The Monty Hall problem continued to short circuit the “intuitive reasoning” part of my maths brain until I saw it explained as “Imagine there are a million goats and one car. You pick a door, Monty opens 999,998 doors. Should you switch?”, to which the answer is clearly “Of course” (you had a one in a million chance the first time). Then reduce from there.
The problem with the Monty Hall is that there is a hidden rule that people never explain. If you put there rule there explicitly, it's reasonably easy to see that the presenter is messing with your up-front probabilities when he opens a door.
I vaguely remember a 3blue1brown video that explains this really well too. Something to do with thinking about it in terms of rotations on the complex number plane which can land you back on the axis of real numbers. That was the most intuitive thing that I remember seeing
Overly pedantic but I think i^i is actually not well-defined (and in general exponentiation of complex numbers is not well defined) since e^(2 * pi * i) = 1 so for instance we would have i = e^(0.5 * pi * i) = e^(2.5 * pi * i) and then i^i = e ^(-0.5 * pi) = e^(-2.5 * i) but those last two numbers are both real and not equal so complex exponentiation only works when you pick a branch of the complex logarithm. Anyway...
By this point you're probably used to the idea that most of the real numbers are irrational numbers. That is, even though there are countably many rational numbers, there are uncountably many real numbers.
Numbers like pi and e, even though they are irrational, can still be described. There is still an algorithm that computes their digits. The same goes for 0.112358132134... or Apery's constant. But the vast majority of real numbers can never even be described.
Proof: Any description of a number (any algorithm that computes its digits) is a finite string of symbols. There are countably many finite strings of symbols, but uncountably many reals. Math goblins.
I still have trouble understanding why people struggle with the Monty Hall problem. It seems so obvious to me how the math works, yet very smart people have been fooled by it even after having it explained. I don’t get it.
I think part of it is that the answer depends on the fact that the host's strategy is to always open a door with a goat behind it that the player didn't choose. If the host instead opened one of the doors the player didn't choose at random, then one third of the time the prize will be revealed. The remaining two thirds of the time, switching and keeping your choice have an equal probability of getting you the prize.
Many descriptions of the problem don't tell you what the host's strategy actually is, so I'd guess that a lot of people who are "fooled" by the problem actually just have a different reasonable interpretation of what it's asking.
It's because the problem is usually stated incorrectly. The character in the comic gets it wrong.
The correct formulation is "Monty knows where all the goats are and is compelled to always reveal a goat." The alternative interpretations are "Monty opens a door at random that happens to reveal a goat" (in which case the 50:50 interpretation is correct), or "Monty reveals a goat only if you picked the car" (in which case you should never switch). All three interpretations are compatible with the version stated in the comic.
I wonder if you could trick people who mistake Monty hall like this.
Create a Casino, let users pick one of 3 doors, and give them 4x the money back if they pick the right one. Seems like bad odds for the casino, right? But the trick is, if they pick the right door you open another and give them the option to choose. Naïve as they are they will think that now they have 2/3 chance to win if they switch. But overall with these rules they have 0% chance to win. It will work as long as you only play the game once per person.
The Monty hall problem is a very nice example of why reasoning with conditional probabilities is hard. If you have studied conditional probabilities and encounter the Monty Hall problem it is easy to think you are in the situation to apply your knowledge and reason like that:
I know that the prize is not behind door A (because the host showed me the goat there).
Given that the prize is not behind door A calculate the probability that it is behind door B.
Well that's obviously 1/2. Why? Just use conditional probabilities that's what they are for.
(The calculation is trivial: Event X="prize behind B", event Y="prize not behind A",
P(X)=1/3, P(Y)=2/3. P(X and Y)=1/3.
P(X|Y)=P(X and Y)/P(Y) = 1/2.)
Exactly why this argument is wrong is the subtle part. Of course the argument doesn't put to use part of the information given - which door I chose originally and what contract I had with the host, so this is normally enough to make one uncomfortable, but still we know that, e.g.
"In probability theory, conditional probability is a measure of the probability of an event occurring, given that another event (...) has already occurred."
(source: wikipedia)
Having thought about it I find that adding some more precision to the intuitive motivation of conditional probabilities might be helpful. I certainly think twice now before using conditional probabilities to model the real world, but realizing that I should has cost me quite a bit of head scratching.
