10 ms·

Probability, Paradox, and the Reasonable Person Principle

169 points by yomritoyj 11y ago

jgrahamc 11y ago
The hand-wavy way I've always thought about the Monty Hall problem is as follows: When the contestant picked a box he was making a random choice between three boxes and had a probability of 1/3 of picking the car. The probability that one of the other two boxes was the car is thus 2/3. When Monty Hall opens a box it's still the case that the probability of the car being in the other two is 2/3 but the contestant can now eliminate the open box and knows that the probability that the car is in the other box is 2/3. Thus it's worth switching.
- mattdeboard 11y ago
  This is by far the most simple, concise explanation of the Monty Hall problem I've ever read: https://www.reddit.com/r/videos/comments/1bcuo8/this_has_been_explained_to_me_over_and_over_again/c95w7h9 https://www.reddit.com/r/videos/comments/1bcuo8/this_has_bee... > Basically, you probably got a goat the first time, so switch. It's so concise that I use more words to re-post it every time Monty Hall comes up than are in the explanation itself.
  - adevine 11y ago
    The other phrasing of the problem that I've seen help people is this (I think was also originally from Marilyn Vos Savant): Suppose there are a thousand doors, and only one is a car. Monty opens 998 of them to reveal goats, leaving door #628 still hidden. Do you switch? That helps people understand that their original probability of picking the car the first time was very small, and didn't change because of what Monty did.
    - darkmighty 11y ago
      Most paradoxes vanish, become obvious this way (when you increase the numbers). The brothers paradox mentioned: 'I have two children. At least one is a boy.'/'I have two children. At least one is a boy born on Tuesday.'; 'What is the probability I have two boys?' drastically changes when you change to 'I have two children. At least one is a boy born at 7h:35m:42s.' -- it becomes clear that the succeeding information exactly specifies the boy as it becomes specific, in the limit turning into 'I have two children. The first is a boy.' (so the probability goes from 1/3 to 1/2).
    - thaumasiotes 11y ago
      I prefer this one: You pick a door. Monty opens another door, revealing a goat. You now get the option to switch from your door to the other two doors, keeping the car if it's behind either one. Whether Monty opens the door or not doesn't mean much.
      - baddox 11y ago
        An even simpler version, which is equivalent: Monty Hall says that you can either choose 1 door, or 2 doors. It can't get more obvious than that.
        dllthomas 11y ago
        That is only equivalent if Monty (as intended but usually not stated clearly) always picks a door without a car using knowledge of where the car is. If, instead, Monty picks randomly from the other two doors and just happens not to have revealed the car this time, odds are the same whether you stick with your door or switch. Even in the case where your restatement is mathematically equivalent, it's obviously not at all equivalent as a puzzle because what is being requested by the puzzle is seeing that the original situation is mathematically equivalent to that one.
- proksoup 11y ago
  Monty chooses a door? I thought he selected it at random. That was always the confusing unstated assumption for me, that makes me think the problem is only confusing because of how the wording usually de-emphasizes that distinction. If he chooses a door, you can benefit from the secret information he reveals sometimes by doing so. If the door was chosen at random, he is not adding any information, so you can't act on it?
  - maxerickson 11y ago
    He never reveals a car when he opens the door.
    - malisper 11y ago
      That actually doesn't matter. What does matter is if he knows he is going to reveal a door that is empty. If he does know you should switch. If he reveals a door arbitrarily, and it just so happens to be empty, it doesn't matter if you switch. Edit: Since I have been downvoted, here is a citation: http://probability.ca/jeff/writing/montyfall.pdf http://probability.ca/jeff/writing/montyfall.pdf (see Monty Fall problem)
      - maxerickson 11y ago
        I agree that my phrasing is not superiorally precise, but the meaning is clear enough, the choice of the door isn't arbitrary.
      - kgwgk 11y ago
        To be fully pedantic, what matters is not "if he knows he is going to reveal a door that is empty". Maybe he has no idea and he just reads his lines off a teleprompter. What matters is if you know that he is going to reveal a door that is empty, because then you can reason according to that information.
      - baddox 11y ago
