[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

(Suggested by Eisenberg and Ghosh) A deck of playing cards can be described as a Cartesian product

[[math]] \mbox{Deck} = \mbox{Suit} \times \mbox{Rank}\ , [[/math]]

where [math]\mbox{Suit} = \{\clubsuit,\diamondsuit,\heartsuit,\spadesuit\}[/math] and [math]\mbox{Rank} = \{2,3,\dots,10,{\mbox J},{\mbox Q},{\mbox K},{\mbox A}\}[/math]. This just means that every card may be thought of as an ordered pair like [math](\diamondsuit,2)[/math]. By a suit event we mean any event [math]A[/math] contained in Deck which is described in terms of Suit alone. For instance, if [math]A[/math] is “the suit is red,” then

[[math]] A = \{\diamondsuit,\heartsuit\} \times \mbox{Rank}\ , [[/math]]

so that [math]A[/math] consists of all cards of the form [math](\diamondsuit,r)[/math] or [math](\heartsuit,r)[/math] where [math]r[/math] is any rank. Similarly, a rank event is any event described in terms of rank alone.

• Show that if [math]A[/math] is any suit event and [math]B[/math] any rank event, then [math]A[/math] and [math]B[/math] are independent. (We can express this briefly by saying that suit and rank are independent.)
• Throw away the ace of spades. Show that now no nontrivial (i.e., neither empty nor the whole space) suit event [math]A[/math] is independent of any nontrivial rank event [math]B[/math]. Hint: Here independence comes down to
  [[math]] c/51 = (a/51) \cdot (b/51)\ , [[/math]]
  where [math]a[/math], [math]b[/math], [math]c[/math] are the respective sizes of [math]A[/math], [math]B[/math] and [math]A \cap B[/math]. It follows that 51 must divide [math]ab[/math], hence that 3 must divide one of [math]a[/math] and [math]b[/math], and 17 the other. But the possible sizes for suit and rank events preclude this.
• Show that the deck in (b) nevertheless does have pairs [math]A[/math], [math]B[/math] of nontrivial independent events. Hint: Find 2 events [math]A[/math] and [math]B[/math] of sizes 3 and 17, respectively, which intersect in a single point.
• Add a joker to a full deck. Show that now there is no pair [math]A[/math], [math]B[/math] of nontrivial independent events. Hint: See the hint in (b); 53 is prime.

\medskip The following problems are suggested by Stanley Gudder in his article “Do Good Hands Attract?” He says that event [math]A[/math] attracts event [math]B[/math] if [math]P(B|A) \gt P(B)[/math] and repels [math]B[/math] if [math]P(B|A) \lt P(B)[/math].

Notes