In the second installment, I investigated completely pure hands, hands where both sides have all the honors in their two longest suits, and calculated the expected number of defensive winners that will cash. The results appeared in Table 2, reprised below:
Table 2: Expected Defensive Winners for Completely Pure Hands
SF |
Length of Longest (Solid) Suit |
Length of Next Longest (Solid) Suit |
Defensive Winners |
14 |
7 |
7 |
4.36 |
14 |
8 |
6 |
4.18 |
15 |
8 |
7 |
3.82 |
15 |
9 |
6 |
3.85 |
16 |
8 |
8 |
3.28 |
16 |
9 |
7 |
3.49 |
16 |
10 |
6 |
3.32 |
17 |
9 |
8 |
2.95 |
17 |
10 |
7 |
2.96 |
17 |
11 |
6 |
3.06 |
18 |
9 |
9 |
2.62 |
18 |
10 |
8 |
2.62 |
18 |
11 |
7 |
2.70 |
19 |
10 |
9 |
2.09 |
19 |
11 |
8 |
2.16 |
I ended that piece by looking at the clusters where the defensive tricks were similar. These appeared, like 9-8, 10-7, and 11-6 where the total number of cards in the long suits were the same. That number is called the second fit number, or SF. When SF is 17, both sides will cash, on average, three tricks on defense, and so there will be 20 total tricks available:
This suggests a simplistic formula for completely pure hands,
a formula that holds fairly well as a first approximation. Of course, such a formula can’t hold universally, or, when the pranksters at the club stack a hand where everyone has a 13 card suit, there would be 29 total tricks. In fact, for completely pure hands, when SF is either 25 or 26, there are 26 tricks available (as a nice exercise, try showing that the set of all hands with SF=25 averages 25.846 total tricks). We expect some tailing off as SF grows, and, looking at Table 2, when SF is 18, we expect around 20.7 winners, and when SF is 19, around 21.8 winners.
For less pure hands, SF + 2 works as a good starting formula, and Colchamiro and I showed that this formula (with some adjustments) predicts total tricks quite well, substantially better than total trumps. These results are in the December 2011 Bridge World, for those interested.
All this raises one obvious question. I am absolutely convinced that the Second Fit number, and not the totality of trumps, determines the number of tricks available, on average. So, Bloom, how come, statistically, total trumps and total tricks line up so well?
The answer is quite simple: Let’s suppose, as I believe, that total tricks align with the SF number. Let’s take an arbitrary hand and modify the distribution to raise the SF total by one, and so, on average, increase the trick total by one. We can make such a change in one of four ways –
These four moves change the total trumps, respectively, by 2 trumps, 1 trump, 1 trump, and 0 trumps[1]. Notice that these trump changes average out to an increase of one trump. So we have two causalities in play:
These two increases move along together in complete lock-step. Yet there is nothing, whatsoever, about total trumps that directly impacts total tricks. The great statistical correlation between trumps and tricks is a direct result of the influence of the second fit number. The Law is anything but! The apparent connection between total trumps and total tricks is, in fact, just one big coincidence.
[1] This simplistic arithmetic only applies when there is really a longest and shortest suit. Things are more complicated when we have two fits of the same length. Here, for example, is the correct analysis when both sides have two eight card fits. SF is 16, so SF+2 gives 18 tricks, but both sides adjust down by half a trick for having only an eight card trump fit. So SF predicts 17 total tricks. Now move a card from either short suit to either long suit. That change gives both sides a nine and an eight card fit, SF = 17, with no adjustments. So total tricks rise to 19, a two trick improvement. That shift also increases total trumps from 16 to 18, another two trick change, with the usual correlation.
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