 The concept of a combination is fundamental and elementary to bridge calculations. Combinations are applied when attempting to calculate the probability of a specified hand pattern or the division of a suit among the four players at the bridge table. Combinations are also often used in calculating the respective probabilities, or the ratio between the division of the combined holding of the defenders in a specified suit or with specified honor holdings. Many mathematicians have used their knowledge in calculating these probabilities, and their results, dry as they may seem, are presented on this web page.    Mathematicians have arrived at a general expression for a combination, and the formula is nCr, which is the number of combinations of n things taken r at a time. Therefore, 4C2 indicates the number of ways in which 2 articles can be select out of a total of 4 articles.

It is noted that if r=n, then the formula nCn can be replaced with nCr. Therefore, it follows that whatever the number n represents, nCn is equal to 1. It stands to logic that there is only 1 way in which n articles can be selected from n articles, and there is only 1 way in which no (0) articles can be selected.

In order to calculate the value of a combination, the mathematicians use the concept of the factorial. The factorial number is the product or result of all numbers from 1 up to and including the specified number. Therefore, 6 factorial is the product or result of 1x2x3x4x5x6. Mathematicians use the short version of 6! to symbolize this, which means the result of 6x5x4x3x2x1. Conventionally, the value of 0! is taken as 1.

When considering the number of ways in which 13 cards can be selected from a pack of 52 cards, the first card can be any one card of 52 cards. The second selected card is then taken from 51 cards. In selecting the remainder of the individual cards, the result is 52x51x50x49x48x47, and so on. Mathematicians have arrived at the formula: The obtained result would be correct if the goal were to determine the order in which 13 cards were selected. However, the order in which the 13 cards were selected is irrelevant. Mathematicians have taken this procedure one step further, and decided to take 1 of the 13 cards at random. This 1 card could have been selected from any 1 of the 13 cards drawn. The second card drawn randomly could be any one of the 12 remaining cards, and so on. This signifies that the selection is based on 13x12x11x10x9, and so on, or 13!, as expressed mathematically, which means the number of ways in which these specific 13 cards could be selected. In other words, the total number of ways in which 13 cards can be selected from a pack of 52 cards can be expressed mathematically in the following formula: The general formula, therefore, for this type of calculation, as determined by the mathematicians, is: As an illustration of the working of these formulas, an example of where the defenders have a combined holding of four cards in a specified suit would be descriptive. The result is that the defenders hold 22 other cards in the other 3 suits. As a result of the formulas, any named player could have a holding in the specified suit of:

 0 cards in: 4C0 x 22C13 ways 1 card in: 4C1 x 22C12 ways 2 cards in: 4C2 x 22C11 ways 3 cards in: 4C3 x 22C10 ways 4 cards in: 4C4 x 22C9 ways

Keeping in mind that 26 cards can be divided between the 2 defenders in 26C13, or 10,400,600, ways, the following mathematical table can be obtained:

 4C0 x 22C13 = 1 x 497,420 % = 497,420 = 4.782 6 4C1 x 22C12 = 4 x 646,646 = 2,586,584 = 24.869 6 4C2 x 22C11 = 6 x 705,432 = 4,232,592 = 40.695 7 4C3 x 22C10 = 4 x 646,646 = 2,586,584 = 24.869 6 4C4 x 22C9 = 1 x 497,420 = 497,420 = 4.782 6 Total 10,400,600 100.000 1

The extra 0.0001% is due to approximation.

It is evident that nCr = nC(n-r). This means that 4C1 is equal to 4C3. This is self-evident, if one player can hold r amount of cards in a specified number of ways, his partner must be able to hold the remainder of the partnership cards in exactly the same number of ways.   In examining the problem of a holding of specified cards, the assumption is made that the 4 cards held by the defense in a named suit consisting of K-Q-x-x. The question is then, what is the probability that a named defender, for example, West, holds both the King and the Queen. West could probably hold:

 K-Q-x-x = 4C4 x 22C9 = 1 x 497,420 = 497,420 K-Q-x = 2C2 x 2C1 x 22C10 = 1 x 2 x 646,646 = 1,293,292 K-Q = 2C2 x 22C11 = 1 x 705,432 = 705,432 Total 2,496,144

The respective percentages are: 4.7826% and 12.4348% and 6.7826%. The total is exactly 24%.

Comparing the probability of the holding by West of the doubleton K-Q, in this example, with the probability of the holding by West of the singleton King, the result would be:

 K-Q = 2C2 x 22C11 = 1 x 705,432 = 705,432 = 6.782 6% K = 1C1 x 22C12 = 1 x 646,646 = 646,646 = 6.2174%

The result is that 705,432 and 646,646 have a highest common factor, or HCF, of 58,786, resulting in a ration of 12:11.

Since 2C2 and 1C1 are both equal to 1, it is the same as comparing 22C11 and 22C12. This comparison can be shown with the following formula. In the above example, n = 22 and r = 11, and therefore: which results in a ration of 22C11 to 22C12 as 12:11. This method may be used to draw other comparisons as well, such as, which has the greater probability, and by how much, that a named player will hold two out of the four missing cards or that this named player will hold three such cards. The comparison is between 4C2 x 22C11 and 4C3 x 22C10.

It has been established that:

22C11 = 22C10 x 12/11

and that:

4C2 = 4C3 x 3/2

and therefore: The ratio is therefore:

4C2 x 22C11 : 4C3 x 22C10 :: 18 : 11

The meaning of this is that the chance that a named player holding 2 of 4 missing cards is higher than the chance that the named player holds 3 of the 4 missing cards. However, the overall chance of a 3-1 or 1-3 break is 22:18, or 11:9, as there are two different and equal ways in which the cards of the defenders can be divided so that one of them holds 3 cards, but only one way in which each of the defenders holds 2 cards.

If you have read this far, then the realization that the mathematicians have given this combination much thought is evident. To what extent it assists the player in determining the percentages of one defender holding a certain and specified card combination or not is questionable, since a mental calculation is practically impossible within the time limit established to play a card to a trick. Therefore, this is only a presentation of the probability of combination devised by mathematicians, who have taken an interest in the probability factor, which is admirable, but hardly a useful tool at the bridge table.   VulkaN...2020