Coin flipping without commitments or random oracles

It's well known that two parties, Alice and Bob, can flip a fair coin using commitments.

1. Alice picks a random number $a \in \mathbb{Z}_q$ and computes $c_a = Com(a, r_a)$ where $r_a \xleftarrow{R} \mathbb{Z}_q$. She then sends to Bob $c_a$.
2. Bob does the same, pick a number $b \in \mathbb{Z}_q$, and compute $c_b$ and send it to Alice.
3. Then, as Bob went second, he's required to open $c_b$ first, revealing $(b, r_b)$. Alice checks the opening.
4. Then Alice does the same, and Bob checks the opening.
5. Each of them just computes $a + b \mod q$ and outputs 1 if the answer is less than $\frac{q}{2}$ and 0 otherwise

Commitments require that Alice and Bob be PPT as commitments cannot be statistically binding and hiding. Is there an impossibility result that says one cannot flip a fair coin without commitment or assuming some form of public key cryptography? Can two unbounded algorithms flip a fair bit?

From https://dl.acm.org/doi/pdf/10.1145/12130.12168, it seems impossible to get an unbiased coin in the two-party protocol when one of the parties is corrupt (and unbounded).

The argument is as follows, let $n$ be the security parameter:

Let $\Pi$ be a protocol between Alice and Bob, such that at the end of the protocol, Alice outputs $a$, and Bob outputs $b$. Without loss of generality, assume that Bob is a rushing adversary (Alice sends the first message in each round).

It's safe to assume after $\Pi$ concludes and both parties have output $a$ and $b$, P[a=b] > 1/2. If P[a=b] = 1/2, then Bob just outputs a random bit regardless of what Alice does. If this is the case, Alice can output $a \oplus b$ as the final answer, and she has an unbiased coin. Bob's entire goal is to prevent this from happening.

So we assume that $\Pi$ is $\epsilon$ consistent i.e. P[a=b] = 1/2 + $\epsilon$ for $\epsilon > 0$. This a very weak assumption that states that B will not output an unbiased random bit. In this case Theorem 2.2 of the cited paper shows that for all polynomial functions $f$ : $Pr[f(a,b) = 1 ] \geq 1/2 + \frac{\epsilon}{4r + 1}$ where $r$ is the number of rounds of messaging in $\Pi$.

As $r$ is a polynomial function of the security parameter $n$ the final output is non-negligibly biased away from 1/2.