5 ms·

The strange non-transitivity of random variables (2014)

brudgers 11y ago
Related, Non-transitive dice: https://en.wikipedia.org/wiki/Nontransitive_dice https://en.wikipedia.org/wiki/Nontransitive_dice
- mtdewcmu 11y ago
  That is neat and a lot easier to understand than the original link. All the notation made my eyes glaze over.
  - hebdo 11y ago
    While Kolmogorov's definition of probability looks a bit spooky at first, I definitely recommend going through it at some point. Not only it feels nice, it also makes understanding probabilistic concepts easier (e.g. understanding that a statistical space is simply a "lifted", or rather parametrized, version of a probabilistic space).
- lifeformed 11y ago
  I like the stories about Warren Buffett's fascination with the dice: > Buffett once attempted to win a game of dice with Bill Gates using nontransitive dice. "Buffett suggested that each of them choose one of the dice, then discard the other two. They would bet on who would roll the highest number most often. Buffett offered to let Gates pick his die first. This suggestion instantly aroused Gates's curiosity. He asked to examine the dice, after which he demanded that Buffett choose first."[1] > In 2010, Wall Street Journal magazine quoted Sharon Osberg, Buffett's bridge partner, saying that when she first visited his office 20 years earlier, he tricked her into playing a game with nontransitive dice that could not be won and "thought it was hilarious".[2]
quietplatypus 11y ago
this isnt paradoxical at all, it's just you are looking at only 2 dice at a time not the full joint probability of a < b < c. i can come up with similarly retarded "paradoxical conclusions" if i took enough samples of different tiny slices of things. dude is just trying to sound smart with yet another "counterintuitive probability post"
- jeffsco 11y ago
  Imagine a tournament to choose the best tennis player. If you imagine these dice as a simple model of a tennis player, there can be no best tennis player. So tennis tournaments make no sense. This is paradoxical if you believe in tennis rankings.
  - mziel 11y ago
    It's not. Let's assume that ">" means probability of winning over 0.5 Federer > Djokovic (well not lately) Djokovic > Nadal Nadal > Federer Everything depends on the draw. Same with the initial choice of dice.
  - wolfgke 11y ago
    > This is paradoxical if you believe in tennis rankings. So the solution is to go the rational way and stop believing in ranking if you claim that tennis ranking is based on a mathematical model that has the non-transitivity property. Where exactly is the problem?
  - CamperBob2 11y ago
    It makes sense to me that the best athlete cannot be determined in some scenarios, or that the term "best" is simply meaningless. Due to training and/or genetics, player A might be strong in areas where player B is weak. Likewise, player B might be strong in areas where player C is weak. Neither of these statements gives us any insight into what will happen if player A faces player C, but a traditional tournament structure might well keep A and C from ever facing each other.
  - Veedrac 11y ago
    A more revealing model might be a race, since one's performance is not obviously linked to another's and there's only one variable being measured (time to completion).
danharaj 11y ago
> Examples of binary relations are binary operations (like +,–,⋅,/) on N,Z,Q,R,C,Rn, … etc. (where defined), since functions are special cases of relations, or, the order relations <,≤,>,≥ on R, for instance. A function of two variables is a ternary relation, not a binary relation. /pedant
- x3n0ph3n3 11y ago
  Last I checked, binary implies 2 and ternary implies 3. There are 2 inputs, therefore it's a binary relation or operator.
  - dietrichepp 11y ago
    This is completely incorrect. There are two inputs and one output, e.g., ℝ⨯ℝ→ℝ. Since there are two inputs, it's a binary operator. Since there are three inputs and outputs combined, it's a ternary relation. Likewise, a unary operator is a binary relation. For example, x ↦ -x, or x ↦ x+1.
  - danharaj 11y ago
    Idk what to tell you other than that you should check the definitions. A unary function is a special case of a binary relation and in general an n-ary function is a special case of a n+1-ary relation. Edit - Maybe this clarifies it: When we talk about a function we designate one of its variables as the 'output' and talk about its arity in terms of the inputs. When we talk about a relation there is no distinguished output: A relation is a fully symmetrical definition. That's why the arity of a function is one less than the arity of that same function thought of as a relation between its inputs and output.
  - deleted 11y ago
    [deleted]

