There&#x27;s no hashing construction that will &quot;completely eliminate the possibility&quot; (see <a href="https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Pigeonhole_principle" rel="nofollow">https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Pigeonhole_principle</a>), so I&#x27;m not sure what answer you&#x27;re looking for. It&#x27;s computationally infeasible to generate collisions (accidental or malicious) if SHA-2 is not broken. If you&#x27;re interested in reading about how combining multiple hashes affects security, see this paper: <a href="https:&#x2F;&#x2F;www.iacr.org&#x2F;cryptodb&#x2F;archive&#x2F;2004&#x2F;CRYPTO&#x2F;1472&#x2F;1472.pdf" rel="nofollow">https:&#x2F;&#x2F;www.iacr.org&#x2F;cryptodb&#x2F;archive&#x2F;2004&#x2F;CRYPTO&#x2F;1472&#x2F;1472....</a>