2021 AMC Fall 10B Problem 20

P16-20~8 minPrint

Conditional Probability & States Basic Probability Split into exhaustive cases and handle each Exploit symmetry to reduce work or pair up objects

In a particular game, each of $4$ players rolls a standard $6{ }$ -sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo's first roll was a $5,$ given that he won the game?

Show solution without answering

Key insight

Hugo wins with chance 1/4 by fairness. After rolling 5 he wins with chance 41/96, splitting ties with other 5s evenly. Then divide.

Solution

"Given that he won" means: out of all the games Hugo wins, what fraction started with a $5$ ? So compute
$\frac{\text{chance Hugo rolls 5 first AND wins}}{\text{chance Hugo wins}}.$

1. Chance Hugo wins. The game is fair, and there are $4$ players, so each wins with chance $\frac14$ .
2. Chance Hugo rolls a $5$ and then wins. First, Hugo rolls a $5$ : chance $\frac16$ . Now each of the other three players rolls one of:
- a $6$ (chance $\frac16$ ): Hugo loses;
- a $5$ (chance $\frac16$ ): a tie with Hugo;
- $1$ to $4$ (chance $\frac46$ ): below Hugo.
3. If nobody rolls a $6$ , some number $k$ of the others tie Hugo with a $5$ . Then $k + 1$ players re-roll in a fresh, fair game, and Hugo wins with chance $\frac{1}{k+1}$ .

others who roll 5	chance of this	Hugo then wins with	product
$0$	$\left(\frac46\right)^3 = \frac{64}{216}$	$1$	$\frac{256}{864}$
$1$	$3 \times \frac16 \times \left(\frac46\right)^2 = \frac{48}{216}$	$\frac12$	$\frac{96}{864}$
$2$	$3 \times \left(\frac16\right)^2 \times \frac46 = \frac{12}{216}$	$\frac13$	$\frac{16}{864}$
$3$	$\left(\frac16\right)^3 = \frac{1}{216}$	$\frac14$	$\frac{1}{864}$

(The $3$ in rows $1$ and $2$ counts which of the three others tie.) Adding the last column:
$\frac{256 + 96 + 16 + 1}{864} = \frac{369}{864} = \frac{41}{96}.$
So once Hugo rolls a $5$ , he wins with chance $\frac{41}{96}$ , and the chance of "rolls $5$ and wins" is $\frac16 \times \frac{41}{96}$ .
4. Divide.
$\frac{\frac16 \times \frac{41}{96}}{\frac14} = \frac{41}{576} \times 4 = \frac{41}{144}.$

The answer is $\boxed{\textbf{(C)}\ \frac{41}{144}}$ .

Why this works

A "given that" question is a ratio: the chance of both things happening divided by the chance of the given thing. Fairness among the players hands you two of the numbers for free. The re-rolls look endless, but a tie among $j$ players is just a new fair game among them, so each wins it with chance $\frac1j$ .

Concise solution

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

We want
$P(\text{first roll } 5 \mid \text{won}) = \frac{P(\text{first roll } 5)\cdot P(\text{won} \mid \text{first roll } 5)}{P(\text{won})}.$

By symmetry among the four players, $P(\text{won}) = \frac14$ , and $P(\text{first roll } 5) = \frac16$ .

Now suppose Hugo rolled a $5$ . Each other player independently rolls a $6$ (probability $\frac16$ , Hugo loses), a $5$ (probability $\frac16$ , tie), or at most $4$ (probability $\frac46$ ). If nobody rolls a $6$ and exactly $k$ others roll $5$ , then $k + 1$ players re-roll from scratch and Hugo wins with probability $\frac{1}{k+1}$ by symmetry. Therefore
$P(\text{won} \mid 5) = \left(\tfrac46\right)^3 + 3\left(\tfrac16\right)\left(\tfrac46\right)^2 \cdot \tfrac12 + 3\left(\tfrac16\right)^2\left(\tfrac46\right)\cdot\tfrac13 + \left(\tfrac16\right)^3\cdot\tfrac14.$
Over the denominator $864 = 4 \cdot 216$ : the terms are $\frac{256}{864} + \frac{96}{864} + \frac{16}{864} + \frac{1}{864} = \frac{369}{864} = \frac{41}{96}$ .

Finally
$P(5 \mid \text{won}) = \frac{\frac16 \cdot \frac{41}{96}}{\frac14} = \frac{4 \cdot 41}{576} = \frac{41}{144}.$

The answer is $\boxed{\textbf{(C)}\ \frac{41}{144}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques (§8.5)
Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)