7. Easter Problem ${ }^{3}$. (From 6th grade, 2 points) Two players are playing an "Easter egg fight." In front of them is a large basket of eggs. They randomly take an egg each and hit them against each other. One of the eggs breaks, the loser takes a new egg, while the winner keeps their egg for the next round (the outcome of each round depends only on which egg has the stronger shell; the winning egg retains its strength). It is known that the first player won the first ten rounds. What is the probability that he will win the next round as well?
Solution. Regardless of how events unfolded in the first ten rounds, in the 10th round, the egg that is the strongest among the first 11 eggs randomly drawn from the basket will be the winner. This means that the strongest egg will break in the 11th round only if the 12th egg drawn is stronger, that is, it is stronger than the previous 11 eggs. The probability of this is 1/12. Therefore, the probability of the opposite event “the winner of the ten rounds will also win the eleventh round” is $1-\frac{1}{12}=\frac{11}{12}$.

Answer: $11 / 12$.