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.

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 $ P_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 $ \frac{1}{6} $.
- If Player 1 rolls any number other than 6, the die is passed to another player. Specifically:
  - With probability $ \frac{1}{6} $, Player 1 rolls a 1 and keeps the die.
  - With probability $ \frac{1}{6} $, Player 1 rolls a 2 and passes the die to Player 2.
  - With probability $ \frac{1}{6} $, Player 1 rolls a 3 and passes the die to Player 3.
  - With probability $ \frac{1}{6} $, Player 1 rolls a 4 and passes the die to Player 4.
  - With probability $ \frac{1}{6} $, Player 1 rolls a 5 and passes the die to Player 5.

### Step 2: Define probabilities for other players
Let $ P_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 $ \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 $ 1 - 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 $ 1 - P_3 $, $ 1 - P_4 $, or $ 1 - P_5 $, respectively.

### Step 3: Write the equation for $ P_1 $
Player 1 can win in two ways:
- They roll a 6 directly (probability $ \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 $ P_1 $ as:
\[
P_1 = \frac{1}{6} \times 1 + \frac{1}{6} P_1 + \frac{1}{6} (1 - P_2) + \frac{1}{6} (1 - P_3) + \frac{1}{6} (1 - P_4) + \frac{1}{6} (1 - P_5)
\]
Simplifying:
\[
P_1 = \frac{1}{6} + \frac{1}{6} P_1 + \frac{1}{6} (1 - P_2 + 1 - P_3 + 1 - P_4 + 1 - P_5)
\]
\[
P_1 = \frac{1}{6} + \frac{1}{6} P_1 + \frac{5}{6} - \frac{1}{6} (P_2 + P_3 + P_4 + P_5)
\]
Multiplying the entire equation by 6 to eliminate the

In 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.

Explore how to calculate the probability that Player 1 wins in a dice game with 5 players, using key probability concepts, formulas, and the Law of Total Probability. This problem demonstrates applications of Markov Chains and expected value, suitable for Grades 10-12.

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