Main Article 7Z70 by Richard Pavlicek
How do you measure the freakness of a bridge hand? To most bridge players, the question is purely subjective, and you would get answers anywhere from flat or square for balanced hands to wild and crazy for fluke hands the latter of which makes me wonder why Steve Martin never became a bridge player.
Years ago (I cant believe its nearly 20) I worked out a method for this. The basic premise was that the flattest pattern (4-3-3-3) is given a freakness of zero (that surely makes sense) and all the other patterns gain points according to this rule:
|Add 1 point for each card over four or under three in each suit.|
The rule seemed to work OK at first; for example, 4-4-3-2 would get 1 point (one suit under three cards) and 5-3-3-2 would get 2 points (one suit over four, and one suit under three). Unfortunately, the scheme also awarded 2 points for 4-4-4-1 (one suit with two cards under three), thereby equating a balanced hand to a hand with a singleton. This had to be fixed. After some more thought I came up with two adjustments, which apply to the hand as a whole:
|Add 1 point if the hand contains a singleton, or|
|Add 2 points if the hand contains a void.|
Note that the above are exclusive; you cannot add both. If a hand contains a void, it gets 2 extra points period, even if it also happens to have a singleton.
So whats the reason for all this? The purpose is to be able to quantify the freakness of a bridge hand on a linear scale, much like point count is used to evaluate its strength. I have used this formula for a number of computer applications, such as my deal database and computer bidding programs. It facilitates decision making based on the shape of a hand. For example, a balanced hand is now simply defined as a hand with a maximum freakness of 2.
Applying the formula to the 39 generic hand patterns gives pleasing results. The patterns are ranked in an order that feels right. When two different shapes produce the same freakness number, it usually is difficult to judge one as more freakish than the other. For example, the patterns 4-4-4-1 and 5-4-2-2 both calculate to 3 points, and its a moot question which should be considered more unbalanced.
Another pleasing aspect of the formula is the 0-to-20 scale, i.e., nice round numbers. Note that the extreme patterns grow quickly from 14 to 20, and the formula yields no freakness of 15, 17 or 19. I thought about the possibility of graduating the high-end values to 20, 19, 18, 17, etc., but this would only complicate the formula for no apparent advantage.
Once the formula for hand freakness is developed, it is easy to quantify the freakness of an entire deal. The nature of a deal is determined by the composition of the four hands, so the obvious conclusion would be:
|The freakness of a deal is equal to the sum of the freaknesses of each hand.|
Hence, the freakness of a deal ranges from 0 to 80, although the vast majority lie on the low end. There also are more unattainable numbers. For example, next to 80, the highest possible deal freakness would be 76 because if any of the 13-0-0-0 patterns were changed to 12-1-0-0, a second hand must also change. But we wont be too worried about deals like that.
A curious puzzle is, What is the lowest freakness value that cannot be obtained for a deal? My calculations show the impossible numbers to be 79, 78, 77, 75, 73, 71 and 69, so the last would be the answer. Hmm 69 was also the year of the moon landing and my wedding; some might say theres a correlation.
The next question that comes to mind is, What is the average freakness of a bridge hand? This is not easily answered. There is no simple way to calculate it (at least I couldnt think of one) so I approached it by brute force the old pickax method.
For each hand pattern, I calculated the total number of bridge hands it could produce. This is done with combinatorial arithmetic. The number of combinations of N items taken R at a time (often abbreviated NcR) is determined by this formula:
The notation N! (N factorial) means the product of all integers from 1 through N. For example, 6! = 720. Also noteworthy is that 0! is defined as 1.
For example, to determine the number of 4-3-3-3 hands (say, with four spades) there are 13c4 ways to choose the spades, and 13c3 ways to choose each of the other suits. Multiplying (13c4 × 13c3 × 13c3 × 13c3) provides the total. But wait! Thats only the hands with four spades, so we must multiply by four to get the total number of 4-3-3-3 hands. This last factor indicates the number of ways the generic pattern can be permuted among the four suits, which will either be 4, 12 or 24, depending on the like digits in the pattern.
The values of the combinatorials (13c0 to 13c13) are shown in the chart below. Note the equality of any two numbers that add to 13, which should be obvious from the formula. (I have eliminated the 13 to save space in the chart and table.)
A calculation for all of the generic hand patterns is shown in the table that follows. To verify its accuracy, consider that the total of the hands column should equal the total number of bridge hands, which is 52c13 or 635,013,559,600. It does. Check it if you wish just pretend you are balancing Bill Gatess checkbook.
All that remains is to multiply the hand total of each pattern by its freakness in the right column (the first one shows that I wasted my time calculating 4-3-3-3 hands), add the results and divide by 635,013,559,600. The final tallies are shown at the end.
