Ted's Bridge World How Many Auctions Are There?
by Ted Muller

The number of possible 52-card bridge deals is:


The number of possible auctions is:


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:

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:

7 basic matrix
+ 21 × 7 7-notrump bids
 – – –
154 Total

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:

7 no further bidding
+ 21 × 7 7-notrump overcall
+ 21 × 154 7-spade overcall
 – – –
3388 Total

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:

7 Notrump opening =  7 =  7 × 220
7 Spades opening =  154 =  7 × 221
7 Hearts opening =  3388 =  7 × 222
 – – –
1 Club opening =  7 × 2234

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  0Rt-Arrowx  of integer n.  It is:  (nx+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:

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!

For the less astronomically minded, I offer this equivalent:

Even simpler yet: That total is approximately equal to the odds against choosing 30 consecutive winners on a roulette wheel.  So the next time your partner suggests that your bidding system is ready, you might refer her/him to this article!


Yes, I appreciate that this mind-boggling concept can be difficult to reconcile; after all, it's just five different denominations on seven levels, right?  I submit that the doubter might glean a better understanding with this empirical test:  Just start at the level of, say, 7 Clubs, and begin listing the possible iterations on paper.  If you can manage to record an average of three auctions per minute amidst all your other activities, you will be nearly finished within a year.  Then go back and start the bidding at 6 Notrump; but plan on spending 22 years this time.

When you're finished with that, you might like to consider some truly large numbers.  Did you know that there are more than 288 billion different ways that the two players can make their first four moves in a chess game?  If you don't believe that, then try counting them.  But don't try to itemize the number of possible 40-move games, because you can forget about grains of sand.  That total far exceeds the number of atoms in the known universe!

Go Back