Grinstead-Snell-Probability · Problem 4.1.54

Conditional Probability

Mathematics · independence-deck-of-cards

Problem

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

\text{Deck} = \text{Suit} \times \text{Rank},

where

\text{Suit} = \{\clubsuit, \diamondsuit, \heartsuit, \spadesuit\}

and

\text{Rank} = \{2, 3, \ldots, 10, J, Q, K, A\}

. This just means that every card may be thought of as an ordered pair like

(\diamondsuit, 2)

. By a suit event we mean any event

A

contained in Deck which is described in terms of Suit alone. For instance, if

A

is ``the suit is red,'' then

A = \{\diamondsuit, \heartsuit\} \times \text{Rank},

so that

A

consists of all cards of the form

(\diamondsuit, r)

(\heartsuit, r)

where

r

is any rank. Similarly, a rank event is any event described in terms of rank alone. (a) Show that if

A

is any suit event and

B

any rank event, then

A

and

B

are independent. (We can express this briefly by saying that suit and rank are independent.) (b) Throw away the ace of spades. Show that now no nontrivial (i.e., neither empty nor the whole space) suit event

A

is independent of any nontrivial rank event

B

. Hint: Here independence comes down to

c/51 = (a/51) \cdot (b/51),

where

a

b

c

are the respective sizes of

A

B

and

A \cap B

. It follows that 51 must divide

ab

, hence that 3 must divide one of

a

and

b

, and 17 the other. But the possible sizes for suit and rank events preclude this. (c) Show that the deck in (b) nevertheless does have pairs

A

B

of nontrivial independent events. Hint: Find 2 events

A

and

B

of sizes 3 and 17, respectively, which intersect in a single point. (d) Add a joker to a full deck. Show that now there is no pair

A

B

of nontrivial independent events. Hint: See the hint in (b); 53 is prime.

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