AlmostSurely
Problem 4.11 Easy Free 3 variations

Marked Cards Before the Joker

A deck of 1414 cards consists of one card of each rank from Ace to King together with a Joker, shuffled into a uniformly random order. You turn the cards face up one at a time; you win $100 if the King, Queen and Jack have all appeared before the Joker does, but if the Joker appears before all three of them have shown up, the game ends immediately and you win nothing. What is the fair price to pay for this game (its expected payout)?

Work it out, then compare with the answer.

SolutionShow the worked solution

Only four cards matter: the King, Queen, Jack and Joker. The other ten cards do not affect the outcome. In a uniformly random shuffle, the relative order of these four cards is uniform over all 4!4! orders.

You win exactly when the Joker comes after all three of the other cards, that is, when the Joker is last among the four. By symmetry each of the four cards is equally likely to be last, so

P(win)=14.\mathbb{P}(\text{win})=\frac14 .

The payout is $100\$100 with probability 14\tfrac14 and 00 otherwise. The expected payout is therefore 100⋅14=25100\cdot\tfrac14=25 dollars, and this is the fair price.

$25\boxed{\$25}

Key idea. Restrict attention to the relevant cards: their relative order is uniformly random, whatever the rest of the deck does.

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 4.11a

A standard 52-card deck is shuffled uniformly at random and dealt one card at a time until the two of hearts is revealed. Find the probability that none of the twelve face cards (kings, queens, jacks) appear before the two of hearts.

Work it out, then compare with the answer.

Solution

The relevant cards are the twelve face cards and the two of hearts, thirteen in all. Their relative order is uniform.

The event happens exactly when the two of hearts comes first among these thirteen cards. Each of the thirteen is equally likely to be first.

113\boxed{\tfrac{1}{13}}
Variation 4.11b

A standard 52-card deck is shuffled uniformly at random and dealt one card at a time until the two of hearts is revealed. Find the probability that exactly one king, exactly one queen, and exactly one jack appear before the two of hearts.

Work it out, then compare with the answer.

Solution

Again restrict attention to the thirteen relevant cards: the twelve face cards and the two of hearts. Their relative order is uniform.

The event requires two things:

  • The two of hearts is fourth among these cards. This has probability 113\tfrac1{13}.
  • The three face cards before it are one king, one queen and one jack. Given the first condition, these three form a uniform 33-subset of the twelve face cards. There are 43=644^3=64 subsets with one card of each rank, out of (123)=220\binom{12}{3}=220.

Hence

P=113⋅64220=16715≈0.0224.\mathbb{P}=\frac1{13}\cdot\frac{64}{220}=\frac{16}{715}\approx 0.0224 .
16715\boxed{\tfrac{16}{715}}
Variation 4.11c

A playlist of 147147 songs contains 44 songs by one composer and 66 songs by another composer (the remaining 137137 songs are by other composers). The playlist is put into a uniformly random order. Find the probability that all 66 songs by the second composer occur before the second song by the first composer does.

Work it out, then compare with the answer.

Solution

Only the 1010 songs by the two composers matter. The set of positions occupied by the first composer’s 44 songs, among these 1010 relative positions, is uniform over all (104)=210\binom{10}{4}=210 subsets.

All six songs by the second composer precede the second song by the first composer exactly when at most one first-composer song comes before the last second-composer song. This means relative positions 8,9,108,9,10 are all by the first composer. The fourth song by the first composer can then be in any of positions 1,…,71,\dots,7, giving 77 favourable subsets. Hence

P=7210=130.\mathbb{P}=\frac{7}{210}=\frac1{30}.
130\boxed{\tfrac{1}{30}}
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