<p>If you are doing pairwise comparisons, then with $n$ individuals there will be $$\dfrac{n(n-1)}{2} $$ such comparisons.  You can write this as $\binom{n}{2}$ or as $\frac12n^2-\frac12n$ if you prefer.  </p>

<p>$1$ for two people, $3$ for three, $6$ for four and so on, known as <a href="https://en.wikipedia.org/wiki/Triangular_number" rel="nofollow noreferrer">triangular numbers</a></p>

CC BY-SA 3.0
2017-06-13T12:58:06.960Z
It makes sense thank you, I somehow thought every new allocation would change the older allocations so that there would a massive number of equilibria before you get to the final one.
Nathan Furnal
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2017-06-13T17:45:41.527Z
@Henry It seems I am somewhat slow today. Having read the answer I still do not understand what the question was. Could you please explain?
Giskard
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2017-06-13T22:31:35.317Z
@denesp Presumably to check Pareto optimality and equilibrium, you need to ensure there are no possible exchanges of allocations between pairs of individuals which make both happier.  I took the question to be how many such pairs there were to be checked, given there are $n$ individuals
Henry
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2017-06-14T06:45:58.953Z
@Henry I see. Thank you. On a side note: It is not enough to check pairs. Consider the case of three goods and three consumers with $U(x,y,z) = \min(x,y,z)$ and the initial distribution
$$
(3,0,0) \hskip 10pt (0,3,0) \hskip 10pt (0,0,3).
$$
Pairwise no improvements are possible but clearly they could all get $(1,1,1)$ which is better.
Giskard
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2017-06-14T07:11:11.437Z
@denesp If you can compare multiples, then the answer is $1$ (just compare everybody together).
Henry
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2017-06-14T08:45:33.517Z
@Henry True, but this does not make pairwise comparison any good. I am voting to close the question as unclear. OP's comment "every new allocation would change the older allocations" also seems to be out of sync with your interpretation.
Giskard
CC BY-SA 3.0
2017-06-13T07:40:34.590Z
2017-06-13T07:48:02.530Z
2017-06-13T07:48:02.530Z
Henry
<p>Say I have a starting situation with Pareto efficient allocations between 2 individuals, now a third one comes in and changes the allocations. Now a fourth one, etc. Up untill a nth individual comes in. Do we now have to resolve (n-1) equations or n^(n-1) equilibria? How does the "difficulty" scale for n individuals? I don't know if that makes sense? Thanks.</p>

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2017-06-13T05:26:17.807Z
It does not, please clarify.
Giskard
<microeconomics><pareto-efficiency>
How does a Pareto efficient equilibrium scale for n individuals?
Nathan Furnal
# How does a Pareto efficient equilibrium scale for n individuals?

(Asked by: username_0 on 2017-06-12)

Say I have a starting situation with Pareto efficient allocations between 2 individuals, now a third one comes in and changes the allocations. Now a fourth one, etc. Up untill a nth individual comes in. Do we now have to resolve (n-1) equations or n^(n-1) equilibria? How does the "difficulty" scale for n individuals? I don't know if that makes sense? Thanks.

**Tags:** microeconomics, pareto-efficiency
> **Comments:**
> **username_1**: It does not, please clarify.  

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## 1 answers

### Answer by username_2 on 2017-06-13. Score: 0 | Accepted answer

If you are doing pairwise comparisons, then with $n$ individuals there will be $$\dfrac{n(n-1)}{2} $$ such comparisons. You can write this as $\binom{n}{2}$ or as $\frac12n^2-\frac12n$ if you prefer.

$1$ for two people, $3$ for three, $6$ for four and so on, known as [triangular numbers](https://en.wikipedia.org/wiki/Triangular_number)

> **Comments:**
> **username_0**: It makes sense thank you, I somehow thought every new allocation would change the older allocations so that there would a massive number of equilibria before you get to the final one.  
> **username_1**: @Henry It seems I am somewhat slow today. Having read the answer I still do not understand what the question was. Could you please explain?  
> **username_2**: @denesp Presumably to check Pareto optimality and equilibrium, you need to ensure there are no possible exchanges of allocations between pairs of individuals which make both happier.  I took the question to be how many such pairs there were to be checked, given there are $n$ individuals  
> **username_1**: @Henry I see. Thank you. On a side note: It is not enough to check pairs. Consider the case of three goods and three consumers with $U(x,y,z) = \min(x,y,z)$ and the initial distribution $$ (3,0,0) \hskip 10pt (0,3,0) \hskip 10pt (0,0,3). $$ Pairwise no improvements are possible but clearly they could all get $(1,1,1)$ which is better.  
> **username_2**: @denesp If you can compare multiples, then the answer is $1$ (just compare everybody together).  
> **username_1**: @Henry True, but this does not make pairwise comparison any good. I am voting to close the question as unclear. OP's comment "every new allocation would change the older allocations" also seems to be out of sync with your interpretation.  

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