How to calculate the probability of at least one "yarborough"

Question

Could I please ask for help with the following:

In a card game for four players, a pack of fifty-two cards is dealt round so that each player receives thirteen cards. A hand that contains no card greater than nine is called a yarborough. How many deals are necessary for the probability of at least one hand to be a yarborough to be greater than 1/2 (Ace ranks high).

The answer given in the book is 1267

Here's my initial reasoning:

If only one hand of 13 cards was being dealt from the pack, then the probability of a yarborough could be calculated as follows:

The sample space has 52 * 51 * … * 40 equally likely outcomes.

The number of cards having a values 9 or lower (remember ace ranks high) is 8 * 4 = 32 and so there are 32 * 31 * … * 20 outcomes for which there is no card higher than a nine among the thirteen cards dealt.

Thus, the probability of a yarborough is (32 * 31 * … * 20) / ( 52 * 51 * … * 40 ) = 0.00054703

Ok, so a yarborough is very unlikely.

So the probability in the above scenario of NOT getting a yarborough is 1 - 0.00054703 = 0.99945297

(I feel happy with the above, but from here on in I'm not so happy with my reasoning):

But in one "deal", four hands are dealt.

So could I say that the probability that every one of the four players does NOT have a yarborough is 0.99945297^4 ? This feels wrong to me as the above argument was for one hand being dealt, and probabilities on subsequent hands are affected by the result of previous hands.

Well, if I could say this then the probability that at least one yarborough is dealt would be 1 - 0.99945297^4 = 0.002186325. So very unlikely that at least one yarborough is dealt.

Then to find the number of "deals" required for the probability of at least one yarborough being dealt to be equal to 0.5 I would have to solve:

(0.99945297^4)^n = 0.5 giving n = 316.691 and so n = 317 deals.

Which is not the given answer. As I say, I feel my reasoning is faulty. Could any one please offer a correct method?

Answer 1

Let $A$ denote the probability that on one dealing of $(13)$ cards to $(4)$ players, none of the $(4)$ players has a yarborough. This implies that $0 < A < 1$. This implies that $\ln(A) < 0.$

Then, the probability of no yarborough to anyone, in the first $n$ deals is $A^n$.

So, once $A$ is calculated, you will be looking for the smallest positive integer $n$ such that $A^n < \dfrac{1}{2}$.

This is equivalent to solving for $n$ such that

$\displaystyle e^{n \times \ln(A)} < e^{-\ln(2)}.$

Therefore, keeping in mind that $n \times \ln(A)$ will be a negative number, as will $-\ln(2)$,

you want $~~~n \ln(A) < -\ln(2) \implies n > \dfrac{-\ln(2)}{\ln(A)}.$

So, you want $n$ to be the smallest positive integer that is greater than $\dfrac{-\ln(2)}{\ln(A)}.$

Therefore, the problem has been reduced to computing $A$.

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Unfortunately, the computation of $A$ is a job for Inclusion-Exclusion.

See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

Let $B = 1 - A$.

So, $B$ represents the probability that on a single deal, at least one of the $4$ players was dealt a yarborough.

I am going to compute $B$ combinatorically, as

$$\frac{N\text{(umerator)}}{D\text{(enominator)}}.$$

Here, $~\displaystyle D = \binom{52}{13} \times \binom{39}{13} \times \binom{26}{13} \times \binom{13}{13}.$

So, $~\displaystyle D = \frac{(52)!}{[(13)!]^4}.$

Therefore, the problem has been reduced to computing $N$.

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Using the syntax in the Inclusion-Exclusion answer that I linked to:

For any set $E$ with a finite number of elements, let $|E|$ denote the number of elements in the set $E$.

Let $S$ denote the set of all possible ways of distributing $(13)$ card each to $(4)$ players. That is, $|S| = D.$

Let $S_k$ denote the subset of $S$ that has player $k$ receiving a yarborough. Here $k \in \{1,2,3,4\}$, since there are $4$ players.

Then, $N = |S_1 \cup S_2 \cup S_3 \cup S_4|$.

Let $T_1 = |S_1| + |S_2| + |S_3| + |S_4|.$

Let $T_2 = |S_1 \cap S_2| + |S_1 \cap S_3| + |S_1 \cap S_4|$
$+ |S_2 \cap S_3| + |S_2 \cap S_4| + |S_3 \cap S_4|$.

Let $T_3 = |S_1 \cap S_2 \cap S_3| + |S_1 \cap S_2 \cap S_4|$
$+ |S_1 \cap S_3 \cap S_4| + |S_2 \cap S_3 \cap S_4|$.

