How many different games are there in bridge?

Question

I know that there are ${52 \choose 13}$ hands. I was thinking maybe it was ${52 \choose 1} {52 \choose 39} {52 \choose 26} {52 \choose 13}$.

Answer 1

There are this many layouts of the four hands:$${52\choose 13,13,13,13}=\frac{52!}{13!13!13!13!}={52\choose13}{39\choose13}{26\choose13}$$ If you then ask how many games, including which cards are played for which trick, it becomes tricky because everyone must follow suit if they can.
EDIT: As Mark says in comments, how many bidding sequences are there?
1. Start with 0,1,2 or 3 passes.
2. At every level a bid may be doubled and redoubled: B,BP,BPP; BD,BDP,BDPP; BPPD,BPPDP,BPPDPP; BDR,BDRP,BDRPP; BDPPR,BDPPRP,BDPPRPP; BPPDR,BPPDRP,BPPDRPP; BPPDPPR,BPPDPPRP,BPPDPPRPP; so 21 possibilities. Or that level might not appear: 22 possibilities.
That would give $22^{35}$ bidding sequences, but I will treat 'passed out' separately.
There are four options of $0,1,2,3$ passes at the start, but 3 passes at the end, so I think there are $1+(4/3)(22^{35}-1)$ possible bidding sequences.
EDIT2:
How many ways can a hand be played? I ran a million hands. Within a hand, I let the players take turns leading a card. I multiplied the number of options each player had at each point, to estimate the number of possible ways to play the hand. The rule about following suit drastically reduces the number of possible ways. A few hands had a much greater number of ways, which distorted the average. I got a these results:
$$\begin{array}{lll} \text{Average}&:& 6.1*10^{26}\\ \text{Median}&:& 2.4*10^{23}\\ \text{Geometric mean}&:& 3.1*10^{23}\\ \text{Max}&:& 3.6*10^{31}\\ \text{Max possible}&:& (13!)^4=1.5*10^{39}\\ \text{Min}&:&8.0*10^{18}\end{array}$$