Math Problem Statement

n a certain game, Players $1$ through $5$ take turns rolling a standard six-sided die, beginning with Player $1.$ If a six is rolled, that player wins. If any other number is rolled, the die is given to the player whose number corresponds to the number rolled, even if that is the player whose turn it currently is. What is the probability that Player $1$ wins? Express your answer as a common fraction.

Solution

We are tasked with finding the probability that Player 1 wins in a game where players roll a six-sided die. A player wins if they roll a 6. If any other number is rolled, the die is passed to the player corresponding to the number rolled.

Step 1: Define the probabilities

Let P1P_1 denote the probability that Player 1 wins the game.

  • On Player 1's turn, they roll a die. The probability that they roll a 6 and win is 16\frac{1}{6}.
  • If Player 1 rolls any number other than 6, the die is passed to another player. Specifically:
    • With probability 16\frac{1}{6}, Player 1 rolls a 1 and keeps the die.
    • With probability 16\frac{1}{6}, Player 1 rolls a 2 and passes the die to Player 2.
    • With probability 16\frac{1}{6}, Player 1 rolls a 3 and passes the die to Player 3.
    • With probability 16\frac{1}{6}, Player 1 rolls a 4 and passes the die to Player 4.
    • With probability 16\frac{1}{6}, Player 1 rolls a 5 and passes the die to Player 5.

Step 2: Define probabilities for other players

Let P2,P3,P4,P5P_2, P_3, P_4, P_5 represent the probabilities that Players 2, 3, 4, and 5 win the game, respectively.

  • The probability that Player 1 rolls a 2 and passes the die to Player 2 is 16\frac{1}{6}, after which Player 2 will play the same game starting from their turn. Therefore, Player 1 wins if Player 2 does not win. Hence, the probability that Player 1 wins in this scenario is 1P21 - P_2.
  • Similarly, if Player 1 rolls a 3, 4, or 5, the die is passed to the respective player, and Player 1's probability of winning is 1P31 - P_3, 1P41 - P_4, or 1P51 - P_5, respectively.

Step 3: Write the equation for P1P_1

Player 1 can win in two ways:

  • They roll a 6 directly (probability 16\frac{1}{6}).
  • They roll a number between 1 and 5, pass the die to another player, and eventually win the game. The combined probability of these events is the sum of the individual probabilities for each outcome.

Thus, we can write the equation for P1P_1 as: P1=16×1+16P1+16(1P2)+16(1P3)+16(1P4)+16(1P5)P_1 = \frac{1}{6} \times 1 + \frac{1}{6} P_1 + \frac{1}{6} (1 - P_2) + \frac{1}{6} (1 - P_3) + \frac{1}{6} (1 - P_4) + \frac{1}{6} (1 - P_5) Simplifying: P1=16+16P1+16(1P2+1P3+1P4+1P5)P_1 = \frac{1}{6} + \frac{1}{6} P_1 + \frac{1}{6} (1 - P_2 + 1 - P_3 + 1 - P_4 + 1 - P_5) P1=16+16P1+5616(P2+P3+P4+P5)P_1 = \frac{1}{6} + \frac{1}{6} P_1 + \frac{5}{6} - \frac{1}{6} (P_2 + P_3 + P_4 + P_5) Multiplying the entire equation by 6 to eliminate the

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Math Problem Analysis

Mathematical Concepts

Probability
Markov Chains
Expected Value

Formulas

P_1 = (1/6) + (1/6)P_1 + (1/6)(1 - P_2 + 1 - P_3 + 1 - P_4 + 1 - P_5)

Theorems

Law of Total Probability
Geometric Series

Suitable Grade Level

Grades 10-12