Monty Hall, Erdos, and Our Limited Minds
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> There are two doors with goats and one with a car. You choose one door from the three. The host selects one of the doors with a goat from the remaining two doors, and opens it. Should you switch doors if given the chance?
has a different answer than this one
> There are two doors with goats and one with a car. You choose one door from the three. The host chooses one of the remaining two doors at random and opens it, showing a goat. Should you switch doors if given the chance?
In the first example the hosts choice is either forced (if you chose a goat initially) or doesn't make a difference (if you chose a car).
In the second example, the host could have opened a door with a car because he was choosing the doors at random. Given that he didn't, your estimate of the probability that you have chosen the car should increase - in fact, it should increase to 1/2, and switching doesn't make any difference.
To me, this is the really interesting thing about the Monty Hall problem - the knowledge and intentions of the host are important in the correct solution to the problem! So there are two ways to come up with the answer "it doesn't matter if you switch" -
1. You understand the problem in the first sense above, and your intuition leads you to an incorrect answer.
2. You understand the problem in the second sense above, and you are correct that it doesn't matter whether you switch or not.
> There are two doors with goats and one with a car. You choose one door from the three. The host may select one of the doors with a goat from the remaining two doors, and open it. Should you switch doors if given the chance?
Now the probability can be anything from 0 to 1, depending on what makes the host tick...
Considering the usual ambiguity in the problem formulation, the only safe reply is "It depends".
Proofs are built with an adversarial mind set. Always assuming the worst possible case and proving that a theorem holds in all conditions. The "hand of god" (or in this case the game show host) doesn't move to make theorems easier to prove. That isn't the universe works!
In the second case, you now know that you are not in one of the states of the world where the host accidentally opened a door that had a car behind it. They could have opened a door with a car, but they didn't. That gives you some information - it stands to reason that since they didn't open a door with a car, it was probably a bit more likely that they would pick a door with a goat, which makes it more likely that there are two goats than that there is one car and one goat, which is equivalent to saying that it's a bit more likely that you have already chosen the car.
If the host picked at random and happened to get a goat, it is 1/2 vs 1/2. If the host picked a door he knew had a goat, it's 2/3 vs 1/3.
Again, write the simulation. It shouldn't take you 5 minutes.
The hidden assumption is that the odds that he opens a door containing a car is zero. That is what changes the odds in the second step. Without that they don't change.
Seriously. Write the simulation.
I understand what you were trying to ask but you worded the second question wrong. You can't tell the listener that the host picks a goat. That defeats the purpose of the "randomness" which is meaningless since we know the host picks a door with a goat.
car = wins = 0
many = 100000
actual = 0
many.times do
choice1 = rand(3)
car = rand(3)
#host_opts = [0, 1, 2] - [choice1, car] # host knows what is behind
host_opts = [0, 1, 2] - [choice1] # host doesn't know what is behind
#choice2 = [choice1] # don't switch
choice2 = [0, 1, 2] - [choice1, host_opts.first] # switch
if host_opts.first == car then
# Discard? Automatic win? It doesn't actually matter!
# It only changes the denominator.
else
# According to the puzzle, it is decision time!
wins += 1 if choice2.first == car
actual += 1
end
end
That is false 100% of the time given the parameters that the outcome of the host's choice is a goat (as the question states)
Edited to add:
I think there may be a different misunderstanding here...
Set aside, for the moment, whether the above implements the originally posed question or the variant. Imagine a gameshow that runs the above code, but the switch/stay code is moved below the "decision time!" line and the decision is made based on a prompt rather than hard coded. When you hit that prompt, "Monty" has already picked his door, and we've checked that it's not a car.
Do you expect your chance to win, on answering "switch" at that prompt, to depend on which line above is commented out?
