If they had the same initial and final states there would be no difference in the entropy calculation. So there has to be a difference between the final state of the irreversible process and the final state of the reversible process.
As an example, consider a quasi-static reversible adiabatic expansion of an ideal gas from Vi to Vf and an irreversible adiabatic expansion (let's say a free expansion - no work done - to make it simple) from Vi to Vf. The free expansion results in no change in internal energy/temperature, since no work is done. The reversible expansion does work so the internal energy/temperature decreases.
To calculate the change in entropy of the irreversible free expansion, you have to find the integral of dQ/T over the reversible
path between the initial and final states. That would be an isothermal reversible expansion in which there is heat flow into the gas (dQ>0), so the integral of dQ/T over that path is > 0.