What is the probability of winning this game? 
A tennis player has a 60% chance of winning any given point in a tennis game. Calculate the probability that she will win the game within the first 6 points, stating any assumptions you make.
(A game is won when a player has won at least 4 points and won at least 2 more points than their opponent.)

So there are 3 choices:

*

*she wins 4 points with the probability 60%$^{4}$

*she wins 4 points and lose 1 point with probability 4 $\times$ 60%$^{4}$ $\times$ 40%

*she wins 4 points and lose 2 points 15 $\times$ 60%$^{4}$ $\times$ 40%$^{2}$
So the total probability of her winning the game is 0.69984. Is this right?
 A: To win in $\leq 6$ points, she must lose at most $2$ points in $6$, so
P(win) = $\binom 6 0\cdot 0.4^0\cdot 0.6^6 + \binom 6 1\cdot 0.4^1\cdot 0.6^5  + \binom 6 2\cdot  0.4^2 \cdot 0.6^4 = 0.54432$

PS
The above formulation is the simplest way to get the answer. However, just for corroboration, let us also solve it by summing up
P(win in exactly $4,5,\;or\; 6$) points
To win in exactly $4$ points, she must win all: $0.6^4 =0.1296$
To win in $5$ points, she must win the fifth point and three of the first four: $\left(\binom 4 3 \cdot 0.6^3\cdot0.4\right)\cdot0.6 = 0.20736$
To win in $6$ points, she must win the sixth point, and three of the first five:$\left(\binom 5 3 \cdot 0.6^3\cdot0.4^2\right)\cdot0.6 = 0.20736$
$0.1296+ 0.20736 +0.20736 = 0.54432$, as before
But error prone method with possibly confusing binomial coefficients
A: Using Stars and Bars, the correct answer is:
$$P(\text{win})=\binom{3}{0}0.6^4+\binom{4}{1}0.6^4\times0.4+\binom{5}{2}0.6^4\times0.4^2$$
$$=0.54432$$
She must win the last point, and so the sums represent 3 wins no losses, 3 wins 1 loss and 3 wins 2 losses in the preceding points of the game.
