The Mathematics of Fits

The other day, my partner held

x
AQxxx
KJx
KQxx

and opened the bidding 1H. My RHO passed, I bid 1NT, and my LHO bid 2S, which was passed out. It made on the nose and we went -110.

After the hand, I told partner that I would have made a takeout double with her hand. She said that she didn't think that our side had a fit.

Stop! Time for some very easy math.

How many spades do I have? Well, I bypassed 1S to bid 1NT. So I shouldn't have more than three.

Next, how many non-spades do I have? At least ten.

Can you arrange those ten cards into three suits in any manner in which we don't have at least an eight-card fit? Well, three spades (from the auction), two hearts (giving us only a seven-card fit there), four diamonds (giving us only a seven-card fit there), three clubs (only seven there)... that leaves one card left over. We have to have an eight-card fit somewhere!

In fact, we can generalize this further. If the opponents have a nine-card fit, that gives us four cards in that suit out of our twenty-six total. Twenty-two cards are left for the other three suits. So our fits in those suits are, if the hands fit as poorly as possible, are 8, 7, 7!

If they have an eight-card fit, that gives us five there and twenty-one in the other three. The worst possible case is that the other three suits split 7-7-7. It's much more likely that we have an eight-card fit or longer, though.

Long story short: If the opponents have a nine-card fit, we definitely have an eight-card or longer fit. If they only have an eight-card fit,we probably have one too. Compete!