Also, "it only changes the denominator" may be misleading, as it does not change the only denominator actually visible in the code. It changes the total numbers of wins and losses, but not the total numbers of wins and losses given that you saw a goat.
open_door = host_opts.sample
choice2 = [0, 1, 2] - [choice1, open_door] # switch
if open_door == car then
(# of possible worlds in which we win after switching) / (all possible worlds in which we are shown a goat after the initial pick)
vs
(# of possible worlds in which we win without switching) / (all possible worlds in which we are shown a goat after the initial pick)
in this case 2/3 vs 1/3. We are only considering the cases where a goat has been shown so what happens when a goat isn't shown or what the host's intentions are when a goat is show is irrelevant.
1. We simulate a million trials where the contestant chooses one of three door, one of which has a car and two a goat
2. We simulate the host randomly choosing one of the remaining doors to open.
3. We discard all trials that resulted in a car. We are left with the number of trials that resulted in a goat and store that number in a variable 'total'
4. In the remaining trails we switch and reveal what was behind the door. We store the number of times we saw a car in a variable called 'wins'.
5. The probability of winning after switching is 'wins' / 'total'
It doesn't matter how many times the host shows a car because those trials are discarded.
If the host picks randomly and you discard rounds with a car, your odds switching are 50/50. If the host uses knowledge of where the car is to definitely reveal a goat, your odds switching are better.
I've written the simulation, and you're right. It is 50% when the host chooses randomly!
Two game shows, two hosts, both reveal a goat. The random host revealed a goat with 1/3 probability; the canonical host revealed a goat with 1 probability.
We know that both hosts revealed a goat. We also know (for the purposes of this problem) that they had different likelihoods of doing so before the fact.
Does it make intuitive sense to you that the prior probability of an event could affect how you react to it? That's all that's going on here.
I understand the math. I know what the op was trying to say. But they worded it wrong. The 2nd riddle is the same as the first.
Are you familiar with the concept of prior probability? The premise of the meta-riddle is that both hosts showed a goat, but one host could have not shown a goat. So there must be a difference in the information conveyed by that goat.
Again, there isn't anything special about this problem, it's just how probabilities work. Respectfully-- there are a number of different people trying to explain this to you in different ways, and you can verify it on Wikipedia. Are you positive you do understand the math?
If "... and shows you a goat" is a part of the procedure of the game show, there is additional information imparted by the host's selection compared to the situation where "... and shows you a goat" is an artifact of the particular play through the game that is being described in the riddle and the procedure by which that situation was arrived at did not incorporate knowledge of where the car was.
If the host picks randomly, and the host picks a goat, and you switch, your odds of a car are 1/3, just as if you'd stuck. There's no point.
If the host always chooses a goat, and the host picks a goat, and you switch, your odds are 2/3 in favor of now having a car. You should always switch.
Intuition suggests this. Statistics proves it. If you implement this in software, you will see it happening to you in black and white, right before your eyes.
1/2, once you see the goat. One of 1/3, 2/3, and 1/2 beforehand, depending on what happens when Monty shows a car: whether you automatically win, automatically lose, or start over from the top, respectively.
Again, write the simulation. Either the switch strategy will win 2/3 of the time or 1/2 of the time, depending on how you write Monty. Only one of those approaches matches the results we saw IRL - Monty never revealed the car (so far as I'm aware) - but either is consistent with many phrasings of the question, and OP correctly specified the other case above.
Specifically, the host in the first formulation never reveals a car, even if she wants to. The host in the second formulation reveals a car roughly 1/3 of the time, even if she'd rather not. Thus, intuitively, it makes sense that the former leaks information about the place of the car, and the latter doesn't. [0]
Of course that's just a rewording of your point, based on my sense that "knowledge and intentions" are concepts that apply readily to game show hosts, but awkwardly to rule-following systems in general.
[0] To the extent that revealing something can not leak information about it...
Edit: To elucidate my quibble further, as a third formulation: "The host chooses one of the two remaining doors at random, which, by coincidence, is always a goat." I assert we should always switch: The intention of the host is the same as in the second case, but the behavior of the system, magically known to us, is that of the first. We make decisions based on that.
But really, it has to do with the information content of the actions of the host.
