Thursday, November 19, 2009

Incremental Raises of Preempts

In my previous blog, I alluded, without proof or justification, to the generally held principle that if you take more than one call to reach a contract that you could have reached in just one call, you must have a more defensive hand.  Let's examine why this should be so in the case of raising of partner's preempt (although similar arguments would apply to all situations).

For the sake of an example, let's assume that partner has opened with a disciplined weak two (spades) in second seat at all white.  You have a relatively balanced hand with some values and some spades.  If he makes his contract, you will be +110 or +140.  You don't really expect him to make game based on your hand (but you can never be sure).  On the basis of just your hand and partner's, there is no point in raising or bidding anything else.  A pass will do just fine.

But now let's suppose that you have exactly three spades and some moderate shape, say a doubleton.  Opponents, being what they are, are quite likely to get into the auction.  They have only four spades between them so that leaves them 22 other cards.  There's no guarantee they have a nine-card fit of their own, but if not they must have at least one eight-card fit (and two seven-card fits).  So, let's take a look at the total trick possibilities (No, I am not going to assert that total tricks equals total trumps!). We assume for now that each side knows when to double, although in practice this is usually not the case.  In the following table of absolute par scores for our side, the left-hand column shows the total number of tricks available (may or may not equal the total number of trumps) and the other columns show the par score when we can take the indicated number of tricks with spades as trumps.  For simplicity, we assume that their best suit is .

Total Tricks
Par (7)
Par (8)
Par (9)
Par (10)
-100 (2♠X-1)
+110 (2♠=)
+140 (2♠+1)
+420 (4♠=)
-140 (3=)
+100 (3X-1)
+140 (2♠+1)
+420 (4♠=)
-420 (4=)
-100 (3♠X-1)
+140 (3♠=)
+420 (4♠=)
-450 (4+1)
-300 (4♠X-2)
+100 (4X-1)
+420 (4♠=)
-980 (6=)
-450 (5=)
-100 (4♠X-1)
+300 (5X-2)

Essentially, the upper-right of the matrix half shows where it is our hand (we can make a plus score) while the lower-left half shows the situations where it is their hand.  The right-most column is there for completeness – as mentioned above, we don't think we have game.  It's also possible that we can take even fewer than 7 tricks, but that would be unusual and even then, such hands are "their hands" and despite having the lower-ranking suit, they will generally be able to control how the auction goes.

As I recall, the most common number of total tricks in practice is 17, with 18 being a bit more likely than 16.  It's also possible of course to have fewer or more total tricks.  Fewer total tricks than 15 would be fairly rare, especially given that partner has a six-card suit.  Having more total tricks would not be so very rare, but such case would require great shape and purity of suits.

Let's see which conditions are favorable for a direct raise to 4♠ over either a pass, double or 3-level bid by RHO:
  • whenever we can take 10 tricks (in all such situations we will double if they bid on);
  • whenever there are 19 total tricks;
  • when there are exactly 18 total tricks and we can take exactly 8 of them.
This is a total of 8 cases out of 20.  The specific case of 18/8 is not easy to diagnose of course.  Now, let's look at the remaining situations where we should want to raise to 3♠:
  • whenever we can take exactly 9 tricks (but see below);
  • when there are exactly 17 total tricks (the most likely number) and we can take 8 of them.
Note that there are only half as many cases (four) where it is right to raise to 3♠ as opposed to 4♠.  Taking the cases where we can take 9 tricks, there are some dangers.  In two situations (15 and 16 total tricks) we are unnecessarily risking a minus score.  The other 9 trick situation is discussed below.  There remain seven situations in which it is appropriate to pass throughout.

There is exactly one case where we may do well to bid 3♠ and later take the push to 4♠:
  • when there are 18 total tricks and we can take exactly 9 tricks (giving us a chance of 300 at the risk of -100).
Can it ever be correct to bid 3♠ and then take the push to 4♠?  Yes, if by doing so we increase our chance of beating the absolute par.  It is reasonable in just this one situation.  We can make 3♠ and they can make 3.  As shown in the table, par on this board is 100 for pushing them to 4 and then doubling.  Assuming we're in a good field, this will give us more or less an average score.  Can we do better?  We might do a bit better if they don't take the push and we get to make 140.  But we hit the jackpot if we push them to 5 and double them for down 2 (300)!  Of course, if they don't take the push, we will be minus 50 (best) or 100 (worst).  So, how does the "slow play" strategy compare with the "fast play"?  Let's say that the opponents have a 50% chance of getting it right (more or less) at each level.

Let's assign the following match-points (on a 51 top like you might find on the second day of an NABC event):
  • 500  (1) 51
  • 420  (2) 49-
  • 300  (3) 47
  • 170  (3) 44
  • 150  (1) 42
  • 140  (3) 40
  • 110  (1) 38
  • 100  (12) 31
  • 50 (5) 23
  • 0 (1) 20
  • -50 (6) 16-
  • -100 (8) 9-
  • -140 (3) 4
  • -300 (1) 2
  • -420 (1) 1
  • -500 (1) 0
Our expectations with the slow play strategy are:
  • they pass us in 3♠ and we beat par with 140 (40 mps) 50%: 20
  • they bid on to 4 and we bid on to 4♠ (50%)
    • they let us go down quietly (16- mps) (25%): 2.06
    • they double us (9- mps) (25%): 1.19
    • they bid on to 5 and we double for 300 (47 mps) (50%): 11.75
The total expectation this way is 35.

With the fast-play strategy, we give up being able to bid and make 3♠.  Our net expectation this time is 30, slightly lower than with the slow-play but still a little above average.

So, what kind of hand do we need for the slow-play strategy to work?  It's a hand we're reasonably confident of making 140 if allowed to play it.  But, at the same time, it has defensive strength too because, combined with partner, we expect to take four tricks on defense.

Granted, I have only examined the white/white situation above.  Naturally, things get even more exciting when one of both sides is vulnerable!

A hand such as ♠J84 Q6 743 ♣KQT92 (the one I discussed in the previous post) is a defense-oriented hand but, combined with partner's hand, is likely to be significantly outgunned.  I would expect us to be able to take 5 spades, perhaps a heart ruff, and maybe two other tricks, in other words we can just about make 2♠.  We probably have around 16-17 hcp between us, which means that our opponents have 23-24.  I would expect them to be able to take 9 or perhaps 10 tricks if they have a decent fit in hearts.  Note that the presence of the Q in our hand tends to reduce our estimate of the total tricks.  As it happened, we were already three tricks too high at the two-level! Opener's hand was a not-very-powerful ♠KQT762 9543 85 ♣7.  And the opponents can actually make 7 (although nobody bid it).  So the number of total tricks was about as expected (18) but the tricks were distributed 13-5!
A hand such as ♠J874 7643 ♣AKT92, however, needs to bid to 4♠ immediately!  It might even make (you have no idea if it will), but it might keep the opponents our of a making slam even.  Given the void in RHO's suit (he did bid 3) and the purity of the layout, the total tricks on this deal could easily be 20 or 21.

I'd like to add some more scenarios and calculate their expectations, too.  I haven't lost sight of the fact that when at the table, you don't know the total tricks, and you don't know how many tricks your side can actually take.  But experience gives you some good indications.  Watch this space.

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