Game theory addendum

The last post started investigating a game theoretic mixed-strategy situation where I speculated that the best strategies for maximizing tricks might not be the best strategy for maximizing boards (at BAM scoring).

      A7
      -
      T
      -
?3             ?8
-              -
J              -
-              6
      QT
      -
      7
      -

Initially spades were 2=4, and we're presuming there's no additional information about the a priori location of the SK. You throw LHO in with a diamond and he leads a spade. How often is RHO supposed to bare the K if he has it? How often is declarer supposed to let the spade return ride to the Q?

Basically, if LHO has the SK, playing SA wins 1 trick, and low wins 2. If RHO has the SK and pitches a club, you always make 1 trick. If RHO has SK and bares it, going up wins 2 tricks and low wins 0.

Jonathan and I suggested that for trick maximization, RHO should bare the SK 1/4 of the time that he holds it, so that when declarer sees a spade discard he expects the SK to be on his right 1/3 of the time -- then going up expects to win 1 trick 2/3 of the time and 2 tricks 1/3 of the time for 4/3 tricks; while playing low wins 2 tricks 2/3 of the time and 0 the rest, also 4/3 tricks. (When declarer sees a club pitch--which happens half the time--declarer always makes 1 trick, so the total expectation is 7/6. If RHO never bares the K declarer can score 4/3 tricks by always playing low.)



But, BAM scoring complicates the analysis. Intuitively, the 1-trick error (flying SA when the endplay was on) always costs a half a board, while the 2-trick error (losing to stiff K instead of dropping it) usually costs you (proportionately) a full board (when the other table just pitches a club holding SK), but only costs you half a board when the other declarer also faces the same problem, which he will some of the time. So the BAM cost of the 2-trick error is not twice the BAM cost of the 1-trick error. Therefore, declarer should be more willing to risk the 2-trick error, and the defense needs to bare the SK more often to be optimal.

Suppose I know my opponents believe that baring 1/4 is optimal (if it really is optimal, I won't be able to exploit this). I will always play a low spade. Meanwhile, at the other table, I plan to bare 30% of the time and to tell them. Believing that this is suboptimally high, their response should be to always play SA (which will in fact win more tricks). If this happens, then:

SK on left (1/3): My team wins the board.
SK on right and my opponent bares (1/6): My team always loses the board.
SK on right and my opponent discards a club (1/2): I always take 1 trick. Meanwhile, at the other table, they take 2 tricks 30% of the time (when my 'mates bare SK) and 1 trick 70% of the time (when my 'mates also discard a club), so we push the board 70% of the time.

In total we score 1/3 + .7/4 for a touch under .51 boards.

What if the other declarer also always plays low? Then:

SK on left (1/3): both take 2 tricks, push, 0.5
SK on right and my opponent bares (1/6): 30% push, 70% lose, 0.15
SK on right, opp discards a club (1/2): 30% win, 70% push, 0.65

Total expectation: .517

So, clearly 1/4 baring does not maximize boards won. By my math, the right frequency of baring the K in this situation is 1-sqrt(0.5), or about 29.3%. I wasn't previously aware that a rational bridge strategy could ever be an irrational frequency.

Finally, note that if spades had initially been 3=3 (or more with LHO) things are much simpler: RHO should always bare the SK and declarer should always play low. Theoretically declarer can adopt any strategy and have the same expectation, but always playing low gains against non-barers without losing expectation to barers (presuming, of course, you have no read on your opponents).

To me, it's a bit interesting that in the 2=4 case RHO wanted P(SK on right | RHO kept club) to be over 0.5, but exploiting that option in the 3=3 case (by not always baring) is wrong.