If you're a mathematician you don't tend to introduce any additional assumptions unless they are stated. Assuming the host has knowledge of what is behind all of the doors (or at least where the car is) changes everything. Some wordings of the question include it, others don't.
Of course a non-mathematician might assume things about the game show host based on their knowledge of game show hosts but a mathematician is entirely correct to simply take the question at face value and not inject additional unstated assumptions. Otherwise what is preventing them from putting in other completely wacky assumptions like "the host wants you to win the car and knows you know about the monty hall problem and is using reverse psychology"?
If I was in a culture where gameshow hosts behaved differently the question would have different priors and a different answer.
What if all gameshow hosts shot contestants immediately after they won? Would you be better of switching? No, you'd be dead, so you didn't "win" the car. Completely silly example but it shows why this can't be taken seriously as a mathematical question with probabilities without injecting additional assumptions (unless of course it is stated clearly but then I bet few mathematicians would get tripped up).
[1] http://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-doo...
Of course, i was wrong :D
The simulation testing the four possible scenarios:
#!/usr/bin/env ruby
def run_games(monty_knows:, change_choice:)
options = [:goat1, :goat2, :car]
total_games = 0
wins = 0
100000.times do
choice = options.sample
monty_choice = (options - [choice, (:car if monty_knows)]).sample
# Don't consider cases where Monty shows a car.
next if monty_choice == :car
if change_choice
choice = (options - [monty_choice, choice]).first
end
total_games += 1
wins += 1 if choice == :car
end
wins / total_games.to_f
end
puts 'Monty knows & keep choice: %.2f' %
run_games(monty_knows: true, change_choice: false)
puts 'Monty knows & change choice: %.2f' %
run_games(monty_knows: true, change_choice: true)
puts 'Monty ignores & keep choice: %.2f' %
run_games(monty_knows: false, change_choice: false)
puts 'Monty ignores & change choice: %.2f' %
run_games(monty_knows: false, change_choice: true)
Monty knows & keep choice: 0.33
Monty knows & change choice: 0.67
Monty ignores & keep choice: 0.50
Monty ignores & change choice: 0.50
After verifying that the simulation makes sense, i tried to figure out why it makes sense.
In an information theory kinda way, it makes sense that if Monty picks at random then he is not providing any new information: you are just being shown what was behind a random door you didn't choose and you can't choose now. But that explanation is not intuitive to my brain.
I found it more intuitive to actually draw a decision tree with the possibilities and their probabilities. The tree is actually very simple:
With the Always Change Door strategy:
car (1/3) -> LOSE (Monty always reveals goat,
/ we change and we always lose)
first choice
\ goat (1/2) -> WIN (we change to the
\ / car door)
goat (2/3) -> Monty choice
\
car (1/2) -> IGNORE
With the Always Keep Door strategy:
car (1/3) -> WIN (doesn't matter what Monty chooses)
/
first choice
\ goat (1/2) -> LOSE (we keep our
\ / goat-hiding door)
goat (2/3) -> Monty choice
\
car (1/2) -> IGNORE
In conclusion, my intuition sucks. Thanks for showing me so!
Self-doubt is a quantity often found wanting among smart people, computer programmers, and especially among the intersection ;)
"Pigeons repeatedly exposed to the problem show that they rapidly learn always to switch, unlike humans (Herbranson and Schroeder, 2010)."
This just blows my mind and, as you might tell from my username, I'm a big fan of the bird that some denigrate as 'rats with wings'; I prefer to think of them as rat angels.
It amazes me how many people believe that human beings are primarily logical beings.
The best way I've come up to explain the Monty Hall problem is that in the first place you have a 2/3 chance of choosing a Goat and in that case switching guarantees you will win the car. Alternatively you have a 1/3 chance you chose the car to start with and switching guarantees you will win a goat. Therefore a strategy of always switching gives you a 2/3 chance of winning the car.
a) because a goat is revealed, swapping will always give you the opposite of what you had
b) you're more likely to pick a goat in the first place, so you're more likely to switch to a car
The chance is 1/3 the car is behind the door chosen at first. The chance is 0 for the opened door. As the three chances must sum to 1, the last door has a chance of 2/3.
