## winning and losing teams probability distribution

Suppose two teams play a series of games, each producing a winner and a loser, until one team has won two more games than the other. Let Y be the total number of games played. Assume each team has a chance of 0.5 to win each game, independent of results of the previous games. Find the probability distribution of Y.

3. The attempt at a solution

At first I thought I'd say that given that team A has had X games and team B has Z games, then Y=X+Z, and since it takes two more games for the winner to get, then Z=X+2, so Y=2x+2. Probability function would be then P(Y<=y)=P(Y<=2x+2) but I am not sure where to go from here...Also, was thinking that it may be a binomial, with (Y choose 2x+2)(0.5)^(2x+2)(0.5)^y where x=1,2,3....y.

Any input is appreciated!

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 Quote by libragirl79 Suppose two teams play a series of games, each producing a winner and a loser, until one team has won two more games than the other. Let Y be the total number of games played. Assume each team has a chance of 0.5 to win each game, independent of results of the previous games. Find the probability distribution of Y. 3. The attempt at a solution At first I thought I'd say that given that team A has had X games and team B has Z games, then Y=X+Z, and since it takes two more games for the winner to get, then Z=X+2, so Y=2x+2. Probability function would be then P(Y<=y)=P(Y<=2x+2) but I am not sure where to go from here...Also, was thinking that it may be a binomial, with (Y choose 2x+2)(0.5)^(2x+2)(0.5)^y where x=1,2,3....y. Any input is appreciated!
Have you studied Markov chains yet? If so, it is a simple exercise to get the generating function of Y, hence to find the distribution P{Y = n}, n = 2,3,4,... .

Here is a hint: let X = #won by A - #won by B. We start at X = 0. X increases by 1 if A wins and decreases by 1 if B wins. The game stops when X reaches ± 2. So, we have a simple 4-state Markov chain with states X = {0,1,-1,'stop'}, and Y = first passage time from state 0 to state 'stop'.

RGV

 No, we haven't done Markov Chains yet... What do you mean by "first passage time"?

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## winning and losing teams probability distribution

 Quote by libragirl79 No, we haven't done Markov Chains yet... What do you mean by "first passage time"?
The first passage time from any state i to state 'stop' is the first time (t = 1,2,3,...) at which state 'stop' is reached. It is exactly the Y that you seek. (Calling it a first-passage time allows you to do a meaningful Google search of that term to find out more about it.)

RGV