In the online card game, Hearthstone, there is a play mode called Arena. Players who enter the arena build decks (commonly referred to as drafting) and play until they win 12 matches or lose 3 matches. Since the decks are built 'randomly,' and the matches are also random, we can assume that each player has a 0.5 chance of winning and a 0.5 chance of losing.
What would be the best way to calculate the probability of ending with exactly 0 wins, 1 win, 2 wins, ..., and 12 wins? Since the arena ends after any 3 loses there are branches of the binomial tree that would never occur.
ie Completing 3 matches have the following possible outcomes. WWW, WWL, WLW, WLL, LLL, LLW, LWL, LWW. Since LLL fulfills the 'lose 3 match requirement' LLLL and LLLW can not occur. Likewise during a 5th round, WLLLW, WLLLL, LWLLW, LWLLL, LLWLW, LLWLL, LLLWW, LLLWL, LLLLWL, LLLLL can not occur.
We look at the probability of $k$ wins, where $k\lt 12$.
This will happen if we play $k+3$ games, lose $2$ of the first $k+2$, and lose the last. The probability is $$\binom{k+2}{2}\left(\frac{1}{2}\right)^{k+2}\cdot\frac{1}{2}.$$
The case $k=12$ is a little different. One can find the probability by subtracting the sum of the above computed probabilities from $1$. Or else (better) we can divide into cases: (i) no losses: (ii) $1$ loss; (iii) $2$ losses. The first is very easy, and the analysis for the other two is as above.
The same idea will take care of situations where we win any game with probability $p$, and lose with probability $1-p$.