In a game of bridge, there are $4$ players, each is dealt a hand of $13$ cards coming from $4$ suits each of which has $13$ cards in that suit. We say that a player has a 4333 distribution if they have $4$ cards in a single suit and $3$ cards in each of the three remaining suits.
There are two parts to the game of bridge. The first is an auction, and the second is the play of the hand. The game is played in pairs, and there is an interesting quirk that the one of the hands of the pair that wins the auction is revealed to everyone, we call this hand the dummy, and the partner of the dummy is called the declarer.
Let's say that I am defending (i.e. not the declarer nor the dummy), and I have a $4333$ distribution, and say that the dummy is revealed to also have a $4333$ distribution, and our four card suits are different (say I have a $4$ card heart suit and dummy has a $4$ card spade suit). Furthermore, let us assume that from the auction that the declarer made a bid that promised that their hand is balanced (say a $1$NT opening). This means that the declare has a distribution of $4333$, $4432$, or $5332$, is it the most probable situation that the declarer is $4333$?
I initially thought that each of the three distributions should be equiprobable, but then I realized that I was being silly. Note that we can rephrase the question in the following way given a deck of $26$ cards of which there are $4$ suits and $6$ cards in two suits and $7$ cards in the other two suits what is the probability of getting dealt a $4333$ hand given that the hand that is dealt is $4333$, $4432$, or $5332$?
If someone could let me know if my analysis is correct, the number of $4333$ hands will be:
$$2\binom{7}{4}\binom{7}{3}\binom{6}{3}\binom{6}{3}+2\binom{7}{3}\binom{7}{3}\binom{6}{4}\binom{6}{3}$$
Note this number is $1715000$
Now the number of $5332$ hands should be $$4\binom{7}{5}\binom{7}{3}\binom{6}{3}\binom{6}{2}+2\binom{7}{5}\binom{7}{2}\binom{6}{3}\binom{6}{3}+2\binom{7}{3}\binom{7}{3}\binom{6}{5}\binom{6}{2}+4\binom{7}{3}\binom{7}{2}\binom{6}{5}\binom{6}{3}$$
Note this number is $1808100$
The number of $4432$ hands should be $$2\binom{7}{4}\binom{7}{4}\binom{6}{3}\binom{6}{2}+4\binom{7}{4}\binom{7}{2}\binom{6}{4}\binom{6}{3}+4\binom{7}{4}\binom{7}{3}\binom{6}{4}\binom{6}{2}+2\binom{7}{3}\binom{7}{2}\binom{6}{4}\binom{6}{4}$$
Note this number is $3050250$.
Thus, we should find that the probability of a $4333$ is $$\frac{1715000}{1715000+1808100+3050250}=\frac{700}{2683}\approx.26$$
The probability of a $5332$ is $$\frac{1808100}{1715000+1808100+3050250}=\frac{738}{2683}\approx .275$$
The probability of a $4432$ is $$\frac{3050250}{1715000+1808100+3050250}=\frac{1245}{2683}\approx .465$$
Thus, we find that it is most likely that the declarer has a $4432$ distribution.
