Calculating odds of someone holding x clubs in a 4-player game given that he holds the club Ace, at least 4 clubs, and you hold 3 clubs

Question

In the context of a Bridge game: I have 3 clubs in my hand and I want to find how many clubs my partner has. Before any card has been played (i.e. from the bidding alone), I know he holds at least 4 clubs, one of which is the club Ace. How do I find the odds of my partner holding, say, x=4 clubs in this case? The value of x doesn't matter but I would like to learn the approach to solving problems like this.

When I came across this question I realised I had a few sub-questions:

1. How does one calculate the odds of the 10 clubs being distributed among 3 players, e.g. such that my partner has 4 clubs? (i.e. the concept behind tackling such distribution questions)

2. How does the fact that I know my partner has one particular card affect my calculation? For instance, can I simply ignore that card so that I only need to find how 9 clubs can be distributed among 3 people such that partner has 3 clubs?

edit: clarity in response to @joriki's post

Answer 1

For $4\le x\le 10$,$\;$let $p(x)$ be the probability that your partner has exactly $x$ clubs, given that

• You know your partner has at least $4$ clubs including the Ace of clubs.$\\[4pt]$
• You have exactly $3$ clubs.

Then $p(x)$ is given by $$ {\small{ {\large{p(x)}} = \frac { {\displaystyle{\binom{9}{x-1}\binom{29}{13-x}}} } { {\displaystyle{ \sum_{k=4}^{10} \binom{9}{k-1}\binom{29}{13-k} }} } }} $$ Explanation:

Excluding the Ace of clubs and the $3$ clubs in your hand, there are $9$ remaining clubs.

Excluding the $10$ non-clubs in your hand, there are $29$ remaining non-clubs.

Hence for $4\le k\le 10$, there are $$\binom{9}{k-1}\binom{29}{13-k}$$ ways to distribute $k-1$ of the remaining clubs and $13-k$ of the remaining non-clubs so that your partner has exactly $k$ clubs (including the Ace of clubs).

Answer 2

I gather from point $1$ (and the term “partner”) that you’re assuming a Bridge-like situation where the deck has $52$ cards and all of them are distributed among $4$ players.

With this kind of question, it’s important to be clear about how you gained the knowledge that you have; the probability distribution depends on this.

The answer that @quasi gives assumes that you’re asking about the conditional probability that your partner has $x$ clubs, conditional on your partner having $4$ clubs and having the ace of clubs. This is the appropriate concept e.g. if you gained your knowledge by asking whether these conditions hold, i.e., you asked your partner, “Do you have at least four clubs?” and “Do you have the ace of clubs?”, and she answered yes to both questions. In that case, all hands that are compatible with these conditions are equally likely, and the resulting probability is the conditional probability given those conditions.

But it doesn’t follow from your question that this is how you gained your knowledge. For instance, you could have asked your partner whether she has at least four clubs and then asked her to tell you the rank of a random one of those clubs, and she told you that one of them is the ace of clubs. In this case, that second piece of information doesn’t change the probabilities, which would just be the conditional probabilities given that she has $4$ clubs:

$$ p(x)=\frac{\binom{10}x\binom{29}{13-x}}{\sum_{k=4}^{10}\binom{10}k\binom{29}{13-k}}\;. $$

There are of course more complicated ways to gain the knowledge that you have. For instance, in Bridge you might have deduced that knowledge from your partner’s play. Then the appropriate probability distribution may depend on the details of how you came to this conclusion. For example: If your partner made a bid that she would only make if she had four clubs including the ace of clubs, the conditional probability given these facts would be appropriate. But suppose she made a bid that she would only make if she had $4$ clubs, and then she led with the ace of clubs. What that tells you would depend on the details of which hands would cause her to lead with the ace of clubs. If she would definitely have led with the ace of clubs if she had it, then that would be like answering a question whether she has it, and the conditional probability in @quasi’s answer would be appropriate. But if (this is probably unrealistic, just as an example) she could equally well lead with any club and she just happened to lead with the ace, then the other conditional distribution above would be appropriate (since leading with a random one of her clubs is like telling you the rank of a random one of her clubs). The actual knowledge to be gained from her leading with the ace of clubs in a real game is likely more complicated than either of these two extremes.