Does the Gale-Shapley stable marriage algorithm give at least one person his or her first choice? Given $x$ males and $x$ females with their table of priorities, does the Gale-Shapley stable marriage algorithm guarantee that at least one person gets his or her first choice? It seems like the answer should be no, but all simple examples I can think of don't give this answer no matter how I change the table of priorities.

From the original article, that is Example 2 in _College Admissions and the Stability of Marriage_ (Gale and Shapley, 1962), this may indeed not always be possible to _find_ (no matter the algorithm) a stable marriage where anyone gets his or her first choice.

The example the authors give is for $n=4$ men and women, where the ranking matrix is reproduced below. $$ \begin{pmatrix} & A & B & C & D \\\ a & (1,3) & (2,3) & \boxed{(3,2)} & (4,3) \\\ b & (1,4) & (4,1) & (3,3) & \boxed{(2,2)} \\\ c & \boxed{(2,2)} & (1,4) & (3,4) & (4,1) \\\ d & (4,1) & \boxed{(2,2)} & (3,1) & (1,4) \\\ \end{pmatrix} $$ The only stable marriage is the one highlighted: $(a,C),(b,D),(c,A),(d,B)$. (In particular, the Gale—Shapley algorithm is bound to find this marriage.)