In bridge there are 4 players; each get dealt 13 cards. In my hand I have 7 clubs, in partners hand they have 2. One of the opponents opens 1 No Trump which means they have at least 2 cards in every suit. Therefore they have 2, 3 or 4 clubs. What is the probability that they have each of these?

Use combinations. There's only one way they can have 4 of 4 clubs, disregarding the 9 random cards we don't care about. How many ways can they get 3 of 4 clubs? How many ways can they get 2 of 4 clubs? Add all those up and you have your denominator.

To follow up on NewOldGuy's post above, clearly you can consider cards "clubs" or "nonclubs". Of course, the original deck has 13 clubs and 39 nonclubs.

You and your partner combined have a certain number of clubs and a certain number of nonclubs.

Essentially, your opponent's cards are then dealt from the "deck stub" created by removing you and your partner's combined cards from the original deck.

In probability terms, the hypergeometric distribution describes the probabilities of obtaining X "successes" and Y "failures" when drawing X+Y items without replacement out of an urn with a total of S successes and F failures. In our case, of course, successes are clubs and failures are nonclubs.

I will put into a spoiler the probabilities that I obtained and how I derived them.

In many real cases, a 1NT opening bid gives you more information than just the absence of voids/singletons. For instance, it gives you also information about high card distribution. Say that your line holds 24 HCP. If your club suit is missing some high card (say you are missing the King and three low cards) and the 1NT opening is in the 15-17 HCP range, you know that the opener must hold the King. So, you infer that also a xxx/K or xx/Kx splits are impossible, besides all the 0-4 and 1-3 splits. In the same way whosenext showed, you should find the probability of those split happening a priori, remove any impossible split and renormalize to get the correct probabilities.

Thanks, I actually got 58% using an approximation based on a priori bridge suit break probabilities (0.41/0.41+0.3) but intuitively I felt that answer was wrong because I was thinking about it incorrectly.

Originally I was thinking that the 1N opener gets dealt 2 clubs, and now there are 2 more to distribute so will break 1-1 most of the time and that the other opponent would only get dealt 2 clubs roughly over a quarter of the time.

But that's the wrong approach; the 1N bidder gets dealt 13 cards, and we exclude the ones in which he doesn't have 2 clubs to get the correct conditional probability, as you showed.

The hand in question was _ AKxxxx _ ATxxxxx. Auction goes P 1N ? at unfavourable, was trying to work out the EV of slam in clubs if partner has 2 clubs.

I thought more about this problem last night. I wish to amend one of my previous statements and the result I posted.

I previously said that the number of cards held by hero and partner in each of the nonclub suits would not affect the probabilities of the no-trump bidding opponent holding a specific number of clubs. After further review, I'd like to back off from that statement.

Clearly, the number of cards in each suit is fixed (at 13). If hero and partner have X cards in a suit, then the two opponents combined necessarily have 13-X. So any imbalance in cards in the nonclub suits available to no-trump bidder can skew the chances that she has a specific number of clubs (skewed relative to ignoring the nonclub suit distribution) since she has exactly 13 cards in her hand.

I just ran a simple brute-force computer program to tally the probs in one specific case. A formula can probably be found that applies to this or any other case.

Suppose that hero+partner are known to have exactly 5 spades, 7 hearts, 5 diamonds, and 9 clubs and that a specific opponent (the no-trump bidder) is known to have at least two cards of each suit. Then, assuming no other information, I find the tally of the number of clubs this specific opponent has is:

2 clubs: 3,817,632 (.6041)
3 clubs: 2,144,240 (.3393)
4 clubs:__ 357,504 (.0566)

These probabilities are slightly different from the ones presented in the spoiler of post #3 above which ignored the distribution of nonclub suits.

The probabilities derived in post #3 are presumably the correct overall probabilities if all we are told is that hero+partner have exactly 9 clubs.

I am now believing that, under any restrictions to the distribution of suits in villain's hand (e.g., the no-trump bid implying that she has at least two cards of every suit), then the distribution of suits in hero+partner's hands can and does affect these probabilities.

Thoughts?

I am not sure a simple formula exists for these probabilities in the "restricted nonclub suit distribution" case.

Let S = the number of spades held collectively by hero and partner.
Let H = the number of hearts held collectively by hero and partner.
Let D = the number of diamonds held collectively by hero and partner.
Let C = the number of clubs held collectively by hero and partner.

Let S13 = 13-S be the number of spades collectively held by the two opponents.
Let H13 = 13-H be the number of hearts collectively held by the two opponents.
Let D13 = 13-D be the number of diamonds collectively held by the two opponents.
Let C13 = 13-C be the number of clubs collectively held by the two opponents.

Let P(X,Y,Z) be the set of ordered partitions of Y integers that add up to X where each of the Y integers is greater than or equal to Z.

Then the number of hands no-trump bidder can have with exactly 2 clubs is given by:
Sum over {[a,b,c] element of P(11,3,2)} C(S13,a)*C(H13,b)*C(D13,c)*C(C13,2)

Then the number of hands no-trump bidder can have with exactly 3 clubs is given by:
Sum over {[a,b,c] element of P(10,3,2)} C(S13,a)*C(H13,b)*C(D13,c)*C(C13,3)

Then the number of hands no-trump bidder can have with exactly 4 clubs is given by:
Sum over {[a,b,c] element of P(9,3,2)} C(S13,a)*C(H13,b)*C(D13,c)*C(C13,4)

Of course, the probabilities are then the respective tallies over the sum of these tallies.

Great post whosenext.

It has to be considered that 1NT opening (aside from high cards) doesn't just tell that the guy has at least 2 cards in each suit. A 1NT opening is very strict and represents just the following shapes (number of cards in each suit):

4-3-3-3
4-4-3-2
5-3-3-2 (provided that the long suit is either clubs or diamonds).

So it's true that you must have at least 2 cards in each suit, but you cannot have two suits with just two cards. A 7-2-2-2 shape, as well as 6-3-2-2 or 5-4-2-2 and others are not a 1NT opening.

In your formulas above, one should restrict the P(x,y,z) in such a way to consider only the above shapes.