Probability of Winning at least 7 times Tried to model a popular game I was playing, but the probabilities seemed off.
A game allows you to have up to 12 wins but only allows 3 losses.
Each win/lose is independent from each other with a 50% probability and assuming we play until 12 wins are hit or 3 losses have occurred.
What is the probability of having at least 7 wins?

Attempt: Let X be number of wins. I know that if X = 7, then there
  were 10 games total. X = 8, 11 games total. . . . For X = 12, there is
  a special case of 12, 13, 14 total games.
I assumed that X was a binomial random variable and summed up the
  possibilities for X = 7, 8, 9, 10, 11, 12 and it comes out to be around
  30.6% which seems plausible, but if I up the win chance to 75%, then the sum of the probabilities becomes greater than one which implies my original logic is flawed.

 A: To have $7$ or more wins in total, you need to have won $7$, $8$ or $9$ of the first $9$ games, which (if the probability of winning each game is $p$) is
$${9 \choose 7}p^7(1-p)^2 + {9 \choose 8}p^8(1-p) + {9 \choose 9}p^9$$
$$= p^7 (28p^2-63p +36)$$ and this goes from $0$ to $1$ as $p$ goes from $0$ to $1$.  For $p=0.5$, this gives about $0.09$.
A: The general "adding up" strategy should work.  But how did you calculate $\Pr(X=7)$? 
We have $X=7$ if there were exactly $2$ losses in the first $9$ games, and then a loss on the $10$-th. The probability of this is $\binom{9}{2}/2^{10}$. Thus $X$ has a distribution that is more like the negative binomial than the binomial, apart from the special stopping condition at $12$ wins.
Calculate the probabilities of $X=8$, $9$, $10$, and $11$ in the same way. The case $X=12$ will have to be handled differently.
A: I think the most difficult part is to visualize the sample space of the game. Let's denote with "W" the event that a game ended with a win (its probability $p$) and with "L" the event that the game ended with a loss (its probability $q,\ p+q=1$).
Then we may have the following cases :
$$
\eqalign{
0. & LLL  \ \ \ \ \ q^3 \cr
1. & \underbrace{W}_{LL}L \ \ \ \ \ {3 \choose 1}p^1 \cdot q^2 \cdot q \cr
2. & \underbrace{WW}_{LL}L \ \ \ \ \ {4 \choose 2}p^2 \cdot q^2 \cdot q \cr
3. & \underbrace{WWW}_{LL}L \ \ \ \ \ {5 \choose 3}p^3 \cdot q^2 \cdot q \cr
4. & \underbrace{WWWW}_{LL}L \ \ \ \ \ {6 \choose 4}p^4 \cdot q^2 \cdot q \cr
5. & \underbrace{WWWWW}_{LL}L \ \ \ \ \ {7 \choose 5}p^5 \cdot q^2 \cdot q \cr
6. & \underbrace{WWWWWW}_{LL}L \ \ \ \ \ {8 \choose 6}p^6 \cdot q^2 \cdot q \cr
7. & \underbrace{WWWWWWW}_{LL}L \ \ \ \ \ {9 \choose 7}p^7 \cdot q^2 \cdot q \cr
8. & \underbrace{WWWWWWWW}_{LL}L \ \ \ \ \ {10 \choose 8}p^8 \cdot q^2 \cdot q \cr
9. & \underbrace{WWWWWWWWW}_{LL}L \ \ \ \ \ {11 \choose 9}p^9 \cdot q^2 \cdot q \cr
10. & \underbrace{WWWWWWWWWW}_{LL}L \ \ \ \ \ {12 \choose 10}p^{10} \cdot q^2 \cdot q \cr
11. & \underbrace{WWWWWWWWWWW}_{LL}L \ \ \ \ \ {13 \choose 11}p^{11} \cdot q^2 \cdot q \cr
12. & WWWWWWWWWWWW  \ \ \ \ \ p^{12} \cr
}
$$
From the above table we may compute the probability "of having at least 7 wins" by summing the probabilities of cases $7-12$.
A: You have to consider certain limitations, e.g. for exactly 7 wins, the pattern can be
7W followed by 3L, 1L in 6W followed W-2L, or 2L in 6W followed by WL, so if p and q are the probabilities of winning /losing, then for k wins, $7 \le k \lt 12$,
$Pr = p^kq^3[1+{k-1\choose 1} + {k-1\choose 2}]$, 
and for k = 12, $Pr = p^{12}[ 1 + {11\choose1}q + {11\choose 2}q^2]$
