9 ms·

Math Goblins

72 points by ColinWright 5y ago

drzoltar 5y ago
Surely there must also be Math Gremlins, who are chipping away at its foundations. CF Gödel and Chaitin’s Incompleteness Theorems.
- hdjjhhvvhga 5y ago
  Actually gremlins are a bit different creatures: https://en.wikipedia.org/wiki/Foo_fighter https://en.wikipedia.org/wiki/Foo_fighter
oceliker 5y ago
The votey (red button to the bottom right of comic) says i^i is a real number. I didn't understand how at first, but the proof seems straightforward: https://math.stackexchange.com/questions/216871/why-is-ii-real https://math.stackexchange.com/questions/216871/why-is-ii-re...
- whatgoodisaroad 5y ago
  That proof uses Euler's formula, which is the real brain melter.
- warent 5y ago
  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
  - nohuck13 5y ago
    https://youtu.be/pq9LcwC7CoY https://youtu.be/pq9LcwC7CoY
- camjw 5y ago
  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...
  - shonenknifefan1 5y ago
    Yes, though regardless of which branch you pick to define i^i, it's value is real.
cortesoft 5y ago
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.
- mrob 5y ago
  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.
  - ketzu 5y ago
    What I personally dislike about the rephrasing is the ambiguity around the states "what happens when Monty selects the prize". I think that's the more important question.
    - Jensson 5y ago
      If he reveals the prize you just lose, the chance to change your option only happens if he picks wrong. It isn't a strange game. You pick one of 3 doors. The host then reveals the correct door and say you lost. Nobody would raise an eyebrow if that happened. Lets take this scenario to show how changing would make you lose: A conman runs a game where you pick one of 3 options and can win a prize. If you pick the correct option the conman will reveal one of the other and ask you if you want to switch. Switching then means you always lose. However if you pick the wrong one then he reveals the correct one and say you lost. So in this case following the rules of the original solutions means you are 100% likely to lose the game. That people have such a hard time understanding this part of the problem is also due to not understanding statistics properly. If what I said above doesn't make sense to you then you don't understand the Monty Hall problem.
archgrove 5y ago
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.
- klipt 5y ago
  That's correct if Monty always opens the doors. But suppose Monty only opens the other doors if you picked the prize, then obviously you shouldn't switch.
  - Deestan 5y ago
    That would be a different problem. In Monty Hall, Monty always opens a door, and you're left to choose between your picked door and one other door.
    - Jensson 5y ago
      This isn't the Monty Hall problem though. The comic doesn't mention the show. It just has a similar setup so people just jump to the same solution even though it is wrong in this case. The correct answer is that it is impossible to know what the chances are if you switch since the problem lacks enough information to tell.
      - Out_of_Characte 5y ago
        How is it wrong in this case? your picked door had a 33% chance to have a price, since its one of three doors, now that a goblin is revealed the other door has a 66% chance of hiding the price. exactly like the monty hall problem
        Jensson 5y ago
        No, you are wrong. The way the comic described it only describes a single instance of the host opening the door in this case. It could very well be that this host doesn't want you to win, and thus reveals another door and tries to goad you into switching only when you have picked the correct door. That is a very reasonable interpretation of a game like this, and in that case the correct answer is to not switch. If you think that this description means that the host always opens a dud after you open the first, then you are wrong. It doesn't say that anywhere, and there is no reason to believe that this game show works that way. The only reason you think it does is that you connect it to the original Monty Hall problem which worked that way, which causes you to make the mistake thinking the same rules apply here.
maxk42 5y ago
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.
westcort 5y ago
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.
codr7 5y ago
Right, good luck with that. 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 5y ago
  The missing piece is: have the doors been allocated before the reveal and allowing the switch? If yes, then the chance is 2/3 for switching.
- ColinWright 5y ago
  If there are two doors, and one prize, and you can choose one door, then you're right, it's a 50% chance. 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.
  - codr7 5y ago
    But that is the situation, there are exactly two doors and one price left.
    - ColinWright 5y ago
      We are either talking past each other, or you are missing the point. In the Monty Hall problem it's not 50:50. Really. It's not. * There are three door, one conceals a prize; * You choose one door; * The host, knowing where the prize is, opens one of the other doors to reveal a non-prize; * You have the option to stick with your choice, or to switch; * If you stick, your probability of having the prize is 1/3; * If you switch, your probability of having the prize is 2/3. There's enough discussion elsewhere, or you can write your own simulation (but make sure you simulate the right thing, and not something different). As others have pointed out, you need to know in advance that (a) the host will always open another door, (b) the host knows where the prize is located, and (c) the host will never reveal the prize. But given that, the odds really are not 50:50.
      - 5y ago
c1ccccc1 5y ago
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.
Jensson 5y ago
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.
- ColinWright 5y ago
  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?
  - mrob 5y ago
    The implications are not clear. I suspect you only think they're clear because you already know the classical problem, or perhaps because you have more faith in the good intentions of strangers than I do. My personal reaction to the scenario described in the comic would be that the host was trying to trick me because it would make for more dramatic television (only revealing the goat because I'd picked the car), and the correct decision would be to refuse the offer. Your belief (a) is compatible with my common sense assumption that the host's goal is to maximize drama by making me lose the car, not to help me. Your belief (b) is incompatible with my assumption, but it is not required by the wording of the Math Goblins problem. Your belief (c) is compatible with my assumption. All of Jensson's comments in this discussion are correct, and they are being unfairly downvoted.
bogosmith 5y ago
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.
jrandm 5y ago
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.
mcguire 5y ago
Also, there are no electrons. (https://www.amazon.com/There-Are-Electrons-Electronics-Earthlings/dp/0962781592 https://www.amazon.com/There-Are-Electrons-Electronics-Earth...)
Jensson 5y ago
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.

