# 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. Commented Dec 31, 2020 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. Commented Jan 1, 2021 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 $$b =_s c$$