, and for you to choose one of the doors. After choosing the host opens one of the other doors and reveals a goat, and then asks if you would want to switch.

So you think: "Hey! I know this! This is the famous Monty Hall problem! There's 1/3 chance that the car is behind my chosen door and a 2/3 chance that it's behind one of the other two. And now I know for sure which door it's not, so the 2/3 chance is behind the other door, so I should definitely switch!"

But then you stop for a moment and ask the host "did you open the door knowing there was a goat behind it?" To which the host responds: "No. I just opened one of the other doors at random, without knowing what's behind it." Which makes you think: "Oh shoot! Then it doesn't matter if I switch or not! Now there's a 50-50 chance."

And this is what call "the Monty Hall paradox" (probably not called that, if it's called anything).

Why would it make a difference what the knowledge of the host was when he opened the door? The setup seems to be completely identical to the original Monty Hall situation: Three doors, your door has a 1/3 chance for the car (before the host does anything), then you get information about the two other doors, so surely the remaining door now has all the 2/3 chances? What exactly makes the 1/3 - 2/3 chances change to 1/2 - 1/2? Why would it make any difference that the host didn't know what's behind the doors? You still got your extra info about the other doors! It shouldn't make any difference what may have happened in other games (past and future). They shouldn't have an effect on this one. Wouldn't it be pretty much the Gambler's Fallacy to think that what happens in other rounds of the game affects the probabilities of this particular round? If we look only at games where the contestant had a choice to make (ie. a goat was revealed by the host), ignoring the ones where a car was revealed, isn't the situation identical to the original Monty Hall setup?

Yet, when you eg. simulate this version of the game, even if you look only at the rounds where the host reveals a goat... the contestant wins in about 50% of those rounds (where the contestant has a choice to make). But why? Shouldn't these rounds, in themselves, be identical to the original setup? It's hard to wrap my head around this seeming paradox.

 

[Halc]

What exactly makes the 1/3 - 2/3 chances change to 1/2 - 1/2?

The odds never change if the host knows where the car is, the odds are always 2/3 that it's the door you didn't pick. A car is never revealed by the host.

If a random door is eliminated then it might be the car, and the 50/50 shot is only if it isn't, no different than if the host wasn't involved at all and it was you picking the door to eliminate. So the revelation of one door will change the odds to either 50% or 0% depending on what's behind the randomly eliminated door.

And yes, it has nothing to do with the outcome of prior rounds.

 

[DaveC426913]

There are countless threads here on PF about the Monty Hall problem. At the bottom of your screen, you should find this:

 

[Ibix]

Yet, when you eg. simulate this version of the game, even if you look only at the rounds where the host reveals a goat... the contestant wins in about 50% of those rounds (where the contestant has a choice to make).

Not the result I'm getting (edit: because there was an error in my understanding of the game - see next post). The game I simulate is:

1. Place the car at random
2. Choose a door
3. Open a door at random
4. Abort if the random door was the one with the car
5. Otherwise, record if the winning choice was switch or stick

Counting only the runs that didn't abort at step 4 I get 2/3 win-if-you-switch, just the same as playing the regular game. Python code below.

import random

OPENED_CAR = 0
WIN_IF_STICK = 1
WIN_IF_SWITCH = 2

def monty():
    # Place the car
    car = random.choice((1, 2, 3))
    # Contestant chooses a door
    choose = random.choice((1, 2, 3))
    # Host opens a random door
    opened = random.choice((1, 2, 3))
    # Bail out if the host opened the door with the car
    if car == opened:
        return OPENED_CAR
    # Return the right choice
    if car == choose:
        return WIN_IF_STICK
    return WIN_IF_SWITCH

# Count up the wins
wins=[0, 0, 0]
for i in range(1000000):
    wins[monty()] += 1
# Show win-if-stick as a fraction of all runs where the host didn't reveal a car
print(wins[WIN_IF_STICK] / (wins[WIN_IF_STICK] + wins[WIN_IF_SWITCH]))

 

[Ibix]

Oh wait - I'm allowing the host to open the door the contestant chose. Never mind - I get 50% when I do that.

The reason for the difference is that when you don't pick the car Monty's hand is forced in the usual rules, and then you have "switch to win". He still has a (random) choice in your version, and this can lead to the abort step or a "switch to win". So the odds of switching being the winning strategy is reduced compared to the standard game.

 

[FactChecker]

But then you stop for a moment and ask the host "did you open the door knowing there was a goat behind it?" To which the host responds: "No. I just opened one of the other doors at random, without knowing what's behind it." Which makes you think: "Oh shoot! Then it doesn't matter if I switch or not! Now there's a 50-50 chance."

And this is what call "the Monty Hall paradox" (probably not called that, if it's called anything).

Why would it make a difference what the knowledge of the host was when he opened the door? The setup seems to be completely identical to the original Monty Hall situation:

It's not identical to the Monty Hall situation and it is not a paradox. The intentional avoidance of opening a door with the car makes all the difference. That filters out a door with no car.

Suppose Monty Hall started with 100 doors, one prize, and opened 98 doors with goats. Then you could be practically certain that the remaining door has the car because he avoided it 98 times.

