John agrees to play chess until he wins one game or loses 3 games. If his probability of winning is 0.6 calculate the expected number of games that will be played.
My Answer
P(win first game)= 0.6
P(lose 3 games)= 0.4 x 0.4 x 0.4 = 0.064
However, the correct answer is
P(win first game)= 0.6
P(win 2nd game)= 0.4 x 0.6= 0.24
P(win 3rd game)= 0.4 x 0.4 x 0.6= 0.096
P(lose 3 games)= 0.4 x 0.4 x 0.4 = 0.064
My question is, why was the probability of winning the 2nd and 3rd game included? Also, why was winning even included? John won only one game and lost 3 games. If anything shouldn't it have been P(losing 2nd game) and P(losing 3rd game)? I'm extremely lost with this question and I have no idea how they got the bolded answers. Any help I can get would be greatly appreciated.
As for calculating the expected number of games played, thankfully I know how to solve for it.