bixby, on Mar 12 2010, 04:48 AM, said:
Hmmm . . . . let's consider this hypothetical case:
732
63
6532
A52
AK4
A54
A8
J10987
The contract is 3NT and the opening lead is the SQ. Declarer wins in hand, despairs of making the contract, and runs the CJ, losing to East's Queen. East returns a spade, which declarer wins in hand. Now declarer holds his breath as he runs the C10 successfully. Declarer plays to the CA, comes back to the DA, and runs the remaining clubs, discarding two diamonds from dummy. On the second diamond discard, however, the CK is discovered under dummy's diamonds!
Now what? Without the original infraction, declarer would have cashed the CAK, successfully dropping the CQ (opponent's clubs were 2-2), and would have made the contract, with 5 clubs, 2 spades, and the red Aces.
I don't see how we can award declarer 9 tricks. Law 12B provides:
"Objectives of Score Adjustment
1. The objective of a score adjustment is to redress damage to a non-offending side and to take away any advantage gained by an offending side through its infraction."
Even though hiding the CK was an infraction, the damage was suffered by the offending side, not the non-offending side. I don't think we adjust to take away the damage suffered by the offending side. There's no law against running the CJ at trick two, and that's what declarer did. We don't give him back that trick. So I say declarer is down one.
I do too, in your case, but your example does not convince me to change my opinion on the original case.
In your case, the result at the table is eight tricks. Declarer has four side suit winners and in clubs he has won three of the first four (revoking in dummy on the fourth) when the missing K
♣ is discovered as the fifth club trick is played. The king of clubs wins the fifth club trick, and declarer is now in dummy with eight tricks in and no more coming. The table result is down one.
There are two infractions, the hiding of the K
♣ by dummy and the revoke in dummy by declarer of the fourth club. Without the first infraction, the declarer would probably have made nine tricks, true, but the table result is eight tricks, so there is no damage to the non-offending side.
In my case, the result at the table was ten tricks. Without the first infraction, the result might have been nine only, in which case I would certainly adjust, but in looking closely at the situation I judged that the overwhelmingly likely result was ten tricks, so I did not adjust.
What I am trying to state as a general principle here is this:
When there are multiple infractions by one side, and it is certain that the first infraction was the direct and only cause of the other ones, equity should be restored by considering the likely result(s) without the first infraction, the departure point from normal.