2007 AMC 10B Problem 22

P21-25~4 minPrint

Expected Value Basic Probability Set up the equation/formula and compute; no special trick needed

A player chooses one of the numbers $1$ through $4$ . After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered $1$ through $4.$ If the number chosen appears on the bottom of exactly one die after it has been rolled, then the player wins $1$ dollar. If the number chosen appears on the bottom of both of the dice, then the player wins $2$ dollars. If the number chosen does not appear on the bottom of either of the dice, the player loses $1$ dollar. What is the expected return to the player, in dollars, for one roll of the dice?

Show solution without answering

Key insight

Of the 16 equal rolls, your number shows on both dice in 1, on exactly one die in 6, and on neither in 9.

Solution

Which number the player picks does not matter; say it is $1$ . Two four-sided dice give $4 \times 4 = 16$ equally likely rolls. Sort them by how many dice show a $1$ .

1. Both dice show $1$ : only $(1,1)$ . That is $1$ roll. Win $\$2$ .
2. Exactly one die shows $1$ : the first is $1$ and the second is $2$ , $3$ or $4$ (3 rolls), or the second is $1$ and the first is $2$ , $3$ or $4$ (3 more). That is $6$ rolls. Win $\$1$ .
3. Neither die shows $1$ : each die has $3$ other faces, so $3 \times 3 = 9$ rolls. Lose $\$1$ .
4. Check the count: $1 + 6 + 9 = 16$ .
5. Imagine playing all $16$ rolls once each. Total money:
$1 \times 2 + 6 \times 1 + 9 \times (-1) = 2 + 6 - 9 = -1.$
6. That is $-1$ dollar over $16$ games, so the average return per game is
$-\frac{1}{16}.$

The answer is $\boxed{\textbf{(B)}\ -\frac{1}{16}}$ .

Why this works

Expected value is the average result per game: add up each payoff times how often it happens. With two dice, the $4 \times 4$ grid of rolls lists every case equally, so counting squares gives exact weights. The answer is very close to zero, so a single miscount (like $3$ instead of $6$ ) or a sign slip changes the choice.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Which number the player picks is irrelevant by symmetry; call it $k$ . Each die independently shows $k$ with probability $\frac14$ and something else with probability $\frac34$ . Out of the $16$ equally likely rolls:

both dice show $k$ : $1 \cdot 1 = 1$ roll, payoff $+2$ ;
exactly one shows $k$ : $1 \cdot 3 + 3 \cdot 1 = 6$ rolls, payoff $+1$ ;
neither shows $k$ : $3 \cdot 3 = 9$ rolls, payoff $-1$ .

The expected return is
$\frac{1}{16}(2) + \frac{6}{16}(1) + \frac{9}{16}(-1) = \frac{2 + 6 - 9}{16} = -\frac{1}{16}.$

The answer is $\boxed{\textbf{(B)}\ -\frac{1}{16}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 11: Expected Value
Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Intermediate Counting & Probability, Chapter 6: Constructive Expectation
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques