In Vívás, the one who reaches 15 hits first wins the bout. Suppose in the competition between $A$ and $B$ (at any point within the bout), $A$ scores the next hit with probability $p$, and $B$ with probability $q=1-p$. (They never score a hit simultaneously.)

Assume that in a bout, $A$ has already scored $14-k$ hits, and $B$ has scored $14-\ell$ hits (where $k$ and $\ell$ are non-negative integers less than 15), and $A$ scores another hit. By how much has this increased the probability that $A$ will ultimately win the bout?
Let $P_{k, \ell}$ denote the probability of player $A$ winning in the described situation, and $p_{k, \ell}$ the desired increment. According to the law of total probability, $P_{k, \ell}=p \cdot P_{k-1, \ell}+q \cdot P_{k, \ell-1}$. Rearranging this, we get $p_{k, \ell}=P_{k-1, \ell}-P_{k, \ell}=$ $q\left(P_{k-1, \ell}-P_{k, \ell-1}\right)$.

We can assume that players $A$ and $B$ continue the duel until they have 29 hits, as this does not change the outcome of the duel. Thus, $P_{k-1, \ell}$ is the probability that $A$ will score at least $k$ hits out of the remaining $k+\ell$ attempts, while $P_{k, \ell-1}$ is the probability that $A$ will score at least $k+1$ hits out of the remaining $k+\ell$ attempts. Therefore, the difference $p_{k, \ell}=P_{k-1, \ell}-P_{k, \ell-1}$ is the probability that $A$ will score exactly $k$ hits out of the remaining $k+\ell$ opportunities, i.e., $p_{k, \ell}=$ $\binom{k+\ell}{k} p^{k} q^{\ell+1}$.