72 points by ColinWright 5y ago

Surely there must also be Math Gremlins, who are chipping away at its foundations. CF Gödel and Chaitin’s Incompleteness Theorems.

hdjjhhvvhga 5y ago
Actually gremlins are a bit different creatures: https://en.wikipedia.org/wiki/Foo_fighter https://en.wikipedia.org/wiki/Foo_fighter

oceliker 5y ago

The votey (red button to the bottom right of comic) says i^i is a real number. I didn't understand how at first, but the proof seems straightforward: https://math.stackexchange.com/questions/216871/why-is-ii-real https://math.stackexchange.com/questions/216871/why-is-ii-re...

whatgoodisaroad 5y ago
That proof uses Euler's formula, which is the real brain melter.
warent 5y ago
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
- nohuck13 5y ago
  https://youtu.be/pq9LcwC7CoY https://youtu.be/pq9LcwC7CoY
camjw 5y ago
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...
- shonenknifefan1 5y ago
  Yes, though regardless of which branch you pick to define i^i, it's value is real.

cortesoft 5y ago

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.

mrob 5y ago
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.
- ketzu 5y ago
  What I personally dislike about the rephrasing is the ambiguity around the states "what happens when Monty selects the prize". I think that's the more important question.
  - Jensson 5y ago
    If he reveals the prize you just lose, the chance to change your option only happens if he picks wrong. It isn't a strange game. You pick one of 3 doors. The host then reveals the correct door and say you lost. Nobody would raise an eyebrow if that happened. Lets take this scenario to show how changing would make you lose: A conman runs a game where you pick one of 3 options and can win a prize. If you pick the correct option the conman will reveal one of the other and ask you if you want to switch. Switching then means you always lose. However if you pick the wrong one then he reveals the correct one and say you lost. So in this case following the rules of the original solutions means you are 100% likely to lose the game. That people have such a hard time understanding this part of the problem is also due to not understanding statistics properly. If what I said above doesn't make sense to you then you don't understand the Monty Hall problem.

archgrove 5y ago

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.

klipt 5y ago
That's correct if Monty always opens the doors. But suppose Monty only opens the other doors if you picked the prize, then obviously you shouldn't switch.
- Deestan 5y ago
  That would be a different problem. In Monty Hall, Monty always opens a door, and you're left to choose between your picked door and one other door.
  - Jensson 5y ago
    This isn't the Monty Hall problem though. The comic doesn't mention the show. It just has a similar setup so people just jump to the same solution even though it is wrong in this case. The correct answer is that it is impossible to know what the chances are if you switch since the problem lacks enough information to tell.
    - Out_of_Characte 5y ago
      How is it wrong in this case? your picked door had a 33% chance to have a price, since its one of three doors, now that a goblin is revealed the other door has a 66% chance of hiding the price. exactly like the monty hall problem
      - Jensson 5y ago
        No, you are wrong. The way the comic described it only describes a single instance of the host opening the door in this case. It could very well be that this host doesn't want you to win, and thus reveals another door and tries to goad you into switching only when you have picked the correct door. That is a very reasonable interpretation of a game like this, and in that case the correct answer is to not switch. If you think that this description means that the host always opens a dud after you open the first, then you are wrong. It doesn't say that anywhere, and there is no reason to believe that this game show works that way. The only reason you think it does is that you connect it to the original Monty Hall problem which worked that way, which causes you to make the mistake thinking the same rules apply here.

maxk42 5y ago

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.

westcort 5y ago

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.

codr7 5y ago

Right, good luck with that. 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 5y ago
The missing piece is: have the doors been allocated before the reveal and allowing the switch? If yes, then the chance is 2/3 for switching.
ColinWright 5y ago
If there are two doors, and one prize, and you can choose one door, then you're right, it's a 50% chance. 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.
- codr7 5y ago
  But that is the situation, there are exactly two doors and one price left.
  - ColinWright 5y ago
    We are either talking past each other, or you are missing the point. In the Monty Hall problem it's not 50:50. Really. It's not. * There are three door, one conceals a prize; * You choose one door; * The host, knowing where the prize is, opens one of the other doors to reveal a non-prize; * You have the option to stick with your choice, or to switch; * If you stick, your probability of having the prize is 1/3; * If you switch, your probability of having the prize is 2/3. There's enough discussion elsewhere, or you can write your own simulation (but make sure you simulate the right thing, and not something different). As others have pointed out, you need to know in advance that (a) the host will always open another door, (b) the host knows where the prize is located, and (c) the host will never reveal the prize. But given that, the odds really are not 50:50.
    - 5y ago

c1ccccc1 5y ago

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.

Jensson 5y ago

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.

ColinWright 5y ago
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?
- mrob 5y ago
  The implications are not clear. I suspect you only think they're clear because you already know the classical problem, or perhaps because you have more faith in the good intentions of strangers than I do. My personal reaction to the scenario described in the comic would be that the host was trying to trick me because it would make for more dramatic television (only revealing the goat because I'd picked the car), and the correct decision would be to refuse the offer. Your belief (a) is compatible with my common sense assumption that the host's goal is to maximize drama by making me lose the car, not to help me. Your belief (b) is incompatible with my assumption, but it is not required by the wording of the Math Goblins problem. Your belief (c) is compatible with my assumption. All of Jensson's comments in this discussion are correct, and they are being unfairly downvoted.

bogosmith 5y ago

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.

jrandm 5y ago

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.

mcguire 5y ago

Also, there are no electrons. (https://www.amazon.com/There-Are-Electrons-Electronics-Earthlings/dp/0962781592 https://www.amazon.com/There-Are-Electrons-Electronics-Earth...)

Jensson 5y ago

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.