There are six different possible ways of choosing two of the loose ends:

Imagine calling the ends of one rope A and B, and the ends of the other rope X and Y.

These are all the different possible pairings:

A & B

A & X

A & Y

B & X

B & Y

X & Y

A & X

A & Y

B & X

B & Y

X & Y

Four of these pairings lead to the ropes being tied together:

A & X

A & Y

B & X

B & Y

A & Y

B & X

B & Y

Therefore the probability that Sam is left holding one piece of rope is $\frac{4}{6}=\frac{2}{3}$.

When Pat picks up one end, there are three possible ends left that she can choose. One of these will be the other end of the same rope and will produce a loop. Either of the other two will produce the desired result.

Therefore the probability that Sam is left holding one piece of rope is $\frac{2}{3}$.

*This problem is taken from the UKMT Mathematical Challenges.*