        The notion that he always randomly opens a door without knowing what's behind it but the door is always a losing door is fairly bizarre and doesn't make much sense when you try to analyze probability in the context of repeated experiments. Something must make him always choose a losing door. Either he knows what's behind the doors, or there's some external force ("fate," "God," or whatever) that causes the door to always be a losing door. Regardless, the problem as stated has Monty Hall open a losing door. Thus it's clear that you should switch after Monty Hall does so, and a very simple computer program can show that you tend to win by switching. Another way to phrase the problem is that you choose one door, then Monty Hall (without revealing anything) gives you the option of taking the door you chose, or taking both the other doors. That's an equivalent problem (assuming the goat has zero utility), and it makes it very obvious that you should switch.
- zcarter 11y ago
  I like to explain it with a wager: pick a star from the night's sky. If you pick the star I'm thinking of, I'll give you a dollar. After you've made your selection, I'll give you the opportunity to switch to one other star, guaranteeing that either your star or this one alternative will be the star I had in mind. Do you want to switch?
- dllthomas 11y ago
  Your argument isn't really wrong as far as it goes, but it doesn't explain why it matters whether Monty deliberately opens a door without a car or picks randomly and just happened to open a car-less door this time.
tawaypdox 11y ago
Another interesting paradox in probability is Bertrand’s Paradox[1]. Given an equilateral triangle inscribed in a circle, what is the probability that chord of the circle chosen at random is longer than the side of the triangle? (Hint: The answer depends on your method of selecting a random chord.) [1] https://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29 https://en.wikipedia.org/wiki/Bertrand_paradox_%28probabilit...
- mamon 11y ago
  For me the most messy part is being allowed to rotate the triangle to match it to one of the described cases. Normally, if you assume triangle does not move, and then draw a random chord, then depending on relative position of the chord and triangle you use one of three methods described by Bertrand to judge whether it is longer or shorter than triangle side. Then you need a way to count probability of each of those cases to happen and then count the weighted average of all three "paradoxical" cases. EDIT: in fact I would just bet on Method 3 as a correct solution, because it is the most general: it does not require rotating the triange. If you are rotating the triangle then you are in fact changing problem definition during solving it, which leads to some hidden assumptions (i.e. Method 2 is only good for chords which are parallel to one side of triangle, and Method 1 is only good for chords that do not cross with triangle sides)
  - kgwgk 11y ago
    Actually, as the wikipedia entry mentions, one can argue that Method 2 is the correct one: it is the only one respecting reasonable translation and scale invariance constraints.
mark_l_watson 11y ago
Great advice on assuming other people are also reasonable, and try to understand the problem they are trying to solve. Years ago, Peter Norvig was a tech editor for an AI book I wrote. I realized two things: he has a clearer way of analyzing things than I do, and also that I was very fortunate that he ended up spending several hours helping me with his review material.
- Denzel 11y ago
  Would you mind expanding upon how he analyzes things in a clearer way? It's something I've felt when reading his blog posts.
  - mark_l_watson 11y ago
    Sorry, that was over ten years ago, and I can't find his review notes so I can't really answer your question except to say that I remember his comments to be very useful.
mamon 11y ago
"Problem 3. One is a boy born on Tuesday. What's the probability both are boys?" - this is not a problem at all. The fact that one boy is born on Tuesday is completely irrelevant as the sex of the child does not depend on day of the week. That information is given only to mess with readers mind. So, if you ignore that Tuesday part then you can reduce Problem 3 to Problem 2.
- oisdk 11y ago
