Q.54 - A First Course in Probability

Question

Compute the probability that a bridge hand is void in at least one suit. Note that the answer is not 

$\frac{\left(\begin{array}{l}4\\ 1\end{array}\right)\left(\begin{array}{l}39\\ 13\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)}$

Step-by-Step Solution

Answer

The probability that a bridge hand is void in at least one suit is $5.1\%$

Step 1: Given Information

Define four events, for $i=1,2,3,4$ :

$E_i=$ event that $i$-th color is absent from the $13$ cards.

Step 2: Explanation

$E_i=$ event that $i$-th color is absent from the 13 cards

These events have equal probabilities, and the wanted probability is

$P\left[\left(E_1\cup E_2\cup E_3\cup E_4\right)\right]$

Remember the Proposition 4.4. (the law of inclusion and exclusion):

Let $A_1,A_2,\dots ,A_n$ be some events.

$$\begin{gathered} P\left(A_1\cup A_2,\dots A_n\right)=\sum _{i\in \{1,2,\dots ,n\}}P\left(A_i\right)+(-1{)}^1\sum _{i_1<i_2}P\left(A_{i_1}\cap A_{i_2}\right) \\ +\dots +(-1{)}^{n-1}\sum _{i_1<i_2<\dots <i_n}P\left(A_{i_1}\cap A_{i_2}\dots \cap A_{i_n}\right) \\ i_1,i_2,\dots i_n\in \{1,2,\dots ,n\} \end{gathered}$$

Step 3: Explanation

The probabilities that occur in the inclusion and exclusion equations are now calculated. The probability of any two or three events intersecting is the same since the events are symmetrical. There are no hands that are devoid of all four hues.

$$\begin{gathered} P\left(E_i\right)=\frac{\left(\begin{array}{l}39\\ 13\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)} \\ P\left(E_{i_1}\cap E_{i_2}\right)=\frac{\left(\begin{array}{c}26\\ 13\end{array}\right)}{\left(\begin{array}{c}52\\ 13\end{array}\right)} \\ P\left(E_{i_1}\cap E_{i_2}\cap E_{i_3}\right)=\frac{\left(\begin{array}{l}13\\ 13\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)} \\ P\left(E_1\cap E_2\cap E_3\cap E_4\right)=0 \\ {i}_1,i_2,i_3\in \{1,2,3,4\} \end{gathered}$$

Step 4: Explanation

And $k$ indexes out of $n$ (which is the same as $k$ ordered indexes) can be any of the $\left(\begin{array}{l}n\\ k\end{array}\right)$combinations.

$$\begin{gathered} P\left(E_1\cup E_2\cup E_3\cup E_4\right)=\left(\begin{array}{l}4\\ 1\end{array}\right)P\left(E_1\right)+(-1{)}^1\left(\begin{array}{l}4\\ 2\end{array}\right)P\left(E_1\cap E_2\right)+(-1{)}^2\left(\begin{array}{l}4\\ 3\end{array}\right)P\left(E_1\cap E_2\cap E_3\right)+(-1{)}^3\left(\begin{array}{l}4\\ 4\end{array}\right)P\left(E_1\cap E_2\cap E_3\cap E_4\right) \\ =4P\left(E_1\right)-6P\left(E_1\cap E_2\right)+4P\left(E_1\cap E_2\cap E_3\right)-P\left(E_1\cap E_2\cap E_3\cap E_4\right) \\ =4\frac{\left(\begin{array}{l}39\\ 13\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)}-6\frac{\left(\begin{array}{l}26\\ 13\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)}+4\frac{1}{\left(\begin{array}{l}52\\ 13\end{array}\right)}-0 \\ \approx 0.051 \end{gathered}$$