Probabilities playing bridge

Question

Being very bad with probabilities, I would greatly appreciate help for the following problem.

The first bid of my partner indicates that, among his $13$ cards, he handles $6$ hearts and a maximum of $3$ spades.

In my hand, I handle $0$ hearts and $6$ spades.

So, my question is : what are the probabilities corresponding to $0$, $1$, $2$ or $3$ spades in my partner's hand ?

Should I need to precise that this is not homework ?

Thnaks for your help.

Answer 1

There are 19 diamonds/clubs and 7 spades available, from which he is dealt 7 cards.

So number of ways of getting $k$ spades is ${7\choose k}{19\choose7-k}$. That gives 50388,189924,244188,135660. Hence the probabilities are 0 spades 8.125%, 1 spade 30.625%, 2 spades 39.375%, 3 spades 21.875%.

Answer 2

First of all, the probability will depend on the numbers of jacks, queens, kings and aces in your hand, since you know how many points your partner has.

But let's forget that extra piece of information, then your hand consists of $6$ high cards, denote that by $H$, and $7$ low cards, $L$. Then we want to find out $P(H_p = i)$ for $i =6,7,8,9$, given that you have $H_m =6$.
So we first find the number of ways to obtain $H_p =i$.
We have $6$ hearts out of a possible $13$ hearts.
We have $i-6$ spades out of a possible $7$ spades.
We have $13-i$ low cards out of a possible $19$ low cards.
So the number of combination wihich gives $H_p=i$ is $$\binom{13}{6}\binom{7}{i-6} \binom{19}{13-i} .$$ As we know that $H_p$ lies between $6$ and $9$, we have $$ P(H_p = i) = \frac{ \binom{13}{6} \binom{7}{i-6} \binom{19}{13-i}}{ \sum_{i=6}^9 \binom{13}{6} \binom{7}{i-6} \binom{19}{13-i}}.$$

This gives the probability on $n$ spades as $\frac{13}{160}$ for $0$, $\frac{49}{160}$ for $1$, $\frac{63}{160}$ for $2$ and $\frac{35}{160}$ for 3.
So let your partner know that you have quite a lot of spades, as in most cases you will have a fit in spades.