  Not really - like Norvig said, that's one interpretation, but there is another valid one. Say you had a bunch of families with two children each. The children are evenly distributed in terms of gender and the the days of the week on which they were born. If you pick one parent from the crowd, the chance that they have at least one boy is 3 / 4, the chance that they have two is 1 / 4, and the chance that they have none is also 1 / 4: | B| G| --|--|--| B |BB|BG| --|--|--| G |GB|GG| Each of the four squares on the above table is equiprobable. However, if the person says they have at least one boy, they must be in either the left column or the top row, so in one of three squares. There is only one square in those three with both boys, so the chance of that parent having two boys is 1 / 3. Now, for the day-of-the-week problem. If you ask a parent if they have a male child born on Tuesday, it is not equiprobable that they're in any of those three squares. In the BG group, they all have male children, so the chance that any parent chosen has a male child born on Tuesday is 1 / 7. Similarly in the GB group. However, in the BB group, either one of their children may be born on a Tuesday to satisfy the condition. The chance that either child is born on a Tuesday is the same as the inverse of neither child being born on a Tuesday, or 1 - (6/7 * 6/7) = 13/49 So the number of parents in the top left group (BB) who satisfy the condition is 13 / 49, whereas the number of parents in the top right (BG) is 1 / 7, and the bottom left (GB) is also 1 / 7. You're looking for the probability that a given parent in that satisfies the condition is in the top left group, which is (13/49) / (1/7 + 1/7 + 13/49) = 13/27
- kgwgk 11y ago
  By that reasoning, in Problem 2 the fact that one child is a boy cannot affect the sex of his sibling so you can reduce Problem 2 to Problem 1. The point that Norvig is making is that there are other reasonable interpretations corresponding to different sample spaces (and therefore to other probabilities).
henrik_w 11y ago
Peter Norvig often covers really interesting stuff. A long time ago here on HN I read his article on the probability of there being no set in the card game SET [1]. I had never heard of SET then, but picked it up based on the article. It's a simple game, but has interesting properties. Analytical answers are hard to find for a lot of the cases, so simulation to the rescue. I ended up running more simulations than Peter did, with some interesting results [2]. Also, the interview with Peter Norvig in the book "Coders at Work" [3] is great - one of my favorites in the book (actually, the whole book is great). [1] http://norvig.com/SET.html http://norvig.com/SET.html [2] http://henrikwarne.com/2011/09/30/set-probabilities-revisited/ http://henrikwarne.com/2011/09/30/set-probabilities-revisite... [3] http://www.amazon.com/gp/product/1430219483/ http://www.amazon.com/gp/product/1430219483/
- norvig 11y ago
  Very nice write-up on SET, Henrik. I hadn'e seen it before, and wasn't aware of your interesting results.
raverbashing 11y ago
Good write up Most paradoxes disappear when the problem is strictly described.
fitzwatermellow 11y ago
Love the George Carlin quote referenced in the section defining The Reasonable Person Principle: "Have you ever noticed when you're driving that anybody driving slower than you is an idiot, and anyone going faster than you is a maniac?" I'll include the canonical quip by Laplace: "The theory of probabilities is at bottom nothing but common sense reduced to calculus; it enables us to appreciate with exactness that which accurate minds feel with a sort of instinct for which ofttimes they are unable to account." -Introduction to Théorie Analytique des Probabilitiés
- JoeAltmaier 11y ago
  Hm. That guy writing the intro was extremely optimistic about human nature. Most folks find statistics very unintuitive. Folks routinely overestimate extremely unlikely events (lotto) and underestimate near-certainties (auto accident). Its the idea that we might get away with something, or maybe it won't happen to me, that guides most human activities. Its this warping of the probability curve that is responsible for major facets of human society: war, witchcraft, gambling, marriage, on and on.
  - darkmighty 11y ago
    I think those problems are a little different. For example, no matter how you approach it, from a rational point of view joining a war for your country makes no sense. You are highly likely to die and highly unlikely to change to outcome of the war. But if you think in an evolutionary sense, if each individual in a tribe acted rationally, we'd get a sort of Tragedy of the Commons: the enemy tribe wins the war, and slaughters your men. So it may happen that if each individual acts rationally the group, each individual included, suffers; so it makes sense individual sense to develop mechanisms to go to war, which by symmetry must include yourself. Even less intuitively even if the probability of death is lower by not going to war (no tragedy of the commons in traditional sense), which means no individual would rationally choose war or develop mechanisms to force going to war, it would be rational for the tribe to choose war, in an evolutionary sense, provided many more of your men are killed than the enemy tribe (so they do better in evolutionary terms).
