If I deal 7 cards what is the probability that there is at least one card from each suit.

Question

I thought that this would be an easy problem, and it seems really obvious to me that the answer would be:

$$ \frac{\binom{13}{1}^4 \binom{48}{3}}{\binom{52}{7}} $$

but for some reason this comes out to be around $3.69$ which is not a valid probability.

So then I thought of a different approach where instead of $\binom{48}{3}$ I used $48 * 47 * 46$ but this is obviously even larger.

I'm so confused why this approach doesn't work because I've done similar problems before in this manner. I'm assuming there has to be something that I am missing.

I then did some searching on here and came to an answer I'm confident in of $0.5696$ using Inclusion-Exclusion and counting the bad hands so my solution looked like:

$$ 1 - \frac{4\binom{39}{7} - 3!\binom{26}{7} + 4\binom{13}{7}}{\binom{52}{7}} = 0.5696...$$

I understand how the Inclusion-Exclusion Principle works but I'm wondering what events we're unioning here that it's taken into account. I'm also wondering if there is a more straightforward way of calculating the number of good hands where there is at least 1 card from each suit out of the 7 dealt.

Answer 1

You might have got confused as the binomial coefficients $\binom41$ etc weren't explicitly mentioned, and in fact $\binom42$ was simplified as factorial $3$ which would make confusion worse confounded

Using the symmetry in the deck, the desired count of "good" hands is better written as

$$ \binom{52}7 - \binom41\binom{39}7 +\binom42\binom{26}7 -\binom43\binom{13}7$$

And, of course, as probability has been asked for, you can write it as

$$ 1 - \dfrac{\binom41\binom{39}7 -\binom42\binom{26}7 +\binom43\binom{13}7}{\binom{52}7}$$

Answer 2

Let $A$ be the set of all $7$-subsets of the deck.

Let $A_i$ be the set of all elements of $A$ that have no card of suit $i.$

Then the inclusion-exclusion is computing:

$$\left|A\setminus(A_1\cup A_2\cup A_3\cup A_4)\right|$$

The key is that the size of $A_i$ are all the same, and the sizes of $A_i\cap A_j$ are all the same when $i\neq j.$

So inclusion-exclusion gives:

$$\binom{52}7-\binom{4}1\binom{39}{7}+\binom{4}2\binom{26}7-\binom43\binom{13}{7}$$

We could, in theory, add $\binom44\binom07,$ but that term is zero, so we just drop it.

That's because $|A_i|=\binom{39}7$ for any $i,$ $|A_{i}\cap A_j|=\binom{26}7$ for each pair $\{i,j\},$ etc.

So this wouldn't work if suits consisted of a different number of cards. You'd have to compute the different sizes for each $A_i,$ each $A_i\cap A_j,$ and each $A_i\cap A_j\cap A_k.$ $A_1\cap A_2\cap A_3\cap A_4$ will always be empty.

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Another way to write this out is to use a generating function.

Let:

$$F(x)=((1+x)^{13}-1)^4=\sum a_ix^i$$

And $a_i$ is the number of ways to deal $i$ cards and get at least one of each suit.

Writing: $$((1+x)^{13}-1)^4=(1+x)^{52}-\binom41(1+x)^{39}+\binom42(1+x)^{26}-\binom43(1+x)^{13}+1$$

We can compute the coefficient of $x^7$ in each term.

This approach might make it easier to compute, say, the number of ways to select $11$ cards where you get at least two cards per suit, giving that we'd want the coefficient of $x^{11}$ in $$((1+x)^{13}-(1+13x))^4$$

This will give a complex formula with $15$ terms. It can be done with inclusion-exclusion, too, but this will be a little cleaner.

Writing $$((1+x)^{13}-(1+13x))^4=\sum_{i=0}^4 \binom{4}i(-1)^i(1+x)^{13(4-i)}(1+13x)^{i}$$

you get the coefficient of $x^n$ is:

$$\sum_{i=0}^4 (-1)^{i}\binom4i\sum_{j=0}^i13^j\binom ij\binom{13(4-i)}{n-j}$$

Answer 3

For a direct count of qualifying hands, consider the possible distribution of suits.

Four cards of one suit and one card from each of the other three suits: $$4\times\binom{13}{4}\times13^3=6,283,420$$

Three cards of one suit, two cards of a second suit, and one card from each of the other two suits: $$4\times\binom{13}{3}\times3\times\binom{13}{2}\times13^2=45,240,624$$

Two cards from each of three suits and one card of the fourth suit: $$\binom{4}{3}\times\binom{13}{2}^3\times13=24,676,704$$

This gives a total of $76,200,748$ qualifying hands and a probability of $0.569578$