Q. 2.34 - A First Course in Probability

Question

The second Earl of Yarborough is reported to have bet at odds $1000$-$1$that a bridge hand of$13$ cards would contain at least one card that is ten or higher. (By ten or higher we mean that a card is either a ten, a jack, a queen, a king, or an ace.) Nowadays, we call a hand that has no cards higher than$9$ a Yarborough. What is the probability that a randomly selected bridge hand is a Yarborough?

Step-by-Step Solution

Answer

There is approximately$1$ in$1800$ or $0.0547\%$chance that a bridge hand does not contain any card that is $10$or higher.

Step 1 Given Information.

The second Earl of Yarborough is reported to have bet at odds$1000$- $1$that a bridge hand of $13$cards would contain at least one card that is ten or higher.

Step 2 Explanation.

If the dealing was fair, all$\left(\begin{array}{l}52\\ 13\end{array}\right)$hands are equally likely.

This is the number of combinations of $13$out of $52$different cards.

By the axioms of probability, the probability of an event $A$that happens in $n$ out of these$\left(\begin{array}{l}52\\ 13\end{array}\right)$hands is:

$P\left(A\right)=\frac{n}{\left(\begin{array}{l}52\\ 13\end{array}\right)}$

In this case:

There are$8$ lower in values in cards than $10$. Accounting for suits $8·4=32$cards from which to choose to create a hand that does not contain a $10$or higher valued card -$\left(\begin{array}{l}32\\ 13\end{array}\right)$possible hands.

$P\left(A\right)=\frac{\left(\begin{array}{l}32\\ 13\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)}=5.47033·{10}^{-4}$

Step 3 Explanation.

There is approximately $1$in$1800$ or $0.0547\%$chance that a bridge hand does not contain any card that is $10$or higher.