In an earlier question, the consensus was that I should answer a takeout double of one heart with one spade (the cheaper bid) rather than two clubs. (I agree with that consensus.) The exact situation was: "Left hand opponent opened 1 heart. Partner doubled for takeout. Right hand opponent passed." I had something like (S) xxx (H) xx (D) Jxxx (C) Jxxx.
Suppose my major suits were reversed, so I had (S) xx (H) xxx (D) Jxxx (C) Jxxx? Would you still bid "one spade" with only two spades, or is "2 clubs" now the lesser evil?
Suppose one of my minors had five cards instead of four: e.g. (S) xxx (H) xx (D) Jxx (C) Jxxxx (three spades), or (S) xx (H) xx (D) Jxxx (C) Jxxxx (only two spades).
In the original question and answer, "one spade" was the preferred bid versus two clubs when I had one fewer spade than clubs. But the above variations hypothesize that I have at least two fewer spades than clubs. Put another way, where is the "tipping point" of spade-club distributions, where you would bid two clubs rather than one spade?