[matmat]

how many losers (according to the LTC) is the suit:
A9xxxxxx?

[effervesce]

Two losers. One loser per A/K/Q missing in a suit. If partner has xx support (a reasonable chance), it's likely to have just one loser so treating it as 1 loser ok.

[P_Marlowe]

2.

But the LTC works not very well for 7-2 fits,
and similar for 8-? fits.
As it is, if you indicate to partner, that you have
a long suit, he should switch to counting cover
cards instead of counting loosers.

With kind regards
Marlowe

[pclayton]

If you are into this kind of voodoo, I'd evaluate it is as 1.5 losers (average 1.7 around the table). If pard raises, then its 1/2.

[helene_t]

An ace is - 1 1/2 loser so it's 3 - 1 1/2 = 1 1/2 loser. Of course common sense says that with such such a long suit you must sometimes subtract from 2 rather than 3. If I were to formalize LTC for 8-cards I would take the probability that no opp has a 3-card into account. I suppose it would be something like 1 LTC for this hand.