Jack and Mark are about to play a game with the following rules. They take turns rolling a standard fair die, starting with Jack. The game ends when someone rolls a 1 or 2 or 3 or 4. That person wins the game. Find the probability that Jack will win the game
Question
Jack and Mark are about to play a game with the following rules. They take turns rolling a standard fair die, starting with Jack. The game ends when someone rolls a 1 or 2 or 3 or 4. That person wins the game. Find the probability that Jack will win the game
Solution
The game ends when someone rolls a 1, 2, 3, or 4. This means that the game continues if a 5 or 6 is rolled.
The probability that Jack wins on his first turn is the probability that he rolls a 1, 2, 3, or 4. Since the die is fair, this probability is 4/6 = 2/3.
If Jack doesn't win on his first turn (which happens with probability 1/3), then Mark gets to roll. The probability that Mark wins on his turn is also 2/3. But if Mark wins, then Jack loses, so we are actually interested in the event that Mark doesn't win. This happens with probability 1/3.
If neither Jack nor Mark wins on their first turn, then the game essentially starts over, but it's Jack's turn again. The probability that Jack wins on this second round of turns is the same as the probability that he wins the game overall. Let's call this probability P(J).
We can write an equation for P(J) based on these observations:
P(J) = P(Jack wins on first turn) + P(Jack doesn't win on first turn, Mark doesn't win on first turn, Jack wins game)
The probability that Jack wins the game given that neither he nor Mark wins on their first turn is just P(J), by the argument above. So we can write:
P(J) = (2/3) + (1/3) * (1/3) * P(J)
Solving this equation for P(J) gives:
P(J) - (1/9) * P(J) = 2/3 (8/9) * P(J) = 2/3 P(J) = (2/3) / (8/9) = 3/4
So the probability that Jack wins the game is 3/4.
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