How Many Auctions Are There? by Ted Muller

The number of possible 52-card bridge deals is:

53,644,737,765,488,792,839,237,440,000

The number of possible auctions is:

128,745,650,347,030,683,120,231,926,111,609,371,363,122,697,557

That is a great number indeed; it seems that auctions outnumber bridge deals by a factor of

2.4 quintillion to one!  (that's 18 zeros)

Let us see how this monster figure is derived.

After a bid of 7 Notrump, there are seven possible continuations:

•  P    P    P
•  P    P    Dbl    P    P    P
•  P    P    Dbl    Rdbl    P    P    P
•  P    P    Dbl    P    P    Rdbl    P    P    P
•  Dbl    P    P    P
•  Dbl    Rdbl     P    P    P
•  Dbl    P    P    Rdbl    P    P    P

These terminal patterns accrue after any bid is made and there is no subsequent overcall.  I refer to them collectively as the Basic Matrix.

After a bid of 7 Spades, any one of the basic follow-ups may occur.  Additionally, within the Basic Matrix, any of the three final passes in any of the seven sequences (shown in red) could be replaced by 7 Notrump, after which the same patterns would apply, with no further variations.  (Replacing earlier passes would be redundant.)  The possible sequences:

This total also can be represented as:   (7 × 1) + (7 × 21)  =  7 × 22

Now consider a bid of 7 Hearts.  Again the Basic Matrix applies, as well as any of the twenty-one final passes that could be replaced by either of two higher-ranking bids:

Substituting from the 7-spade calculation, this totals:   (7 × 22) + 21 × (7 × 22)  =  22 × (7 × 22)  =  7 × 22²

By continuing in this manner, a nice pattern emerges which can be extrapolated to all thirty-five opening bids:

The combined total for all opening bids can be summed as:   7 × (1 + 22 + 22². . . + 22³⁴)

There is a handy-dandy formula for summing the powers  0x  of integer n.  It is:  (n^x+1 − 1) ÷ (n − 1).

Application of this formula to our summation yields:   (22³⁵ − 1) / 3.

We're not quite finished.  The above calculation must be multiplied by 4 to accommodate any number of initial passes.  Finally, the unique case of a passout is included.

Our grand total becomes (22³⁵ − 1) × (4 /3) + 1.  Rearranged for simplicity, we have:

(4 × 22³⁵ − 1) ÷ 3

Who would have guessed that the number of bridge auctions would be a function of the powers of 22?

Note:  In case you still lack a full appreciation of the size of that number, consider this hypothetical comparison:

• The number of grains of sand on the earth is estimated as  10²⁰  to  10²⁴.  Assuming the maximum, that's  100 million × 100 million × 100 million  grains.
   
• There are an estimated  10¹¹  stars in our galaxy, and an estimated  10¹¹  galaxies in the visible universe.
   
• Speculating that our own galaxy is of average size, that's  10 million × 10 million × 100 million  stars.
   
• Envision an average of  12  earth-like planets orbiting every star in every galaxy.

Under those conditions, if you were to count all the grains of sand in the entire universe, the total still would fall short of the number of possible bridge auctions!

So the next time your partner suggests that your bidding system is ready, you might refer her/him to this article!

counting bridge auctions
number of bridge auctions
total bridge deals