I think most of the Monty Hall confusion comes from focusing too much on the question and the math and missing the important part: The confusing setup is part of the game! Essentially it works like this:
Host: Contestant, please choose one of the 3 doors
Contestant: I choose door #1
Host: Confusing jibber-jabber to raise tension and keep this fun to watch
Contestant: I don't know what to do!
Host: Do you think the prize is behind your selection, door #1, or behind any of the others?
Expressed this way there's no confusion but it's the exact same problem. The contestant first selected 1/3 doors and the host is offering to let Contestant keep that 1/3 or choose the other 2/3. Any obfuscation in between is intentional for the sake of the game.
Let’s say 1 in 50 people have a genius-level IQ. What are the odds that a random group of 50 people doesn’t have one such person? The answer is 37%, or 1/e. Why? Because the chances a given person isn’t a genius is 49/50. For 50 people, the odds that none are geniuses is (49/50)^50, which is about 37%. And limit as n approaches infinity of ((n-1)/n)^n is 1/e. Math goblins.
The way he presented it there's a 50% chance that opening another door reveals a goat. I've of the other doors will always contain a goat The key piece of information that is merely implied by the Monty Hall problem but not explicitly defined is that the host knows which door is the winning door and deliberately opened a losing door. If the host does this purposefully then you should switch your answer because two thirds of the time you will have chosen a goat and the host will be forced to reveal the other losing door. However if the host has opened a door at random and it happens to reveal a goat then changing your answer will not improve your odds. There is a 50% chance that the remaining door is a goat rather than a two thirds chance. He could have just as easily revealed the prize. Explained in this way the Monty Hall problem is not confusing. The wording is what was confusing.
The correctly described monty hall problem has 2/3 chance if you switch. This isn't the correctly described monty hall problem, so there is no way to know what the chance is if you switch. It doesn't even say it is the monty hall show, so you can't even patch it up using external knowledge. Instead these are just some loose meme bits that perpetuates a lie because the author didn't fully understand the Monty Hall problem.
You are being repeatedly downvoted because you are repeatedly saying that the description is not of the classical Monty Hall problem. Here's the description in the comic:
* Suppose you can pick door A, door B, or door C.
* Two doors hide a goat. One hides a prize.
* You pick one, then the host reveals that one of the other doors hides a goat.
* Should you switch?
Let me say that I think that in this description it is fairly clearly implied that:
a) The host reveals the presence of a goat by opening one of the other doors;
b) The host will open a door, regardless of whether you picked the prize or not;
c) The host knows where the prize is, and specifically opens a door to reveal a goat.
I believe (a) because we know by counting that one of the other doors hides a goat. If saying "the host reveals that one of the other doors hides a goat" is not content free, then it must imply that the host actually does some revealing.
I believe (b) because we are told that the host does reveal a goat in (a). With no conditioning mentioned I believe it is fair to assume that the host will always open a door.
I believe (c) because if the host does not know where the prize is, then they must open a door at random, and would not be guaranteed to reveal a goat. I believe the statement as it stands precludes that.
So, can you tell us exactly why you think this is an incorrect description? Can you tell us exactly why the probability of winning the prize upon switching are not improved to 2/3?
[+] [-] archgrove|4 years ago|reply
[+] [-] marcosdumay|4 years ago|reply
[+] [-] klipt|4 years ago|reply
But suppose Monty only opens the other doors if you picked the prize, then obviously you shouldn't switch.
[+] [-] oceliker|4 years ago|reply
[+] [-] warent|4 years ago|reply
[+] [-] camjw|4 years ago|reply
[+] [-] whatgoodisaroad|4 years ago|reply
[+] [-] c1ccccc1|4 years ago|reply
Numbers like pi and e, even though they are irrational, can still be described. There is still an algorithm that computes their digits. The same goes for 0.112358132134... or Apery's constant. But the vast majority of real numbers can never even be described.
Proof: Any description of a number (any algorithm that computes its digits) is a finite string of symbols. There are countably many finite strings of symbols, but uncountably many reals. Math goblins.
[+] [-] cortesoft|4 years ago|reply
[+] [-] c1ccccc1|4 years ago|reply
Many descriptions of the problem don't tell you what the host's strategy actually is, so I'd guess that a lot of people who are "fooled" by the problem actually just have a different reasonable interpretation of what it's asking.