tunesmith 11y ago
Man, that's just beautiful - I love reading his notebooks. Just skimming them yields unexpected little gems like "the value of money is roughly logarithmic", which I had never heard before and gets me thinking about progressive taxation. The final step of the St. Petersburg paradox, which he took far further than I had ever seen before, might be implementing the Kelly Criterion, where the amount you bet is related to the size of your own personal bankroll, but I don't know offhand how to relate that to a probability distribution since the Kelly Criterion is normally calculated with one odds value and one payoff value. At the least, I'd like to see a calculation of how many times you'd have to place the bet before you could be reasonably sure of a positive payout - you could do that easily enough with a monte carlo simulation I guess.
- norvig 11y ago
  Good point, tunesmith, I should define a version of `util` that lets you start at your current personal bankroll, rather than at 0, and show how the amount you are willing to pay to play depends on how much you already have.
nkurz 11y ago
I recently came across the "Three Prisoners Problem": https://en.wikipedia.org/wiki/Three_Prisoners_problem https://en.wikipedia.org/wiki/Three_Prisoners_problem If you feel comfortable that the answer to the Monty Hall problem is "switch", you might want to test your understanding to be sure you reach the correct answer here as well. As Wikipedia states, Three Prisoners "is mathematically equivalent to the Monty Hall problem with car and goat replaced with freedom and execution respectively". But mathematically equivalent may not make it intuitively equivalent, and the answer may feel like it contradicts the correct Monty Hall answer. Wikipedia points out that it's originally from a 1950's Martin Gardner column, but I came across it in the textbook "Stastistical Inference" by Casella and Berger. Here's C&B's phrasing: Three prisoners, A, B, and C, are on death row. The governor decides to pardon one of the three and chooses at random the prisoner to pardon. He informs the warden of his choice but requests that the name be kept secret for a few days. The next day, A tries to get the warden to tell him who had been pardoned. The warden refuses. A then asks which of B or C will be executed. The warden thinks for a while, then tells A that B is to be executed. Warden’s reasoning: Each prisoner has a 1 in 3 chance of being pardoned. Clearly, either B or C must be executed, so I have given A no information about whether A will be pardoned. A’s reasoning: Given that B will be executed, then either A or C will be pardoned. My chance of being pardoned has risen to 1 in 2. Who is right? In what I found to be a parody of textbook tropes, C&B begin their explanation "It should be clear that the warden's reasoning is correct..." While it's true that the warden's reasoning is correct, leading off with "it should be clear" seems cruel. Here's a slight variation of the Monty Hall problem, one of the most famous "paradoxes" of popular statistics, and you are going to start with a paraphrase of "it should be obvious to the reader" without the slightest sense of irony, even though this variation produces an answer superficially incompatible with the better known problem? Ah, the strange humor of textbook authors! The remainder of their answer (which is solid) can be found here on Section 1.3 page 22: http://people.unica.it/musio/files/2008/10/Casella-Berger.pdf http://people.unica.it/musio/files/2008/10/Casella-Berger.pd...
powera 11y ago
My TLDR summary: all probabilities are conditional. If you have two probabilities for the same event and they aren't the same, you're just assuming different priors in the statement of the probability. (and if you can't find any different priors, maybe you've discovered something new)
- norvig 11y ago
  Good summary. If I were in charge, I wouldn't even have a notion of unconditional probability -- everything is conditional on something.
r4pha 11y ago
It's always great to see these notebooks from Peter Norvig. There's also some really interesting implementations (in several languages) of probabilistic algorithms on the website [0] of the book he co-wrote on artificial intelligence. Great book and great material. [0] http://aima.cs.berkeley.edu/code.html http://aima.cs.berkeley.edu/code.html
raymondh 11y ago
The Python code is mostly clear and beautiful. Just one small wish: please use collections.deque() instead of showing an O(n) insertion into a list using "deck[0:0] = [card]".
- norvig 11y ago
  You're right! 2.4 was out when I wrote that, so I could/should have used dequeue. But I wanted to use things that I thought readers would be familiar with, and timing wasn't a big issue.

