Probability of winning games at tournament 
I know it's a simple problem but apparently I am doing something wrong:
  The probability of winning every single game at a tournament is 0.4. There is only win and lose - no draw. Find the probability of winning exactly 2 games by playing at most 6 games.

Since winning and losing are mutually exclusive, the probability of losing a game is 0.6.
The required probability is:
Probability of playing 2 games and winning both, +
Probability of playing 3 games and winning 2, +
Probability of playing 4 games and winning 2, +
Probability of playing 5 games and winning 2, +
Probability of playing 6 games and winning 2.
First one is $(0.4)^2$
Second is ${3\choose 2}(0.4)^2(0.6)^1$
then
${4\choose 2}(0.4)^2(0.6)^2$
${5\choose 2}(0.4)^2(0.6)^3$
${6\choose 2}(0.4)^2(0.6)^4$
I am getting a total of 1.45024 and obviously it is wrong. 
Correct answer is 0.76688 but I don't know what I am doing wrong!
Many thanks!
 A: The error in your solution is that you are counting some cases several times. So, the playing 2 and winning 2 is included in all of your other cases.
Split the cases up as
Win first 2
Win 1 out of the first 2 then win the third
Win 1 out of the first 3 then win the fourth
...
See what you get now.
A: Well, my solution is $0.76672$. That is $0.00016$ less than yours :) 
Obviously, there is no point continuing playing if you have already won 2 games.
But in your answer, for example, "probability of playing 3 games and winning 2" you have the probability of winning the first two games and continuing, losing the 3rd game:
${3\choose 2}(0.4)^2(0.6)^1$ is $3\cdot(0.4)^2(0.6)^1$ that means $(0.4)(0.4)(0.6)+(0.4)(0.6)(0.4)+(0.6)(0.4)(0.4)$ but of course we shouldn't add  $(0.4)(0.4)(0.6)$ as we have already won 2 games.
So, 
Probability of playing 2 games and winning 2 is $1\cdot(0.4)^2$
Probability of playing 3 games and winning 2 is $2\cdot(0.4)^2(0.6)^1$
Probability of playing 4 games and winning 2 is $3\cdot(0.4)^2(0.6)^2$
Probability of playing 5 games and winning 2 is $4\cdot(0.4)^2(0.6)^3$
Probability of playing 6 games and winning 2 is $5\cdot(0.4)^2(0.6)^4$
We add all these and we have $0.76672$.
I don't know the distribution but, to understand why, for example, I multiplied by 4 in "probability of playing 5 games and winning 2", these are the outcomes: (W: Win, L: Lose)
W L L L W, L W L L W, L L W L W, L L L W W. (Notice that in any case, we always win the last game)
Same approach is for every case. See for yourself! Great problem!
