F.4 maths probability

[This is a typical question.]
[This kind of question is popular again because of the HKDSE 2014.]

(a) Pr(Kelvin will win the game)
= Pr(The game ends after the 1st throw) + Pr(The game ends after the 3rd throw) + Pr(The game ends after the 5th throw) + ...
= (1/6) + (5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(5/6)(1/6) + ...
= (1/6) + (5/6)²(1/6) + (5/6)⁴(1/6) + ...
= (1/6) / [1 - (5/6)²]
= (1/6) / (1 - 25/36)
= 6 / (36 - 25)
= 6/11

(b) The required probability is
Pr(Daisy did not win the game before her second throw | Daisy has won the game)
= Pr(Daisy did not get "4" in her first throw | Daisy has won the game)
= Pr(Daisy did not get "4" in her first throw and Daisy has won the game) / Pr(Daisy has won the game)

Note that the denominator is Pr(Daisy has won the game)
= 1 - Pr(Daisy has not won the game)
= 1 - Pr(Kelvin has won the game)
= 1 - 6/11 (by (a))
= 5/11

The numerator is Pr(Daisy did not get "4" in her first throw and Daisy has won the game)
= Pr(The game ends after the 4th throw) + Pr(The game ends after the 6th throw) + Pr(The game ends after the 8th throw) + ...
= (5/6)(5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(5/6)(5/6)(5/6)(5/6)(1/6) + ...
= (5/6)³(1/6) + (5/6)⁵(1/6) + (5/6)⁷(1/6) + ...
= (5/6)³(1/6) / [1 - (5/6)²]
= (125/1296) / (1 - 25/36)
= (125/36) / (36 - 25)
= (125/36) / 11
= 125/396

Thus, the required probability is
Pr(Daisy did not win the game before her second throw | Daisy has won the game)
= (125/396) / (5/11)
= 25/36

2014-06-15 08:14:45 補充：
Sum of geometric series to infinity

　a + ar + ar² + ar³ + ...
= a/(1 - r) if -1 < r < 1

In your case, (1/6) + (5/6)²(1/6) + (5/6)⁴(1/6) + ...
= (1/6) / [1 - (5/6)²]

a = 1/6
r = (5/6)²

智術超群 無與論比

敬佩 [ 加 雙 ]