16. Fifty cards (from 9th grade, 3 points). On the table, in a row, 50 cards numbered from 1 to 50 from left to right were laid out. The bottom of the cards is blue, and the top is red. Vasya randomly selects 25 consecutive cards (a segment of the row) and flips them. Then Asey approaches the table and does the same - selects a random segment of 25 consecutive cards and flips them, independently of what Vasya did earlier. For which card (or cards) is the probability of ending up red side up after all the flips less than for the others?
Solution. Randomly selecting a fragment from 25 cards means randomly selecting the first (leftmost) card in this fragment. With equal chances, this card can have a number from 1 to 26.

Consider card No. 1. The probability that Vasya will flip it is \( \frac{1}{26} \) (only if the fragment starts with card No. 1). The probability that Asya will flip it is the same. Therefore, the first card will end up with the red side up (not flipped or flipped twice) with the probability

\[
p_{1}=\left(1-\frac{1}{26}\right)^{2}+\left(\frac{1}{26}\right)^{2}=\frac{2}{26^{2}}-\frac{2}{26}+1=\frac{1}{338}-\frac{1}{13}+1
\]

Now consider card No. 2 and reason similarly: Vasya will flip it with a probability of \( \frac{2}{26} \) (if the fragment starts with card No. 1 or card No. 2). Asya will flip it with the same probability. Therefore, card No. 2 will end up with the red side up with the probability

\[
p_{2}=\left(1-\frac{2}{26}\right)^{2}+\left(\frac{2}{26}\right)^{2}=\frac{1}{338} \cdot 2^{2}-\frac{1}{13} \cdot 2+1
\]

This reasoning is valid for all cards up to the 25th: for \( 1 \leq k \leq 25 \)

\[
p_{k}=\frac{1}{338} k^{2}-\frac{1}{13} k+1
\]

For cards numbered from 26 to 50, the probabilities are the same but in reverse order: card No. 50 has the same probability as No. 1, card No. 49 has the same probability as No. 2, and so on. This is easily seen if the cards are numbered in reverse order.

We need to find the smallest of the probabilities \( p_{k} \). The function

\[
f(x)=\frac{1}{338} x^{2}-\frac{1}{13} x+1
\]

attains its minimum value at the point \( x_{0}=\frac{1}{13}: \frac{2 \cdot 1}{338}=\frac{26}{2}=13 \). Therefore, it is least likely that the 13th card from the left and the 13th card from the right (38th from the left) will be lying with the red side up.

Note. The conclusion of the solution can be based not only on the study of the function but also on other considerations, for example, on the theorem of means:

\[
\sqrt{\frac{p_{k}}{2}}=\sqrt{\frac{\left(1-\frac{k}{26}\right)^{2}+\left(\frac{k}{26}\right)^{2}}{2}} \geq \frac{1-\frac{k}{26}+\frac{k}{26}}{2}=\frac{1}{2}
\]

therefore, the smallest of the probabilities \( p_{k}=\left(\frac{1}{2}\right)^{2} \cdot 2=\frac{1}{2} \) under the condition \( 1-\frac{k}{26}=\frac{k}{26} \), from which \( k=13 \).

Answer: the 13th and 38th.