Let $T_4 = |S_1 \cap S_2 \cap S_3\cap S_4|$.

Then, in accordance with Inclusion-Exclusion, you have that

$\displaystyle N = |S_1 \cup S_2 \cup S_3 \cup S_4| = T_1 - T_2 + T_3 - T_4.$

Therefore, the problem has been reduced to the computation of $T_1, T_2, T_3, T_4$.

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$\underline{\text{Computation of} ~T_1}$
The number of elements in the set $S_1$ equals the number of ways of Player-1 receiving no card higher than a $9$, while the other $(3)$ players can then receive any of the remaining $(39)$ cards. This means that Player-1 received $(13)$ cards from the reduced deck of $(32)$ cards, since it is presumed that Player-1 did not receive any of the $(20)$ cards, $(10)$ or higher.

The number of such possible distributions is

$~\displaystyle |S_1| = \binom{32}{13} \times \binom{39}{13} \times \binom{26}{13} \times \binom{13}{13}$

$\displaystyle =~ \frac{[(32)!] \times [(39)!]}{[(19)!] \times [(13)!]^4}.$

By symmetry, $|S_1| = |S_2| = |S_3| = |S_4|.$

Therefore,

$$T_1 = 4 \times \frac{[(32)!] \times [(39)!]}{[(19)!] \times [(13)!]^4}.$$

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$\underline{\text{Computation of} ~T_2}$

Calculating $|S_1 \cap S_2|$ is very similar to the computation of $|S_1|$ in the previous section. That is, in order for both Player-1 and Player-2 to receive a yarborough, first Player-1 must receive a yarborough from the reduced deck of $32$ cards. Then, there will be $(32 - 13) = 19$ cards left in the reduced deck, from which Player-2 receives his cards. Then, Player-3 and Player-4 can each receive any of the $(13)$ cards left from the $(26)$ cards in the deck.

Therefore,
$\displaystyle |S_1 \cap S_2| = \binom{32}{13} \times \binom{19}{13} \times \binom{26}{13} \times \binom{13}{13}$

$\displaystyle =~ \frac{[(32)!] \times [(26)!]}{[(6)!] \times [(13)!]^4}.$

Again, by symmetry,

$$T_2 = 6 \times \frac{[(32)!] \times [(26)!]}{[(6)!] \times [(13)!]^4}.$$

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$\underline{\text{Computation of} ~T_3}$
Finally, I come to the easy part.
In order for players $1,2,$ and $3$ to each receive a yarborough, they would each have to receive $13$ cards from the reduced deck of $32$ cards.

This is impossible, since $(3 \times 13) > 32.$

Therefore, $|S_1 \cap S_2 \cap S_3| = 0.$

Similarly, by symmetry,

$$T_3 = 0.$$

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$\underline{\text{Computation of} ~T_4}$
The same analysis of the previous section applies here.

$$ T_4 = 0.$$

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Final Computation(s)

$$T_1 = 4 \times \frac{[(32)!] \times [(39)!]}{[(19)!] \times [(13)!]^4}.$$

$$T_2 = 6 \times \frac{[(32)!] \times [(26)!]}{[(6)!] \times [(13)!]^4}.$$

$$N = T_1 - T_2.$$

$$D = \frac{(52)!}{[(13)!]^4}.$$

$$B = \frac{N}{D}.$$

$$A = 1 - B.$$

$$n = ~\text{the smallest positive integer}~ > \frac{-\ln(2)}{-\ln(A)}.$$

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Addendum
Responding to the comment of Michael Seifert, which follows this answer.

I agree. In fact, approximating $N$ by $T_1$ greatly simplifies the computations and analysis, since it allows you to avoid any consideration of Inclusion-Exclusion.

Also, with $N$ approximated by $T_1$, the computation of $~\displaystyle B = \frac{N}{D}$ becomes much simpler.

Answer 2

Here is a computation of the probability by inclusion-exclusion.

The probability that some player has a Yarborough (deliberately over-counting the cases where more than one player has a Yarborough--we will compensate later) is $$\binom{4}{1} \frac{\binom{32}{13}}{\binom{52}{13}}$$ The probability that two players have a Yarborough is $$\binom{4}{2} \frac{\binom{32}{26}}{\binom{52}{26}}$$ It is not possible for three or more players to have a Yarborough. So, by inclusion-exclusion, the probability that at least one player has a Yarborough is $$\binom{4}{1} \frac{\binom{32}{13}}{\binom{52}{13}} - \binom{4}{2} \frac{\binom{32}{26}}{\binom{52}{26}} = \boxed{0.00218812}$$