First of all, I would like to underline that I am very bad in probabilities and that I am a fanatic bridge player.
Yesterday, after a tournement, we had a no-end discussion my partner and I; so the question.
Let me note $n_S,n_H,n_D,n_C$ the number of spades, hearts, diamonds and clubs in a hand.
Suppose that my hand is $n_S=4$, $n_H=4$, $n_D=3$, $n_C=2$.
What are the probabilities that my partner has in his/her hand
- $0<n_S \leq 3$
- $0<n_H \leq 3$
- $n_C \geq 6$
- $n_D < n_C$
- $n_D=13-(n_S+n_H+n_C) \qquad$ (by definition)
The probality of each possible case is important for me to know.
I hope and wish that this question will not be closed because missing context.
Of the 39 remaining cards, there are $39!$ possible hands your partner may have, each having equal probability.
Given a specific shape for your partner, we can also count the number of hands of that shape:
$$ \binom{11}{n_\clubsuit} \cdot \binom{10}{n_\diamondsuit} \cdot \binom{9}{n_\heartsuit} \cdot \binom{9}{n_\spadesuit}$$
because the binomial coefficient
$$\binom{11}{n_\clubsuit} = \frac{11!}{n_\clubsuit! (11 - n_\clubsuit)!}$$
counts the number of ways to choose $n_\clubsuit$ cards from the 11 remaining clubs.
There are enough cases that the simplest approach is to use a computer to simply exhaust over all possible shapes your partner may have.
Of all possible hands your partner may have, roughly 6.1% will meet your constraints. (that number jumps to 6.6% if you made a typo and your partner is allowed to have voids in the major suits)
Among those hands, the most common are:
Of particular note, roughly 44% of the hands of the shape you describe will be semibalanced.
I've posted below the total list of data I've computed. The format is:
where the two different kinds of odds are the proportion assuming your partner's hand satisfies your constraints, and the proportion among all hands.