Each year, 64 college basketball teams compete in the NCAA tournament . Sandbox . com recently offered a prize of $10 million to anyone who could correctly pick the winner in each of the tournament games.
A. How many games are required to get one championship team from the field of 64 teams?
B. If someone makes random guesses for each game of the tournament, find the probability of picking the winner in each game.
C. In an article about the $10 million prize, The New York Times wrote that "Even a college basketball expert who can pick games at a 70 percent clip has a 1 in _____ chance of getting all the games right." Fill in the blank.
I'm confused as far as the statistical data... Help!.
A. 32+16+8+4+2+1 = ?
B. Probability of picking winner in each game = 1/2
C. 1/2^63 = 1/2 to the 63rd power
A. To get one championship team from a field of 64 teams, a total of 63 games are required. In each round, half of the teams are eliminated, until only one team remains as the champion. Since each game eliminates one team, there will be 63 games played in total.
B. If someone makes random guesses for each game of the tournament, the probability of picking the winner in each game is 1/2. This is because there are only two possible outcomes in each game: either team A wins or team B wins. Assuming the teams are equally matched, the chances of picking the winner correctly are 1/2.
C. The probability of picking all the games correctly can be calculated by multiplying the probability of making a correct pick in each game. Since the probability of picking the winner in each game is 1/2, the probability of picking all the games correctly is (1/2)^63, which is approximately equal to 1.08420217e-19 (a very small probability).
To fill in the blank in the New York Times article, we need to calculate the probability of getting all the games right given a success rate of 70 percent. To convert the success rate to a probability, we divide the success rate by 100, which gives us 0.7. The probability of getting all the games right is (0.7)^63, which is approximately equal to 1.81530984e-8.
Therefore, the "1 in _____ chance of getting all the games right" is 1.81530984e-8, or approximately 1 in 55,000,000.