You pick. The doors are now in two groups: your door (1/3 chance) and the other two doors (2/3 chance).
The host picks. The second group is now reduced to a single door, still with 2/3 chance.
Now you decide - stick with the door with 1/3 chance, or switch to the door with 2/3 chance?
This is because the way the problem is structured gives you six choices, not three: [A, A], [A, switch], [B, B], [B, switch], [C, C], [C, switch]. 3 of these are wins and three of these are losses. If you choose from these six options randomly you will win 50% of the time, but if you only choose from the "switch" options you will win 2/3 of the time.
Your variant is a valid analogy too, but the probabilities in your scenario between sticking and switching are so close to each other, it's not going to trigger any new intuition in someone who is already confused.
Don't think of the 100 door puzzle as an analogy with the 3 door puzzle, think of it as a class of puzzle where there are N doors, you can pick one, and then the host (who knows what's behind the doors) reveals N-2 fails. It's clear that the 3 door puzzle is a specific instance of that, and if you have understood something intuitively about that puzzle class through considering N being a large number, that will still hold for N being a small number.
The class of puzzles that you describe is M doors, where you pick one and then the host (who knows what's behind the doors) reveals one fail. The three door puzzle is also a specific instance of that puzzle, but I doubt that you've learned anything from that, since examining M at other numbers probably hasn't given you any insight (still, if it did, those insights are indeed applicable to 3 door Monty Hall).
I've tried a number of ways of explaining this problem to people, and empirically the 100 doors one is the way that seems to work best. Interested to hear any versions you tell that seem to help people understand whats going on.
Let's play a similar game with three doors, two goats and a car. But you get to pick one of two rules to play by:
A) You get to pick one door you want and get what's behind that door.
B) You get to pick one door you don't want and get the better prize behind the other two doors.
Pretty much everyone is going to pick rule B. So after picking the door you don't want I will open the other doors. If the car is behind one of those doors I will always open that door last. Does the last sentence make you wish you picked rule A to play by? Heck no. We already know that one of those two will be a goat. The order I open the doors won't change anything.
Even if I let you retroactively switch to rule A after opening the first door? This is what trips people up. They think they're getting new information and that their original door choice may be incorrect.
But in this version of the game it's easy to see that we're not changing our choice of door, we're changing from the better game rule B to the worse rule A.
So the correct choice is to always stick with rule B.
Now we map the rules onto the real Monty Hall game. Rule A is pick the door you want and stick with it. Rule B is pick the door you don't want and switch.
So the correct choice is to always switch.
You're allowed to pick 1 door out of 100. Your chance of being correct is 1%. Then, the host gives you the option of swapping to the entire other set of 99 doors. Of course, your chance of hitting the car would be 99%, so you swap.
Opening 98 doors with goats behind them is expected, and so just because the host chooses to open 98 goats in a row doesn't mean that the chance of the last 2 doors (your final door out of the 99, and your original 1 door) both have an equal chance. Your set of 99 doors was opened in a specific order and that's what actually matters. The 99th door still has a 99% chance to contain the car, because your set had a 99% chance to contain the car. The order is everything!
There are only 2 possible paths in the problem: Path 1: you originally picked a goat (p = 2/3), now you swap and get the prize
Path 2: you orinally picked the prize (p = 1/3), now you swap and get nothing
This depends in no way on the host having knowledge.
So now, suppose you get to bet on who will win.
Its clear that the chance of the player winning is 1/3. Thus the chance of Monty winning is 2/3. So bet on Monty.
To change it back to the original problem, you are also the player in the game. So now, do you bet on yourself or on Monty?
The chance for the host getting the car is of course also 1/3. Then the complement 2/3 chance of the host getting a goat splits the car chance evenly between the two remaining doors as 1/3 for each door.