Rounding to two decimal places, the average freakness of a bridge hand is 2.98, and the average freakness of a full deal is four times that, or 11.93.
|4-3-3-3||4 × c4 × (c3)³||66,905,856,160||0|
|4-4-3-2||12 × (c4)² × c3 × c2||136,852,887,600||1|
|5-3-3-2||12 × c5 × (c3)² × c2||98,534,079,072||2|
|4-4-4-1||4 × (c4)³ × c1||19,007,345,500||3|
|5-4-2-2||12 × c5 × c4 × (c2)²||67,182,326,640||3|
|5-4-3-1||24 × c5 × c4 × c3 × c1||82,111,732,560||4|
|6-3-2-2||12 × c6 × c3 × (c2)²||35,830,574,208||4|
|6-3-3-1||12 × c6 × (c3)² × c1||21,896,462,016||5|
|5-4-4-0||12 × c5 × (c4)² × c0||7,895,358,900||6|
|5-5-2-1||12 × (c5)² × c2 × c1||20,154,697,992||6|
|6-4-2-1||24 × c6 × c4 × c2 × c1||29,858,811,840||6|
|7-2-2-2||4 × c7 × (c2)³||3,257,324,928||6|
|5-5-3-0||12 × (c5)² × c3 × c0||5,684,658,408||7|
|6-4-3-0||24 × c6 × c4 × c3 × c0||8,421,716,160||7|
|7-3-2-1||24 × c7 × c3 × c2 × c1||11,943,524,736||7|
|6-5-1-1||12 × c6 × c5 × (c1)²||4,478,821,776||8|
|7-3-3-0||12 × c7 × (c3)² × c0||1,684,343,232||8|
|7-4-1-1||12 × c7 × c4 × (c1)²||2,488,234,320||8|
|6-5-2-0||24 × c6 × c5 × c2 × c0||4,134,297,024||9|
|7-4-2-0||24 × c7 × c4 × c2 × c0||2,296,831,680||9|
|8-2-2-1||12 × c8 × (c2)² × c1||1,221,496,848||9|
|8-3-1-1||12 × c8 × c3 × (c1)²||746,470,296||9|
|8-3-2-0||24 × c8 × c3 × c2 × c0||689,049,504||10|
|6-6-1-0||12 × (c6)² × c1 × c0||459,366,336||11|
|7-5-1-0||24 × c7 × c5 × c1 × c0||689,049,504||11|
|8-4-1-0||24 × c8 × c4 × c1 × c0||287,103,960||11|
|9-2-1-1||12 × c9 × c2 × (c1)²||113,101,560||11|
|9-2-2-0||12 × c9 × (c2)² × c0||52,200,720||12|
|9-3-1-0||24 × c9 × c3 × c1 × c0||63,800,880||12|
|7-6-0-0||12 × c7 × c6 × (c0)²||35,335,872||13|
|8-5-0-0||12 × c8 × c5 × (c0)²||19,876,428||13|
|9-4-0-0||12 × c9 × c4 × (c0)²||6,134,700||13|
|10-1-1-1||4 × c10 × (c1)³||2,513,368||13|
|10-2-1-0||24 × c10 × c2 × c1 × c0||6,960,096||14|
|10-3-0-0||12 × c10 × c3 × (c0)²||981,552||14|
|11-1-1-0||12 × c11 × (c1)² × c0||158,184||16|
|11-2-0-0||12 × c11 × c2 × (c0)²||73,008||16|
|12-1-0-0||12 × c12 × c1 × (c0)²||2,028||18|
|13-0-0-0||4 × c13 × (c0)³||4||20|
|Total number of hands (above)||635,013,559,600|
|Total hands × freakness points||1,894,153,566,372|
|Average hand freakness||2.9828553071609|
|Average deal freakness||11.9314212286436|
Most bridge players know there are 39 generic hand patterns. But what about generic deal patterns? How many possible deal shapes exist? I doubt that anyone could answer this myself included until I wasted some time recently. It is easy to calculate the number of possible deals, but enumerating the distinct patterns eludes any simple formula.
First, it is necessary to define the term generic for a deal pattern, as different interpretations are possible. Assume any one hand, say North, has a specific shape, say 5-3-3-2. Which other hand patterns could combine with this to produce a deal? One such possibility would be 5-3-3-2 | 3-4-2-4 | 4-1-4-4 | 1-5-4-3. The suits must align with the original pattern so each adds to 13, but the order of the hands doesnt matter. Also, we are not concerned with permutations of the original 5-3-3-2, as this would create the same permutations in the other hands, i.e., the same generic deal.
Now, for some real number crunching. My first step was to calculate the number of deal shapes that could be produced by each of the 39 generic hand patterns. For example, if Hand 1 has 4-3-3-3 pattern, there are 96,537 possible deal shapes, but these are not generic. They include duplications from rotations or flip-flops between other hands, as well as many match-ups from other groups. I expanded each of the 96,537 shapes to find the number of deals it could produce, and this total is shown in the last column. [See table at end.]
To verify the calculations, I also included the permutation factor (4, 12 or 24) in the deal total, so in theory the grand total of the last column should equal the total number of bridge deals:
Needless to say, I was quite pleased when the totals agreed. Thats the good news. The bad news was that I still had no answer. Somehow, I had to weed out the duplicates among 2,571,711 shapes. This could only be done by sorting, so I encoded each hand pattern as an index from 1-560 (thats the number of hand patterns including permutations) and packed the three highest, in order, into a 32-bit value (the missing pattern could always be recalculated). Even with this coding, it produced a file of 10+ megabytes, which was tough to sort; but Im persistent. After finally sorting the file, it was an easy matter to determine sound the trumpets the number of generic deal patterns:
And thats my final answer.
|Hand 1||Shapes||Total Number of Deals|
© 2000 Richard Pavlicek