72*. Two hunters $A$ and $B$ went duck hunting. Suppose each of them hits a duck they encounter as often as they miss. Hunter $A$ encountered 50 ducks during the hunt, and hunter $B$ encountered 51 ducks. What is the probability that hunter $B$'s catch will exceed hunter $A$'s catch?
72. Since each of the hunters $A$ and $B$ hits and kills a duck as often as they miss, the probability of killing a duck for each of these hunters with each individual shot is $1 / 2$, i.e., the same as the probability of a coin landing heads. Thus, our problem is equivalent to the following: $A$ tosses a coin 50 times, and $B$ tosses it 51 times; what is the probability that heads will appear more times for $B$ than for $A$? It is not difficult to calculate the number of equally probable outcomes of this experiment; however, calculating the number of favorable outcomes here is relatively complex. We will not perform this calculation, but instead use the following simple artificial reasoning.

The set of all outcomes of the experiment, where $A$ tosses a coin 50 times and $B$ tosses it 51 times, can be divided into the following two parts: 1) the set of all outcomes where $B$ gets more heads than $A$, and 2) the set of all outcomes where $B$ gets no more (i.e., fewer or the same number of) heads than $A$. Note now that if $B$ gets the same number of heads as $A$ or even fewer, then it means that the tails for $B$ appeared more times than for $A$, and vice versa, if the tails for $B$ appeared more times than for $A$, then it means that the heads for $B$ appeared either the same number of times as for $A$ or even fewer (since $B$ tosses the coin one more time than $A$). Therefore, the set of outcomes where $B$ gets no more heads than $A$ coincides with the set of outcomes where $B$ gets more tails than $A$. But in the case of complete equality of heads and tails on the coin, the number of outcomes where $B$ gets more heads than $A$ should be equal to the number of outcomes where $B$ gets more tails than $A$. Thus, in our problem, the number of favorable outcomes of the experiment is equal to the number of unfavorable outcomes, from which it immediately follows that the desired probability is $1 / 2$.

Therefore, the probability that $B$'s catch will exceed $A$'s catch is $1 / 2$.

Note. The provided solution remains valid in the case where $A$ encounters $n$ ducks during the hunt, and $B$ encounters $n+1$ ducks. And in this more general case, the probability that $B$'s catch will exceed $A$'s catch is also $1 / 2$.