[+] [-] mrob|4 years ago|reply
The correct formulation is "Monty knows where all the goats are and is compelled to always reveal a goat." The alternative interpretations are "Monty opens a door at random that happens to reveal a goat" (in which case the 50:50 interpretation is correct), or "Monty reveals a goat only if you picked the car" (in which case you should never switch). All three interpretations are compatible with the version stated in the comic.
[+] [-] riffraff|4 years ago|reply
So if I switch door and pick one at random it should be 50% right even if I pick the same one I had picked before.
I trust the smart people on this, but I'm going with the math goblins explanation from now on.
[+] [-] Jensson|4 years ago|reply
Create a Casino, let users pick one of 3 doors, and give them 4x the money back if they pick the right one. Seems like bad odds for the casino, right? But the trick is, if they pick the right door you open another and give them the option to choose. Naïve as they are they will think that now they have 2/3 chance to win if they switch. But overall with these rules they have 0% chance to win. It will work as long as you only play the game once per person.
[+] [-] bogosmith|4 years ago|reply
I know that the prize is not behind door A (because the host showed me the goat there). Given that the prize is not behind door A calculate the probability that it is behind door B. Well that's obviously 1/2. Why? Just use conditional probabilities that's what they are for. (The calculation is trivial: Event X="prize behind B", event Y="prize not behind A", P(X)=1/3, P(Y)=2/3. P(X and Y)=1/3. P(X|Y)=P(X and Y)/P(Y) = 1/2.)
Exactly why this argument is wrong is the subtle part. Of course the argument doesn't put to use part of the information given - which door I chose originally and what contract I had with the host, so this is normally enough to make one uncomfortable, but still we know that, e.g.
"In probability theory, conditional probability is a measure of the probability of an event occurring, given that another event (...) has already occurred." (source: wikipedia)
Having thought about it I find that adding some more precision to the intuitive motivation of conditional probabilities might be helpful. I certainly think twice now before using conditional probabilities to model the real world, but realizing that I should has cost me quite a bit of head scratching.
[+] [-] jrandm|4 years ago|reply
Host: Contestant, please choose one of the 3 doors
Contestant: I choose door #1
Host: Confusing jibber-jabber to raise tension and keep this fun to watch
Contestant: I don't know what to do!
Host: Do you think the prize is behind your selection, door #1, or behind any of the others?
Expressed this way there's no confusion but it's the exact same problem. The contestant first selected 1/3 doors and the host is offering to let Contestant keep that 1/3 or choose the other 2/3. Any obfuscation in between is intentional for the sake of the game.
[+] [-] westcort|4 years ago|reply
[+] [-] drzoltar|4 years ago|reply
[+] [-] hdjjhhvvhga|4 years ago|reply
[+] [-] maxk42|4 years ago|reply
[+] [-] mcguire|4 years ago|reply
[+] [-] codr7|4 years ago|reply
2 doors to choose from, 1 price, 50% chance of winning, period.
And I don't care if there were initially a million doors to choose from.
[+] [-] the_gipsy|4 years ago|reply
[+] [-] ColinWright|4 years ago|reply
That's not the situation, and the chance is not 50%. I don't care if you agree or not, you are mistaken if you think the chances are 50:50.
[+] [-] Biganon|4 years ago|reply
Or maybe I missed the joke
[+] [-] Jensson|4 years ago|reply
[+] [-] ColinWright|4 years ago|reply
* Suppose you can pick door A, door B, or door C.
* Two doors hide a goat. One hides a prize.
* You pick one, then the host reveals that one of the other doors hides a goat.
* Should you switch?
Let me say that I think that in this description it is fairly clearly implied that:
a) The host reveals the presence of a goat by opening one of the other doors;
b) The host will open a door, regardless of whether you picked the prize or not;
c) The host knows where the prize is, and specifically opens a door to reveal a goat.
I believe (a) because we know by counting that one of the other doors hides a goat. If saying "the host reveals that one of the other doors hides a goat" is not content free, then it must imply that the host actually does some revealing.
I believe (b) because we are told that the host does reveal a goat in (a). With no conditioning mentioned I believe it is fair to assume that the host will always open a door.
I believe (c) because if the host does not know where the prize is, then they must open a door at random, and would not be guaranteed to reveal a goat. I believe the statement as it stands precludes that.
So, can you tell us exactly why you think this is an incorrect description? Can you tell us exactly why the probability of winning the prize upon switching are not improved to 2/3?