169 points by yomritoyj 11y ago

The hand-wavy way I've always thought about the Monty Hall problem is as follows: When the contestant picked a box he was making a random choice between three boxes and had a probability of 1/3 of picking the car. The probability that one of the other two boxes was the car is thus 2/3. When Monty Hall opens a box it's still the case that the probability of the car being in the other two is 2/3 but the contestant can now eliminate the open box and knows that the probability that the car is in the other box is 2/3. Thus it's worth switching.

mattdeboard 11y ago
This is by far the most simple, concise explanation of the Monty Hall problem I've ever read: https://www.reddit.com/r/videos/comments/1bcuo8/this_has_been_explained_to_me_over_and_over_again/c95w7h9 https://www.reddit.com/r/videos/comments/1bcuo8/this_has_bee... > Basically, you probably got a goat the first time, so switch. It's so concise that I use more words to re-post it every time Monty Hall comes up than are in the explanation itself.
- adevine 11y ago
  The other phrasing of the problem that I've seen help people is this (I think was also originally from Marilyn Vos Savant): Suppose there are a thousand doors, and only one is a car. Monty opens 998 of them to reveal goats, leaving door #628 still hidden. Do you switch? That helps people understand that their original probability of picking the car the first time was very small, and didn't change because of what Monty did.
  - darkmighty 11y ago
    Most paradoxes vanish, become obvious this way (when you increase the numbers). The brothers paradox mentioned: 'I have two children. At least one is a boy.'/'I have two children. At least one is a boy born on Tuesday.'; 'What is the probability I have two boys?' drastically changes when you change to 'I have two children. At least one is a boy born at 7h:35m:42s.' -- it becomes clear that the succeeding information exactly specifies the boy as it becomes specific, in the limit turning into 'I have two children. The first is a boy.' (so the probability goes from 1/3 to 1/2).
  - thaumasiotes 11y ago
    I prefer this one: You pick a door. Monty opens another door, revealing a goat. You now get the option to switch from your door to the other two doors, keeping the car if it's behind either one. Whether Monty opens the door or not doesn't mean much.
    - baddox 11y ago
      An even simpler version, which is equivalent: Monty Hall says that you can either choose 1 door, or 2 doors. It can't get more obvious than that.
      - dllthomas 11y ago
        That is only equivalent if Monty (as intended but usually not stated clearly) always picks a door without a car using knowledge of where the car is. If, instead, Monty picks randomly from the other two doors and just happens not to have revealed the car this time, odds are the same whether you stick with your door or switch. Even in the case where your restatement is mathematically equivalent, it's obviously not at all equivalent as a puzzle because what is being requested by the puzzle is seeing that the original situation is mathematically equivalent to that one.
proksoup 11y ago
Monty chooses a door? I thought he selected it at random. That was always the confusing unstated assumption for me, that makes me think the problem is only confusing because of how the wording usually de-emphasizes that distinction. If he chooses a door, you can benefit from the secret information he reveals sometimes by doing so. If the door was chosen at random, he is not adding any information, so you can't act on it?
- maxerickson 11y ago
  He never reveals a car when he opens the door.
  - malisper 11y ago
    That actually doesn't matter. What does matter is if he knows he is going to reveal a door that is empty. If he does know you should switch. If he reveals a door arbitrarily, and it just so happens to be empty, it doesn't matter if you switch. Edit: Since I have been downvoted, here is a citation: http://probability.ca/jeff/writing/montyfall.pdf http://probability.ca/jeff/writing/montyfall.pdf (see Monty Fall problem)
    - maxerickson 11y ago
      I agree that my phrasing is not superiorally precise, but the meaning is clear enough, the choice of the door isn't arbitrary.
    - kgwgk 11y ago
      To be fully pedantic, what matters is not "if he knows he is going to reveal a door that is empty". Maybe he has no idea and he just reads his lines off a teleprompter. What matters is if you know that he is going to reveal a door that is empty, because then you can reason according to that information.
    - baddox 11y ago
      The notion that he always randomly opens a door without knowing what's behind it but the door is always a losing door is fairly bizarre and doesn't make much sense when you try to analyze probability in the context of repeated experiments. Something must make him always choose a losing door. Either he knows what's behind the doors, or there's some external force ("fate," "God," or whatever) that causes the door to always be a losing door. Regardless, the problem as stated has Monty Hall open a losing door. Thus it's clear that you should switch after Monty Hall does so, and a very simple computer program can show that you tend to win by switching. Another way to phrase the problem is that you choose one door, then Monty Hall (without revealing anything) gives you the option of taking the door you chose, or taking both the other doors. That's an equivalent problem (assuming the goat has zero utility), and it makes it very obvious that you should switch.
zcarter 11y ago
I like to explain it with a wager: pick a star from the night's sky. If you pick the star I'm thinking of, I'll give you a dollar. After you've made your selection, I'll give you the opportunity to switch to one other star, guaranteeing that either your star or this one alternative will be the star I had in mind. Do you want to switch?
dllthomas 11y ago
Your argument isn't really wrong as far as it goes, but it doesn't explain why it matters whether Monty deliberately opens a door without a car or picks randomly and just happened to open a car-less door this time.

