<p>Allowing indifference will not solve the problem of the effect of the independence of irrelevant alternatives (combined with the unanimity and transitivity requirements).</p>
<p>You have said <span class="math-container">$b &gt;_1 c$</span> and <span class="math-container">$c &gt;_2 b$</span> should lead to <span class="math-container">$b =_s c$</span></p>
<p>Now consider  <span class="math-container">$a &gt;_1 b$</span> and <span class="math-container">$a &gt;_2 b$</span>, which should lead to  <span class="math-container">$a &gt;_s b$</span> by unanimity</p>
<ul>
<li>so <span class="math-container">$a &gt;_1 b &gt;_1 c$</span> and <span class="math-container">$c &gt;_2 a &gt;_2 b$</span> should lead to  <span class="math-container">$a &gt;_s b =_s c$</span> by transitivity</li>
<li>and thus <span class="math-container">$a &gt;_1  c$</span> and <span class="math-container">$c &gt;_2 a $</span> should lead to   <span class="math-container">$a &gt;_s c$</span> by the independence of irrelevant alternatives</li>
</ul>
<p>And also consider  <span class="math-container">$c &gt;_1 a$</span> and <span class="math-container">$c &gt;_2 a$</span>, which should lead to  <span class="math-container">$c &gt;_s a$</span></p>
<ul>
<li>so <span class="math-container">$b &gt;_1 c &gt;_1 a$</span> and <span class="math-container">$c &gt;_2 a &gt;_2 b$</span> should lead to   <span class="math-container">$b =_s c &gt;_s a$</span></li>
<li>and thus <span class="math-container">$b &gt;_1  a$</span> and <span class="math-container">$a &gt;_2 b $</span> should lead to   <span class="math-container">$b &gt;_s a$</span></li>
</ul>
<p>Combining these means <span class="math-container">$b &gt;_1 a &gt;_1 c$</span> and <span class="math-container">$c &gt;_2 a &gt;_2 b$</span> should lead to <span class="math-container">$b &gt;_s a &gt;_s c$</span></p>
<ul>
<li><p>but that implies <span class="math-container">$b &gt;_1 c$</span> and <span class="math-container">$c &gt;_2  b$</span> should lead to <span class="math-container">$b &gt;_s  c$</span></p>
</li>
<li><p>inconsistent with your <span class="math-container">$b &gt;_1 c$</span> and <span class="math-container">$c &gt;_2 b$</span> leading to <span class="math-container">$b =_s c$</span></p>
</li>
</ul>

CC BY-SA 4.0
2021-01-01T16:24:08.390Z
2021-01-01T16:24:08.390Z
Henry
<p>My question relates to my understanding of Arrow's Impossibility Theorem and ranked choice.  It seems to me that the requirements on social choice functions are too strict. A social choice function cannot result in ties. This sounds unreasonable to me.</p>
<p>Let two voters with preferences <span class="math-container">$a &gt; b &gt; c &gt; d$</span> and <span class="math-container">$a &gt; c &gt; b &gt; d$</span>. It seems that any &quot;reasonable&quot; choice function would choose <span class="math-container">$a &gt; b = c &gt; d$</span>. Any other output would be making an arbitrary distinction between <span class="math-container">$b$</span> and <span class="math-container">$c$</span>. So my question is: if we relax the output of our social choice functions to allow for ties, is there a social choice function that satisfies Arrow's criteria? Is there proof that no such function exists?</p>
<p>As an addendum, I tried a &quot;trivial&quot; such function but it doesn't seem to work.
Let <span class="math-container">$f$</span> a social choice function which if every voter prefers <span class="math-container">$c_1 &gt; c_2$</span> then in the final output <span class="math-container">$c_1 &gt; c_2$</span>; otherwise <span class="math-container">$c_1 = c_2$</span>.</p>
<p>This seems to fail by the transitivity of ties. So, for two preferences <span class="math-container">$a &gt; b &gt; c$</span> and <span class="math-container">$c &gt; a &gt; b$</span> then <span class="math-container">$a &gt; b$</span> but <span class="math-container">$c = a$</span> and <span class="math-container">$c = b$</span> which is a contradiction. If we were to choose, <span class="math-container">$a &gt; b = c$</span> we would be violating the independence of irrelevant alternatives. As, if we were to delete <span class="math-container">$b$</span> the result &quot;should&quot; be <span class="math-container">$a = c$</span>. Is there a way to work around this?</p>

