# 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!

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.

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!