Another extreme example: Suppose there is a bucket of sand with a diamond in it and you take a pinch out but do not look to see if you have the diamond. You know that there is very little chance that you got the diamond and that it is almost certainly still in the bucket. Now suppose you strain all of the remaining bucket sand through a sieve until there is only a pinch left. You are practically certain that the diamond is in the remaining pinch and that you should switch. That is completely different from the case where someone just takes another pinch from the bucket and offers to switch with you. In that case, his pinch also has almost no chance of containing the diamond. It has the same chance as yours. The filtering makes all the difference.

 

[Warp]

Suppose Monty Hall started with 100 doors, one prize, and opened 98 doors with goats. Then you could be practically certain that the remaining door has the car because he avoided it 98 times.

That's if the host knowingly opened only goat doors, explicitly avoiding the car door. But suppose that the host opens 98 doors at random (without knowing what's behind them), and it just so happens that this time he opened only goat doors. Very unintuitively, and paradoxically (at least apparently paradoxically), we are now once again in a 50-50 situation! It seems to make absolutely no sense, yet it's true. The randomness of opening the doors makes a difference. Intuitively it shouldn't matter if the choice was random or not, yet it does matter.

 

[Warp]

Thinking about it, I think I get it.

For starters, while it might at first feel that what the host knows and how he acts has no effect (on the rounds where the contestant has a choice), and the only thing that matters is the current door situation at hand, that's actually not true, which becomes evidently clear in a slightly different situation: Suppose the host knows what's behind the doors and acts against the contestant by always opening a car door if it wasn't chosen by the contestant (to make the contestant lose immediately). If he opens a goat door that means that the contestant chose a car door, so he most definitely shouldn't switch! So while outwardly the situation seems identical to the classical Monty Hall problem, how the host behaves actually affects the outcome.

But why does the host opening an unchosen door randomly make the situation different?

What's happening is that, perhaps somewhat unintuitively, when the host always opens an unchosen door at random (and a goat is revealed), it actually converts the game into a two-door game, rather than a three-door one. That's because one third of the possible paths of the three-door game is removed from the equation. This becomes clearer when we examine all possible outcomes, and take the ones where the contestant has a choice. Let's assume the contestant always chooses the first door (it doesn't make a difference). I'll mark the host's random choice with a capital letter and the other doors with a lower-case letter:

c G g -> switching loses (*)
c g G -> switching loses (*)
g C g -> no choice
g c G -> switching wins (*)
g G c -> switching wins (*)
g g C -> no choice

Even if we discard all the games where the contestant loses outright and thus doesn't even have a choice, we are left with four possibilities, marked with (*), two of them winning, two losing, so it's a 50-50 situation.

It becomes clearer that this is just a disguised two-door situation by noticing that two pairs of possibilities in that list are just symmetric and thus make no difference, so what we are essentially just left with is:

c g -> switching loses
g c -> switching wins

 

[FactChecker]

That's if the host knowingly opened only goat doors, explicitly avoiding the car door. But suppose that the host opens 98 doors at random (without knowing what's behind them), and it just so happens that this time he opened only goat doors. Very unintuitively, and paradoxically (at least apparently paradoxically), we are now once again in a 50-50 situation! It seems to make absolutely no sense, yet it's true. The randomness of opening the doors makes a difference. Intuitively it shouldn't matter if the choice was random or not, yet it does matter.

In the case that you know the host opened 98 goat-doors randomly regardless of where the car was, you are forced to accept that you just witnessed a very unlikely event and that your door is just as likely to have the car as the remaining closed door.

 

[Ibix]

I think that the easiest way to understand regular Monty Hall is to note that there are two outcomes: either you guessed right first time ($p=1/n$ for $n$ doors), in which case sticking wins, or you guessed wrong ($p=(n-1)/n$), in which case switching wins.

But the variant in this thread has three outcomes.

• You guess right ($p=1/n$), in which case sticking wins
• You guess wrong ($p=(n-1)/n$) and Monty successfully opens $n-2$ doors without revealing the car ($p=1/(n-1)$, for an overall probability of $p=1/n$), in which case switching wins.
• You guess wrong ($p=(n-1)/n$) and Monty reveals a car ($p=(n-2)/(n-1)$, for an overall probability of $p=(n-2)/n$), in which case we abandon the game.

So we can see that the third outcome is conditional on you guessing wrong and "takes away" some of the probability that, in the usual formulation, leads to switching being the better strategy. Particularly with large $n$, a very large fraction of trials where you picked the wrong door will abort because Monty accidentally reveals a car (I tihnk this is the same point @FactChecker is making). The "leftover" probability is the same as the probability you guessed right in the first place, so the outcome of switching in non-aborted games is 50:50.

 

[FactChecker]

The randomness of opening the doors makes a difference. Intuitively it shouldn't matter if the choice was random or not, yet it does matter.

I disagree. Intuitively, knowing that the host may have intentionally avoided the door with the car should make a great difference and should change the odds.

 

[Ibix]

My intuition is that no two people agree about what is intuitive in statistics.

 

[Office_Shredder]

This is actually known as the Monty Fall problem when he picks randomly - because he falls, and grabs into a door and luckily it's a door that leaves the game in a valid state.I independently constructed both the problem and the name, then found out that my naming convention was shockingly unoriginal.