The question is:
From Mark Whelan, shuffle up and deal. And count very, very high:
Bridge, that venerable card game, begins with an auction. There are four players around a table, each sitting across from his or her partner. At its simplest, the auction goes like this: Beginning with the dealer and orbiting around the table, players can place a bid or pass. Players who’ve passed can re-enter the bidding later. A bid is comprised of a number (one through seven) and a suit. Every legal bid must be higher than the one that came before, meaning either that its number is higher or that its number is the same and its suit is higher (from high to low, the suits go no-trump, spades, hearts, diamonds, clubs).
The auction ends if the other three players pass after a bid, or if all four players pass right away.
How many different legal bridge auctions are there?
Now, things are much simpler. There are only 4 possible auctions that end in 1♣ (anyone can open 1♣, and then everyone else has to pass). For each of these, there are 4 auctions that end in 1♦ (replace any of the final 4 bids with 1♦, and then everyone has to pass, and so on. So with 35 bids available, the total number of auctions in this situation is 'just':
4 +4^2 + 4^3 + ... + 4^35)
= 4*(1 + 4 + ... + 4^34)
or roughly 4*10^20
Compare this to the 'full' case:
28 + 28*22 + 28*22*22 + 28*22*22*22 + .... + 28*22^34
= 28*(1 + 22 + ... + 22^34)
In case you don't feel like counting the digits, that's approximately 1.29*10^47.
We hinted at this last time (you can try the naive system where the player with the ♣2 bids 1♣, the player with the ♣3 bids 1♦, etc., but this only manages to let you place 35 cards, so you definitely need the doubles and redoubles), and now we can see why - there are 'only' ~5*10^28 deals, so without using the doubles and redoubles, there aren't enough auctions for it to even be possible to place every card (regardless of the issue of whether everyone knows which bid to make at which point.