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# Counting and Probability-Halp!

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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.

Jul 14, 2024

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Understanding the Problem

There are 5 players.

Player 1 starts.

A player wins by rolling a 6.

If a number other than 6 is rolled, the die passes to the corresponding player.

Solution

Let P be the probability that Player 1 wins.

There are two ways Player 1 can win:

Player 1 wins on their first roll: This happens with probability 1/6.

Player 1 wins on a subsequent roll: This happens if the die cycles through all players and returns to Player 1 without a 6 being rolled, and then Player 1 rolls a 6.

The probability of the die cycling through all players without a 6 being rolled is (5/6)^4 (since each player has a 5/6 chance of not rolling a 6). Then, Player 1 has a 1/6 chance of winning on their return roll. So the probability of Player 1 winning on a subsequent roll is (5/6)^4 * (1/6).

Therefore, the total probability of Player 1 winning is:

P = (1/6) + (5/6)^4 * (1/6)

To simplify this fraction, we can find a common denominator:

P = (6^4 + 5^4) / (6^5)

Calculating the numerator:

6^4 + 5^4 = 1296 + 625 = 1921

So, the probability that Player 1 wins is:

P = 1921/7776

Therefore, the probability that Player 1 wins is 1921/7776.

Jul 14, 2024