Question

Two snooker players, Player A and Player B, play a match of a ‘best-of-seven’ frames of snooker, i.e. the first player to win 4 frames wins the match. The probability Player A wins a frame is 0.55 and the probability Player B wins a frame is therefore 0.45.

(a) What is the probability that Player A wins the match by a score of 4-2?
(b) What is the overall probability Player B wins the match?
(c) What is the expected number of frames played in a match?

Answer 1

P(Chances of A wininng a frame) = P(A) = 0.55

P(Chances of B winning a frame) = P(B) = 0.45

(a) P(Player A wins the match by a score of 4-2) that means from starting 5 matches, A should win 3 matches and 6th match should be won by A

P(Player A wins the match by a score of 4-2) = ⁵C₂ (0.55)³ (0.45)² * 0.55

= 0.1853

(b) Probability that player B will win this Match =

P(B win intiatil 4 matches) + P(B win 4 out of 5 starting matches) + P(B win 4 out of 6 starting matches) + P(B win 4 out of 7 matches)

= 0.45⁴ + ⁴C₃ (0.55)¹ (0.45)³ * 0.45 + ⁵C₃ (0.55)² (0.45)³ * 0.45 + ⁶C₃ (0.55)³ (0.45)³ * 0.45

= 0.45^4 [ 1 + 4 * 0.55 + 10 * 0.55² + 20 * 0.55³]

= 0.3917

(c) Here there are 4 possibilites of number of frames i.e. 4,5,6 and 7

if x is the number of frames to be played than

P(x = 4) = P(A win in 4 straight frames) + P(B win in 4 straight frames)

= 0.55⁴ + 0.45⁴ = 0.1325

P(x = 5) = P(A win in 5 frames) + P(B win in 5 frames)

= ⁴C₃ (0.55)³ (0.45)¹ * 0.55 + ⁴C₃ (0.55)¹ (0.45)³ * 0.45 = 0.2549

P(x = 6) = P(A win in 6 frames) + P(B win in 6 frames)

= ⁵C₃ (0.55)³ (0.45)² * 0.55 + ⁵C₃ (0.55)² (0.45)³ * 0.45= 0.3093

P(x = 7) = P(A win in 7 frames) + P(B win in 7 frames)

= ⁶C₃ (0.55)³ (0.45)³ * 0.55 + ⁶C₃ (0.55)³ (0.45)³ * 0.45= 0.3032

so here

E[X] =

= 4 * 0.1325 + 5 * 0.2549 + 6 * 0.3093 + 7 * 0.3032

= 5.78