XXXIV OM - I - Problem 1

$ A $ tosses a coin $ n $ times, $ B $ tosses it $ n+1 $ times. What is the probability that $ B $ will get more heads than $ A $?
Consider the following events. $ Z_0 $ - the event that player $ B $ has flipped more heads than player $ A $, $ Z_r $ - the event that $ B $ has flipped more tails than player $ A $. These events are equally probable: if the coin is fair, it does not matter which side is called heads and which is called tails. Thus, $ P(Z_0) = P(Z_r) $. Now let $ a_0 $, $ a_r $, $ b_0 $, $ b_r $ be the number of heads and tails flipped by $ A $ and the number of heads and tails flipped by $ B $, respectively. According to the conditions of the problem, $ a_0+ a_r = n $, $ b_0+b_r=n+1 $. Therefore, by subtracting the equations, $ (b_0-a_0)+ (b_r-a_r) = 1 $. From this equation, it follows that one and only one of the numbers $ b_0-a_0 $ and $ b_r-a_r $ is positive, meaning that exactly one of the events $ Z_0 $ and $ Z_r $ must occur. In other words, these events are mutually exclusive and cover all possibilities. Therefore, $ P(Z_0)=P(Z_r)= 1/2 $.