How Many Auctions Are There?

The number of possible 52card 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:
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
followups 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  7notrump 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 twentyone final passes that could be
replaced by either of two higherranking bids:
7  no further bidding  
+  21 × 7  7notrump overcall 
+  21 × 154  7spade overcall 
– – –  
3388  Total 
Substituting from the 7spade calculation, this totals: (7 × 22) + 21 × (7 × 22) = 22 × (7 × 22) = 7 × 22^{2}
By continuing in this manner, a nice pattern emerges which can be extrapolated to all thirtyfive opening bids:
7 Notrump opening  = 7  = 7 × 22^{0} 
7 Spades opening  = 154  = 7 × 22^{1} 
7 Hearts opening  = 3388  = 7 × 22^{2} 
– – –  
1 Club opening  = 7 × 22^{34} 
The combined total for all opening bids can be summed as: 7 × (1 + 22 + 22^{2}. . . + 22^{34})
There is a handydandy 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: (22^{35} − 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: (22^{35} − 1) × (4 /3) + 1. Rearranged for simplicity, we have:
(4 × 22^{35} − 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 mindboggling 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 in a year's
time. 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 40move games, because you can
forget about grains of sand. That total far exceeds the number of
atoms in the known universe!