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Law of Total Tricks - a parity oddity

I was using my program, Deal, to do some experiments pertaining to The Law of Total Tricks, when I came across something that took me a bit of time to diagnose.

Larry Cohen says to add an adjustment of one trick to the total tricks if both sides have two fits of eight cards or longer. I decided to determine how often such double-double fits (DDFs) occur.

How often does a double-double fit occur when the total trumps is n, for n=14...23?

The table is surprising:

Total trumps % with DDFits
14 0.0%
15 0.0%
16 11.9%
17 0.0%
18 28.3%
19 10.8%
20 25.2%
21 17.5%
22 27.4%
23 12.9%

I stared at this table for a long time. I thought I had done the simulation wrong. None of the hands with 17 trumps had a double double fit? After finding nothing wrong with my code, I tried to think about it combinatorially.

To have a DDF with 17 total trumps, one of the partnerships has to have an 8-card fit, the other has to have a 9-card fit.

Say east-west have the 8-card fit. Then their second fit also would have to be an eight-card fit, because we are assuming their 8-card fit was the longest fit they had. So their combined shape must be 8-8-X-Y where X+Y=10 (X>=Y).

Since NS have a 9-card fit, that means Y must be 4, which in turn means that X must be 6, which means that NS do not have a double fit.

So it is impossible to have a double-double fit when the total trumps is seventeen.

In fact, looking at the table above, it is significantly less likely that a DDF will occur whenever the total trumps is an odd number. The graph of the above data is a zig-zag.

What does this mean, bridge-wise? Well, it means that after the DDF adjustment, the total tricks value skews towards the odd values.

Here is a table of total tricks, before and after adjustment for DDFs:

Before AdjustmentAfter Adjustment
1410.5%10.5%
1510.8%10.8%
1626.5%23.2%
1722.8%26.1%
1815.5%11.3%
198.7%12.0%
203.5%3.5%
211.3%2.0%
220.3%0.5%

There is a perversity at this table - the adjustment makes the odd values of 17 and 19 more likely than the even values on either side them. The shape of the graph before adjustment is pretty straightforward, but after adjustment, it becomes a more complex beast.

Should this have any affect on your bidding? It's not clear to me, except that, perhaps, if you think the total tricks is either 16 or 17, you ought to expect 17, without other information to guide you, but "without other information..." is a useless bridge phrase - there is always other information.

Larry Cohen's other adjustments are much harder to determine automatically, so I haven't been able to throw them into my simulation. What is a rigorous definition of hand purity, for example?


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Copyright 1996-2005. Thomas Andrews (thomas@thomasoandrews.com) .