Question

A player of a video game is confronted with a series of opponents and has an 85% probability of defeating each one. Success with an opponent is independent of previous encounters. Until defeated, the player continues to contest opponents.

1. What is the probability mass function of a number of opponents contested in a game?
2. What is the probability that a player defeats at least three players in a game?
3. What is the expected number of opponents contested in a game?
4. What is the probability that a player contests 2 or more opponents?
5. What is the expected number of game plays until a player contests 2 or more opponents?

Answer 1

a)this is geometric distribution with parameter p=0.15

here pmf of X ; number of oppoents played: P(X=x)=(0.85)^x-1*0.15

b)

P(X>=3)=P(defeat first three opponents)=(0.85)³ =0.614125

c) expected number of opponents contested in a game =1/p=1/0.15=6.6667

d) probability that a player contests 2 or more opponents =P(defeat 1st opponent)=0.85

e)

expected number of game plays until a player contests 2 or more opponents

=1*0.85+2*0.15*0.85+3*0.15*0.15*0.85+..=1/0.85 =1.176