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Probability Level 4

Tim, Jim and Slim play another card game. This time they each have a deck of 15 colored cards. Each deck has five red, five blue and five green cards. The boys all shuffle their decks so that the cards are randomly ordered.

There are 15 rounds, in which they all reveal the color of the top card in their respective decks. (i.e. In round 1 the reveal the first card, in round 2 the second card etc. )

A three-way match is (as you might expect) a round in which all three turn up the same color card.

Let $E$ be the expected number of three-way matches. Which of the following is true?

Inspiration

Part of a collection on Matching Cards

0< E \le 0.5

0.5 <E \le 1

1< E \le 1.5

1.5 < E \le 2

2 < E \le 2.5

2.5 < E \le 3

E>3

1 solution

Varsha Dani
Oct 24, 2018

Let $X_i$ be the indicator random variable for the $i$ th round having a 3-way match. That is, $X_i = 1$ if the $i$ th round has a 3-way match and $X_i = 0$ otherwise.

Since the decks are shuffled, the $i$ th card in each deck is random. Also, the probability of any card from any of the three decks being a particular color is 1/3.

$\begin{aligned} \mathbb{E}[X_i] = \mathrm{Prob}(X_i =1) &= \mathrm{Prob}(\mbox{red match on round } i) + \mathrm{Prob}(\mbox{blue match on round } i) + \mathrm{Prob}(\mbox{green match on round } i)\\ &= 3 \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \\ &=\frac{1}{9} \end{aligned}$

Let $X$ be the total number of 3-way matches. Then $X = \sum_{i=1}^{15} X_i$ .

By Linearity of Expectation we have $E = \mathbb{E}[X] = \sum_{i=1}^{15} \mathbb{E}[ X_i] = \sum_{i=1}^{15} \frac{1}{9} = \frac{15}{9} \approx 1.667$

Thus $\fbox{ \$1.5 < E \le 2\$}$ .

Note: The random variables $X_i$ are not independent, so it is fortunate that they do not need to be. Linearity of Expectation applies to any linear combination of random variables.

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