Exercise 1.12

Question

Four players, named A, B, C, and D, are playing a card game. A standard, well-shuffled
deck of cards is dealt to the players (so each player receives a 13-card hand).
\begin{enumerate}[label=(\alph*)]
\item How many possibilities are there for the hand that player A will get? (Within a
hand, the order in which cards were received doesn’t matter.)
\item How many possibilities are there overall for what hands everyone will get, assuming
that it matters which player gets which hand, but not the order of cards within a hand?
\item Explain intuitively why the answer to Part (b) is not the fourth power of the answer
to Part (a).
\end{enumerate}

fifthist x · Accepted Answer

\begin{enumerate}[label=(\alph*)]
\item $${52 \choose 13}$$

\item The number of ways to break $52$ cards into $4$ groups of size $13$ is 
$$\frac{{52 \choose 13}{39 \choose 13}{26 \choose 13}{13 \choose 13}}{4!}.$$
  
  The reason for division by $4!$ is that all permutations of specific $4$ 
  groups describe the same way to group $52$ cards.
  
  Since we do care about the order of the $4$ groups, we should not divide by 
  ${4!}$. The final answer then is 
  $${52 \choose 13}{39 \choose 13}{26 \choose 13}{13 \choose 13}$$

\item The key is to notice that the sampling is done without replacement. 
${52 \choose 13}^{4}$ assumes that all four players have ${52 \choose 13}$ 
choices of hands available to them. This would be true if sampling was done 
with replacement.
\end{enumerate}

Exercise 1.12

Answers

Comments

Exercise 1.12

Answers

Comments

Add answer