Related, Non-transitive dice: https://en.wikipedia.org/wiki/Nontransitive_dice https://en.wikipedia.org/wiki/Nontransitive_dice

mtdewcmu 11y ago
That is neat and a lot easier to understand than the original link. All the notation made my eyes glaze over.
- hebdo 11y ago
  While Kolmogorov's definition of probability looks a bit spooky at first, I definitely recommend going through it at some point. Not only it feels nice, it also makes understanding probabilistic concepts easier (e.g. understanding that a statistical space is simply a "lifted", or rather parametrized, version of a probabilistic space).
lifeformed 11y ago
I like the stories about Warren Buffett's fascination with the dice: > Buffett once attempted to win a game of dice with Bill Gates using nontransitive dice. "Buffett suggested that each of them choose one of the dice, then discard the other two. They would bet on who would roll the highest number most often. Buffett offered to let Gates pick his die first. This suggestion instantly aroused Gates's curiosity. He asked to examine the dice, after which he demanded that Buffett choose first."[1] > In 2010, Wall Street Journal magazine quoted Sharon Osberg, Buffett's bridge partner, saying that when she first visited his office 20 years earlier, he tricked her into playing a game with nontransitive dice that could not be won and "thought it was hilarious".[2]

quietplatypus 11y ago

this isnt paradoxical at all, it's just you are looking at only 2 dice at a time not the full joint probability of a < b < c. i can come up with similarly retarded "paradoxical conclusions" if i took enough samples of different tiny slices of things. dude is just trying to sound smart with yet another "counterintuitive probability post"

jeffsco 11y ago
Imagine a tournament to choose the best tennis player. If you imagine these dice as a simple model of a tennis player, there can be no best tennis player. So tennis tournaments make no sense. This is paradoxical if you believe in tennis rankings.
- mziel 11y ago
  It's not. Let's assume that ">" means probability of winning over 0.5 Federer > Djokovic (well not lately) Djokovic > Nadal Nadal > Federer Everything depends on the draw. Same with the initial choice of dice.
- wolfgke 11y ago
  > This is paradoxical if you believe in tennis rankings. So the solution is to go the rational way and stop believing in ranking if you claim that tennis ranking is based on a mathematical model that has the non-transitivity property. Where exactly is the problem?
- CamperBob2 11y ago
  It makes sense to me that the best athlete cannot be determined in some scenarios, or that the term "best" is simply meaningless. Due to training and/or genetics, player A might be strong in areas where player B is weak. Likewise, player B might be strong in areas where player C is weak. Neither of these statements gives us any insight into what will happen if player A faces player C, but a traditional tournament structure might well keep A and C from ever facing each other.
- Veedrac 11y ago
  A more revealing model might be a race, since one's performance is not obviously linked to another's and there's only one variable being measured (time to completion).

danharaj 11y ago

> Examples of binary relations are binary operations (like +,–,⋅,/) on N,Z,Q,R,C,Rn, … etc. (where defined), since functions are special cases of relations, or, the order relations <,≤,>,≥ on R, for instance. A function of two variables is a ternary relation, not a binary relation. /pedant

x3n0ph3n3 11y ago
Last I checked, binary implies 2 and ternary implies 3. There are 2 inputs, therefore it's a binary relation or operator.
- dietrichepp 11y ago
  This is completely incorrect. There are two inputs and one output, e.g., ℝ⨯ℝ→ℝ. Since there are two inputs, it's a binary operator. Since there are three inputs and outputs combined, it's a ternary relation. Likewise, a unary operator is a binary relation. For example, x ↦ -x, or x ↦ x+1.
- danharaj 11y ago
  Idk what to tell you other than that you should check the definitions. A unary function is a special case of a binary relation and in general an n-ary function is a special case of a n+1-ary relation. Edit - Maybe this clarifies it: When we talk about a function we designate one of its variables as the 'output' and talk about its arity in terms of the inputs. When we talk about a relation there is no distinguished output: A relation is a fully symmetrical definition. That's why the arity of a function is one less than the arity of that same function thought of as a relation between its inputs and output.
- deleted 11y ago
  [deleted]