Probability of A staying ahead of B Alice and Bob toss a coin 11 times. Heads are a point for Alice, tails are a point for Bob. After 11 tosses, Alice wins 7-4.
What is the probability that Alice was never behind in the score (ties allowed)?
This is not homework and I have an answer, but I would like an unbiased second opinion.
Thanks in advance!
 A: This is a typical case the Ballot Theorem (see Feller Vol. 1), though we need to adapt a little bit.  Let us set notations: let $ \epsilon_n $ be the result of the $ n $-th toss. We assign $ \epsilon_n = +1 $ if Alice wins while $ \epsilon_n = - 1 $ if Bob wins. So, set $ S_0 = 0 $ and for $ n \geq 0 $, $ S_n = \epsilon_1 + \dotsb + \epsilon_n $. This represents the lead of Alice over Bob at time $n$. We are given that $ S_{11} = 3 $. We are interested in finding
\begin{equation*}
\mathbb{P} ( S_1 > 0, S_2 > 0 , \dotsc, S_{11} > 0 \mid S_{11} = 3 ) .
\end{equation*}
This will imply that Alice has always lead over Bob (not even ties).
This is exactly what is given  by Ballot Theorem. The above conditional probability :
\begin{equation*}
\mathbb{P} ( S_1 > 0, S_2 > 0 , \dotsc, S_{11} > 0 \mid S_{11} = 3 ) = \frac{3}{11}. 
\end{equation*}
If you want the unconditional probability, then just multiply by $ \mathbb{P} ( S_{11} = 3 ) $, which is given by
\begin{equation*}
 \mathbb{P} ( S_{11} = 3 )  = \binom{11}{4} 2^{-11}
\end{equation*}
as you have place to $ 4 $ tosses among $11$ tosses for Bob and rest for Alice.
A: There are $\binom{11}{7}=\binom{11}{4}=330$ possible sequences.
There are $29$ sequences where Alice is not behind Bob:


*

*$1$ sequence  after the $ 1$st toss ($1:0$)

*$2$ sequences after the $ 2$nd toss ($2:0,1:1$)

*$2$ sequences after the $ 3$rd toss ($3:0,2:1$)

*$3$ sequences after the $ 4$th toss ($4:0,3:1,2:2$)

*$3$ sequences after the $ 5$th toss ($5:0,4:1,3:2$)

*$4$ sequences after the $ 6$th toss ($6:0,5:1,4:2,3:3$)

*$4$ sequences after the $ 7$th toss ($7:0,6:1,5:2,4:3$)

*$4$ sequences after the $ 8$th toss ($7:1,6:2,5:3,4:4$)

*$3$ sequences after the $ 9$th toss ($7:2,6:3,5:4$)

*$2$ sequences after the $10$th toss ($7:3,6:4$)

*$1$ sequence  after the $11$th toss ($7:4$)


Hence the probability that Alice is never behind Bob is $\frac{29}{330}$.