CC BY-SA 4.0
2020-12-31T18:57:00.673Z
I'm pretty confident Arrow's result says even if we allow ties in the social choice function, it is impossible.
Walrasian Auctioneer
CC BY-SA 4.0
2021-01-01T23:18:48.917Z
Notation varies in social choice theory and the term "social choice function" is often used for a function from preference profiles to alternatives. What Arrow looked at were functions from preference profiles to preference orderings over alternatives. These preference orderings can have nontrivial indifferences.
Michael Greinecker
<decision-theory><voting>
Ranked choice preference with ties - Arrow's Impossibility Theorem
Giannis Tyrovolas
# Ranked choice preference with ties - Arrow's Impossibility Theorem

(Asked by: username_0 on 2020-12-31)

My question relates to my understanding of Arrow's Impossibility Theorem and ranked choice. It seems to me that the requirements on social choice functions are too strict. A social choice function cannot result in ties. This sounds unreasonable to me.

Let two voters with preferences $a > b > c > d$ and $a > c > b > d$. It seems that any "reasonable" choice function would choose $a > b = c > d$. Any other output would be making an arbitrary distinction between $b$ and $c$. So my question is: if we relax the output of our social choice functions to allow for ties, is there a social choice function that satisfies Arrow's criteria? Is there proof that no such function exists?

As an addendum, I tried a "trivial" such function but it doesn't seem to work.
Let $f$ a social choice function which if every voter prefers $c\_1 > c\_2$ then in the final output $c\_1 > c\_2$; otherwise $c\_1 = c\_2$.

This seems to fail by the transitivity of ties. So, for two preferences $a > b > c$ and $c > a > b$ then $a > b$ but $c = a$ and $c = b$ which is a contradiction. If we were to choose, $a > b = c$ we would be violating the independence of irrelevant alternatives. As, if we were to delete $b$ the result "should" be $a = c$. Is there a way to work around this?

**Tags:** decision-theory, voting

> **Comments:**
> **username_1**: I'm pretty confident Arrow's result says even if we allow ties in the social choice function, it is impossible.  
> **username_2**: Notation varies in social choice theory and the term "social choice function" is often used for a function from preference profiles to alternatives. What Arrow looked at were functions from preference profiles to preference orderings over alternatives. These preference orderings can have nontrivial indifferences.  

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## 1 answers

### Answer by username_3 on 2021-01-01. Score: 2 | Accepted answer

Allowing indifference will not solve the problem of the effect of the independence of irrelevant alternatives (combined with the unanimity and transitivity requirements).

You have said $b >\_1 c$ and $c >\_2 b$ should lead to $b =\_s c$

Now consider $a >\_1 b$ and $a >\_2 b$, which should lead to $a >\_s b$ by unanimity

* so $a >\_1 b >\_1 c$ and $c >\_2 a >\_2 b$ should lead to $a >\_s b =\_s c$ by transitivity
* and thus $a >\_1 c$ and $c >\_2 a $ should lead to $a >\_s c$ by the independence of irrelevant alternatives

And also consider $c >\_1 a$ and $c >\_2 a$, which should lead to $c >\_s a$

* so $b >\_1 c >\_1 a$ and $c >\_2 a >\_2 b$ should lead to $b =\_s c >\_s a$
* and thus $b >\_1 a$ and $a >\_2 b $ should lead to $b >\_s a$

Combining these means $b >\_1 a >\_1 c$ and $c >\_2 a >\_2 b$ should lead to $b >\_s a >\_s c$

* but that implies $b >\_1 c$ and $c >\_2 b$ should lead to $b >\_s c$
* inconsistent with your $b >\_1 c$ and $c >\_2 b$ leading to $b =\_s c$


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