# Ranked choice preference with ties - Arrow's Impossibility Theorem

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?

• I'm pretty confident Arrow's result says even if we allow ties in the social choice function, it is impossible. Dec 31 '20 at 18:57
• 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. Jan 1 '21 at 23:18

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 $$b =_s c$$