You have purchased a ticket for the 7th game in a 7-game series. Team A has a probability p of winning each game. Find the probability that the series will be decided before the 7th game (that is one of the two teams wins 4 of the first six games). Plot the probability for various values of p in the range 0 to 1.
The series ends after 4 games if one team wins all four games; that is AAAA or BBBB. By independence, P(AAAA) = P(A)P(A)P(A)P(A) = (.5)^4 = .0625. Since the same is true for BBBB, the probability of a 4-game series is .0625 +.0625 = .125.
A 5-game series means that it was 4-1 for one of the teams. Say A wins the series; this can only happen via AAABA, AABAA, ABAAA, or BAAAA, each of which has probability .5^5 = .03125, giving a probability of (4)(.03125) = .125. The same is true if B wins the series, so the probability of a 5-game series is .125 + .125 = .25.
Thus, the probability of a 4 or 5 game series is .25 + .125 = .375. Given this information, we know that the probability of a 6 or 7 game series is (1-.375)/2 = 0.3125 respectively.
Since it is unrealistic to assume that each team has an equal probability of winning, the following simulation varies, “p”, the probability of Team A winning each game.
Joe drives a bus with forty seats. On average, 10% of the people who buy tickets in advance do not show up. The ticket price is $10. If one can’t get on the bus with a purchased ticket because the bus is full, $25 is paid as the compensation. How many seats shall Joe sell in order to maximize his expected revenue?
Based on the results below, Joe should sell 44 seats to maximize his revenue.
Please calculate the expected completion time of the project.
The expected completion time of the project is approximately 66 minutes.
## [1] 66.03916
Suppose you’re on a game show, and you’re given the choice of 33 doors: Behind one door is a car; behind the others, goats.
Use simulation to calculate the probability of winning when you decide to stick and the probability of winning when you decide to switch.
## [1] 0.038
## [1] 0.03
It is a tradition to give Lucky Money in Chinese Lunar New Year. Right now, $100 lucky money is shared within 10 people in the following way. The first person will get U1×100, where U1 is a [0,1] uniform distribution. The second person will get U2 of the remaining money, where U2 is also a [0,1] uniform distribution. So on and so forth. The last person will get all the remaining. Use simulation to calculate the probabilities that the ith person gets the largest amount where i = 1, …, 10.
## [1] 50.5603215 24.5685256 12.3440530 6.3206242 3.0654406 1.5780483
## [7] 0.7801141 0.3983676 0.1919756 0.1925295