Question

4) Professor Arnold practices putting in golf. He starts at the same location and aims for the same hole each shot
(i.e. he puts from the same place regardless of where the ball stopped). Assume that his skill level is constant throughout,
and every shot is independent. His first shot does not make it into the hole.

(Note, since the distribution of distance is continuous,
the probability of two shots having the same distance from the hole is infinitesimal; hence, you may ignore the probability of a tie).

a). His second shot lands further from the hole than his first. What is the probability that the third shot is further than the first?

[Hint: Let X1,X2,X3 be the distance from the hole of the first, second and third shot. You want
P(X3>X1|X2>X1) = P(X3 and X2 > X1)/P(X2 > X1). Observe that P(X3 and X2 > X1) is the probability that X1 is the
best shot (recall all shots are independent and have the same distribution. ]

b). His puts 67 times, and unfortunately it happens that each subsequent shot is further from the hole than the first.
He decides to try one more time. What is the probability that he still fails to get closer to the hole than the first?

Answer 1

Solution:

Given: Professor Arnold's skill level is constant and every shot is independent.

Hence his performance at every shot is independent of previous one and will not be superior or inferior. Hence his prformance for second shot is independent of first shot and performance for third shot is independent of first and second shot and so on. Hence probability of landing shot further from the hole than his first shot and closer than first is same i.e 1/2

Hence

a). His second shot lands further from the hole than his first. What is the probability that the third shot is further than the first?

P(X2>X1|X1)=P(X2>X1)=1/2

and P(X3>X1|X2>X1)=P(X3 and X2 > X1)/P(X2 > X1)

=P(X3>X1)P(X2>X1)/P(X2>X1)

b). His puts 67 times, and unfortunately it happens that each subsequent shot is further from the hole than the first.He decides to try one more time. What is the probability that he still fails to get closer to the hole than the first?

Similarly, as in (a).

P(X68>X1|X2,X3,....X67>X1)=P(X68 > X1)

= 1/2