tawaypdox 11y ago

Another interesting paradox in probability is Bertrand’s Paradox[1]. Given an equilateral triangle inscribed in a circle, what is the probability that chord of the circle chosen at random is longer than the side of the triangle? (Hint: The answer depends on your method of selecting a random chord.) [1] https://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29 https://en.wikipedia.org/wiki/Bertrand_paradox_%28probabilit...

mamon 11y ago
For me the most messy part is being allowed to rotate the triangle to match it to one of the described cases. Normally, if you assume triangle does not move, and then draw a random chord, then depending on relative position of the chord and triangle you use one of three methods described by Bertrand to judge whether it is longer or shorter than triangle side. Then you need a way to count probability of each of those cases to happen and then count the weighted average of all three "paradoxical" cases. EDIT: in fact I would just bet on Method 3 as a correct solution, because it is the most general: it does not require rotating the triange. If you are rotating the triangle then you are in fact changing problem definition during solving it, which leads to some hidden assumptions (i.e. Method 2 is only good for chords which are parallel to one side of triangle, and Method 1 is only good for chords that do not cross with triangle sides)
- kgwgk 11y ago
  Actually, as the wikipedia entry mentions, one can argue that Method 2 is the correct one: it is the only one respecting reasonable translation and scale invariance constraints.

mark_l_watson 11y ago

Great advice on assuming other people are also reasonable, and try to understand the problem they are trying to solve. Years ago, Peter Norvig was a tech editor for an AI book I wrote. I realized two things: he has a clearer way of analyzing things than I do, and also that I was very fortunate that he ended up spending several hours helping me with his review material.

Denzel 11y ago
Would you mind expanding upon how he analyzes things in a clearer way? It's something I've felt when reading his blog posts.
- mark_l_watson 11y ago
  Sorry, that was over ten years ago, and I can't find his review notes so I can't really answer your question except to say that I remember his comments to be very useful.

mamon 11y ago

"Problem 3. One is a boy born on Tuesday. What's the probability both are boys?" - this is not a problem at all. The fact that one boy is born on Tuesday is completely irrelevant as the sex of the child does not depend on day of the week. That information is given only to mess with readers mind. So, if you ignore that Tuesday part then you can reduce Problem 3 to Problem 2.

