27.6. (Australia, 82). Player $A$ tosses a coin $n+1$ times, while player $B$ tosses it $n$ times. What is the probability that in the end, player $A$ will have more "heads" than player $B$?
27.6. Let players $A$ and $B$ get $m$ and $k$ "heads" respectively. Then the desired probability $p$ of the event $m>\boldsymbol{k}$ is equal to the probability $q$ of the event $(n+1)-m>n-k$, i.e., the probability that player $A$ gets more "tails" than player $B$ (since "heads" and "tails" appear with equal probability on each coin toss). On the other hand, the event $m>\boldsymbol{k}$ occurs if and only if $n-m<n-k$, i.e., when $(n+1)-m \leq n-k$ (since $n-m$ and $n-k$ are integers). Therefore, $p=1-q$, from which we have $p=q=1 / 2$.