Solving Probability Questions with Friends Jerry, George and Cosmo

[vlash]

Hi,
I guess it will be an easy question for those who are familiar with probability. I am just getting started with probability and have not understand it good. I have some basic problems to solve. Can anybody help?

The problem is:
Three friends, Jerry, George and Cosmo are playing the following game: each one throws a fair coin (i.e. p(“heads”)= p(“tails”) ). The player which got a result different than the other two players looses. The game ends when there is a looser.
a. what is the probability of ending the game at a given round?
b. what is the probability if Jerry looses at the fourth round?
c. 1. what is the probability of the game lasting at least three rounds?
2. Exactly three rounds?

I tried to solve the problem, but I am not sure I'm right. My solution is:
a. 6/8 --> There are 8 possibilities to end the round, 2 of them does not lead to loser. Therefore there are 6 possibilities to loose.
b. I got no clue.
c. 2/6 --> The probabilitie not to have loser in first round.
(2/6)*4 --> The answer.

Is it correct?

Thank you.

 

[daveb]

a) correct
b) Well, it requires the first 3 rounds to have no loser, and the fourth requires Jerry to lose. What is the probability that Jerry loses given a single round?
c.1.) To last at least 3 rounds, the requirement is to have the first 3 rounds with no loser, the 4th is irrelevant.
c.2.) The first 3 rounds reuqire no loser, and the 4th requires a loser.

Can you figure it from here?

 

[vlash]

Hi,
Thank you for you reply.

I quess it will look like this:
b. 2/8 --> The probability to have no looser in sertain round.
(2/8)*(2/8)*(2/8) --> The probability to have no looser in three rounds.
(6/8)/3 --> The probability that Jerry looses at certain round
(2/8)*(2/8)*(2/8)*((6/8)/3) --> The answer

c.1. (2/8)*(2/8)*(2/8)
c.2. (2/8)*(2/8)*(2/8)*(6/8)

Is it correct?
Thank you.

 

[daveb]

Looks good!