Here is an 11-voter 3-candidate election example proving that Condorcet schemes with ranking-equalities allowed suffer from favorite-betrayal, and that is true whether you use "winning-votes" or "margins" – doesn't matter – same example kills both:
| #voters | their vote |
|---|---|
| 2 | A>B>C |
| 3 | C>A>B |
| 4 | C=B>A |
| 2 | A>B>C |
Defeats are A>B by 7:4, B>C by 4:3, and C>A by 7:4. There is an A>B>C>A cycle in which B>C is the weakest defeat (measured by either winning votes or by margins), so that C is elected.
Notice that the two A>B>C voters shown in blue on the bottom line can turn the "lesser evil" B into the Condorcet Winner by "betraying" their favorite "third party" candidate A and voting B>A>C or B>A=C or B>C>A.
However, changing their vote instead to A=B>C or A>B=C or A=C>B or A>C>B or A>B>C or B=C>A or A=B=C (or C>B>A or C>A>B or C>A=B) does not suffice: then C still uniquely wins in all cases. Hence favorite-betrayal of A was strategically necessary.
All this is true regardless of whether you use "margins" or "winning votes." However I have cheated a bit by assuming the Condorcet method is such that it elects the candidate with the weakest defeat in a 3-cycle.
You may also see the other example with ranking-equalities forbidden and Venzke's original careful proof (which doesn't cheat).