oisdk 11y ago
Not really - like Norvig said, that's one interpretation, but there is another valid one. Say you had a bunch of families with two children each. The children are evenly distributed in terms of gender and the the days of the week on which they were born. If you pick one parent from the crowd, the chance that they have at least one boy is 3 / 4, the chance that they have two is 1 / 4, and the chance that they have none is also 1 / 4: | B| G| --|--|--| B |BB|BG| --|--|--| G |GB|GG| Each of the four squares on the above table is equiprobable. However, if the person says they have at least one boy, they must be in either the left column or the top row, so in one of three squares. There is only one square in those three with both boys, so the chance of that parent having two boys is 1 / 3. Now, for the day-of-the-week problem. If you ask a parent if they have a male child born on Tuesday, it is not equiprobable that they're in any of those three squares. In the BG group, they all have male children, so the chance that any parent chosen has a male child born on Tuesday is 1 / 7. Similarly in the GB group. However, in the BB group, either one of their children may be born on a Tuesday to satisfy the condition. The chance that either child is born on a Tuesday is the same as the inverse of neither child being born on a Tuesday, or 1 - (6/7 * 6/7) = 13/49 So the number of parents in the top left group (BB) who satisfy the condition is 13 / 49, whereas the number of parents in the top right (BG) is 1 / 7, and the bottom left (GB) is also 1 / 7. You're looking for the probability that a given parent in that satisfies the condition is in the top left group, which is (13/49) / (1/7 + 1/7 + 13/49) = 13/27
kgwgk 11y ago
By that reasoning, in Problem 2 the fact that one child is a boy cannot affect the sex of his sibling so you can reduce Problem 2 to Problem 1. The point that Norvig is making is that there are other reasonable interpretations corresponding to different sample spaces (and therefore to other probabilities).

henrik_w 11y ago

Peter Norvig often covers really interesting stuff. A long time ago here on HN I read his article on the probability of there being no set in the card game SET [1]. I had never heard of SET then, but picked it up based on the article. It's a simple game, but has interesting properties. Analytical answers are hard to find for a lot of the cases, so simulation to the rescue. I ended up running more simulations than Peter did, with some interesting results [2]. Also, the interview with Peter Norvig in the book "Coders at Work" [3] is great - one of my favorites in the book (actually, the whole book is great). [1] http://norvig.com/SET.html http://norvig.com/SET.html [2] http://henrikwarne.com/2011/09/30/set-probabilities-revisited/ http://henrikwarne.com/2011/09/30/set-probabilities-revisite... [3] http://www.amazon.com/gp/product/1430219483/ http://www.amazon.com/gp/product/1430219483/

norvig 11y ago
Very nice write-up on SET, Henrik. I hadn'e seen it before, and wasn't aware of your interesting results.

raverbashing 11y ago

Good write up Most paradoxes disappear when the problem is strictly described.

fitzwatermellow 11y ago

Love the George Carlin quote referenced in the section defining The Reasonable Person Principle: "Have you ever noticed when you're driving that anybody driving slower than you is an idiot, and anyone going faster than you is a maniac?" I'll include the canonical quip by Laplace: "The theory of probabilities is at bottom nothing but common sense reduced to calculus; it enables us to appreciate with exactness that which accurate minds feel with a sort of instinct for which ofttimes they are unable to account." -Introduction to Théorie Analytique des Probabilitiés

JoeAltmaier 11y ago
Hm. That guy writing the intro was extremely optimistic about human nature. Most folks find statistics very unintuitive. Folks routinely overestimate extremely unlikely events (lotto) and underestimate near-certainties (auto accident). Its the idea that we might get away with something, or maybe it won't happen to me, that guides most human activities. Its this warping of the probability curve that is responsible for major facets of human society: war, witchcraft, gambling, marriage, on and on.
- darkmighty 11y ago
  I think those problems are a little different. For example, no matter how you approach it, from a rational point of view joining a war for your country makes no sense. You are highly likely to die and highly unlikely to change to outcome of the war. But if you think in an evolutionary sense, if each individual in a tribe acted rationally, we'd get a sort of Tragedy of the Commons: the enemy tribe wins the war, and slaughters your men. So it may happen that if each individual acts rationally the group, each individual included, suffers; so it makes sense individual sense to develop mechanisms to go to war, which by symmetry must include yourself. Even less intuitively even if the probability of death is lower by not going to war (no tragedy of the commons in traditional sense), which means no individual would rationally choose war or develop mechanisms to force going to war, it would be rational for the tribe to choose war, in an evolutionary sense, provided many more of your men are killed than the enemy tribe (so they do better in evolutionary terms).

