In the finals of a rugby tournament, two teams play a best of 5 series. Each team has a probability of 1/2 of winning the first game. For each subsequent game, the team that won the previous game has a 7/10 chance of winning, while the other team has a 3/10 chance of winning. If p is the probability that the series lasts exactly 4 games, what is ⌊1000p⌋?
I am not familiar with your notation of ⌊1000p⌋
but I would do it this way:
after a W prob(W) = 7/10 , prob(L) = 3/10
after a L, Prob(W) = 3/10, prob(L) = 7/10
To have a 4-game series in a best of 5 setup, the winning team could win in these ways
LWWW --- (1/2)(3/10)(7/10)(7/10) = 147/2000
WLWW --- (1/2)(3/10)(3/10)(7/10) = 63/2000
WWLW --- (1/2)(7/10)(3/10)(3/10) = 63/2000
Prob = (147+63+63)/2000 = 273/2000
To find the probability that the series lasts exactly 4 games, we need to consider all the possible ways that the first four games can be played.
There are three scenarios where the series lasts exactly 4 games:
1. Team A wins the first two games, and then Team B wins the next two games.
2. Team B wins the first two games, and then Team A wins the next two games.
3. Team A wins the first game, then Team B wins the next game, then Team A wins the third game, and then Team B wins the fourth game.
Let's calculate the probabilities for each scenario:
1. The probability that Team A wins the first two games is (1/2) * (7/10).
The probability that Team B wins the next two games is (3/10) * (3/10).
So, the probability of this scenario is (1/2) * (7/10) * (3/10) * (3/10).
2. The probability that Team B wins the first two games is (1/2) * (3/10).
The probability that Team A wins the next two games is (7/10) * (7/10).
So, the probability of this scenario is (1/2) * (3/10) * (7/10) * (7/10).
3. The probability that Team A wins the first game is 1/2.
The probability that Team B wins the next game is 3/10.
The probability that Team A wins the third game is 7/10.
The probability that Team B wins the fourth game is 3/10.
So, the probability of this scenario is (1/2) * (3/10) * (7/10) * (3/10).
Finally, we add up the probabilities of all three scenarios to find the probability that the series lasts exactly 4 games.
p = (1/2) * (7/10) * (3/10) * (3/10) + (1/2) * (3/10) * (7/10) * (7/10) + (1/2) * (3/10) * (7/10) * (3/10)
Now, let's calculate ⌊1000p⌋ (the floor of 1000 times the value of p).