Is it safe to say that a XOR with an unknown input provides a random output? I am wondering if you have an input $x_{1}$ and you XOR it with an unknown random second input $x_{2}$, is it safe to say that the resulting $x_{1}\oplus x_{2}$ will be random ?
The final idea would be to first fix all the $x_{i}$ for $i&gt;0$ so that one cannot change its $x_{i}$ afterwards, then set $x_{0}$ to a random value.
Then, person 1 $p_{1}$ gets a value $a_{1}=x_{0}\oplus x_{1}$, $p_{2}$ gets $a_{2}=x_{2}\oplus a_{1}$...
In the end person $p_{i}$ gets the value $a_{i}=x_{i}\oplus a_{i-1}$ $\forall i$
The point of this is to choose someone at random among them so we could pick $p_{min}$ where $a_{min} = min_{i}(a_{i})$.
Do you think it would be safe to choose someone at random in this way ? We probably need to make sure that two different people won't have the same $x_{i}$ otherwise there could be two people with the same $a_{i}$ value but apart from that I don't see any weakness where one could improve its chances at being the chosen one, do you see one?
Edit: I have to add that all of the $x_{i}$ will also be picked at random otherwise there would be a clear link between one $x_{i}$ and the previous $a_{i-1}$.
I am wondering if you have an input $x_{1}$ and you XOR it with an unknown random second input $x_{2}$, is it safe to say that the resulting $x_{1}\oplus x_{2}$ will be random ?

Yes, this is literally what proofs of the security of one-time pads come down to.  Informally it comes down to the following observations:

If $x_2$ is random, then it means that each of its bits is 0 or 1 with equal probability and independent of all other bits in $x_2$;
If $x_2$ has those properties, then $x_1 \oplus x_2$ also must.

And this really comes down that after the XOR, for each bit in $x_1$ you get with equal probability, either the original bit from $x_1$ or the flipped bit, and that you XOR all of the bits independently of all other ones.