Imagine you are the proposer. You will start at the top of your list and work your way down only if rejected. So you are guaranteed to the best you could possibly do.

Imagine A ranks X,Y,Z in that order B ranks Y,X,Z in that order C ranks X,Y,Z

X, Y, Z rank A,B,C in that order.

A will propose to X and match. B will propose to Y and match. C will propose to X and get rejected. C will propose to Y and get rejected. C will propose to Z and match.

Y will never get a proposal from A. Z will never get a proposal from A or B.

Edit: I think I fixed it.

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jb1991 an hour ago | parent [-]

Why will B propose to X first if their first choice is Y?

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2muchcoffeeman an hour ago | parent [-]

You’re absolutely right. I’m on my phone trying to imagine the scenarios in my head and trying to find the simplest possible example.

Maybe 4 participants is too few to clearly see what happens.

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computably 39 minutes ago | parent [-]

(A,X),(B,Y) as the proposals is fine for the point being made. The issue is if we reverse which group is the proposer, we still end up with the same matching, because (A,X) strictly prefer each other.

The minimal example with 4 participants is if each person P has unique preferences and P's 1st choice has P as 2nd choice. So in this case just flip X:

A:XY B:YX X:BA Y:AB

If A,B propose you get AX,BY. If X,Y propose you get BX,AY. Both are stable because the proposers are getting their first choice.

	▲	jb1991 11 minutes ago \| parent [-]
		I find this algorithm to have a curious definition of the word "stable."