tunesmith 11y ago

Man, that's just beautiful - I love reading his notebooks. Just skimming them yields unexpected little gems like "the value of money is roughly logarithmic", which I had never heard before and gets me thinking about progressive taxation. The final step of the St. Petersburg paradox, which he took far further than I had ever seen before, might be implementing the Kelly Criterion, where the amount you bet is related to the size of your own personal bankroll, but I don't know offhand how to relate that to a probability distribution since the Kelly Criterion is normally calculated with one odds value and one payoff value. At the least, I'd like to see a calculation of how many times you'd have to place the bet before you could be reasonably sure of a positive payout - you could do that easily enough with a monte carlo simulation I guess.

norvig 11y ago
Good point, tunesmith, I should define a version of `util` that lets you start at your current personal bankroll, rather than at 0, and show how the amount you are willing to pay to play depends on how much you already have.

nkurz 11y ago

I recently came across the "Three Prisoners Problem": https://en.wikipedia.org/wiki/Three_Prisoners_problem https://en.wikipedia.org/wiki/Three_Prisoners_problem If you feel comfortable that the answer to the Monty Hall problem is "switch", you might want to test your understanding to be sure you reach the correct answer here as well. As Wikipedia states, Three Prisoners "is mathematically equivalent to the Monty Hall problem with car and goat replaced with freedom and execution respectively". But mathematically equivalent may not make it intuitively equivalent, and the answer may feel like it contradicts the correct Monty Hall answer. Wikipedia points out that it's originally from a 1950's Martin Gardner column, but I came across it in the textbook "Stastistical Inference" by Casella and Berger. Here's C&B's phrasing: Three prisoners, A, B, and C, are on death row. The governor decides to pardon one of the three and chooses at random the prisoner to pardon. He informs the warden of his choice but requests that the name be kept secret for a few days. The next day, A tries to get the warden to tell him who had been pardoned. The warden refuses. A then asks which of B or C will be executed. The warden thinks for a while, then tells A that B is to be executed. Warden’s reasoning: Each prisoner has a 1 in 3 chance of being pardoned. Clearly, either B or C must be executed, so I have given A no information about whether A will be pardoned. A’s reasoning: Given that B will be executed, then either A or C will be pardoned. My chance of being pardoned has risen to 1 in 2. Who is right? In what I found to be a parody of textbook tropes, C&B begin their explanation "It should be clear that the warden's reasoning is correct..." While it's true that the warden's reasoning is correct, leading off with "it should be clear" seems cruel. Here's a slight variation of the Monty Hall problem, one of the most famous "paradoxes" of popular statistics, and you are going to start with a paraphrase of "it should be obvious to the reader" without the slightest sense of irony, even though this variation produces an answer superficially incompatible with the better known problem? Ah, the strange humor of textbook authors! The remainder of their answer (which is solid) can be found here on Section 1.3 page 22: http://people.unica.it/musio/files/2008/10/Casella-Berger.pdf http://people.unica.it/musio/files/2008/10/Casella-Berger.pd...

powera 11y ago

My TLDR summary: all probabilities are conditional. If you have two probabilities for the same event and they aren't the same, you're just assuming different priors in the statement of the probability. (and if you can't find any different priors, maybe you've discovered something new)

norvig 11y ago
Good summary. If I were in charge, I wouldn't even have a notion of unconditional probability -- everything is conditional on something.

r4pha 11y ago

It's always great to see these notebooks from Peter Norvig. There's also some really interesting implementations (in several languages) of probabilistic algorithms on the website [0] of the book he co-wrote on artificial intelligence. Great book and great material. [0] http://aima.cs.berkeley.edu/code.html http://aima.cs.berkeley.edu/code.html

raymondh 11y ago

The Python code is mostly clear and beautiful. Just one small wish: please use collections.deque() instead of showing an O(n) insertion into a list using "deck[0:0] = [card]".

norvig 11y ago
You're right! 2.4 was out when I wrote that, so I could/should have used dequeue. But I wanted to use things that I thought readers would be familiar with, and timing wasn't a big issue.