On a birch tree, there were white and black ravens - in total, there were 50. There were definitely white ones, and there were no fewer black ones than white ones. On an oak tree, there were also white and black ravens, and there were 50 in total. On the oak, there were also no fewer black ones than white ones, or the same number, or possibly even one less. One random raven flew from the birch to the oak, and after some time, another (possibly the same) random raven flew from the oak to the birch. What is more likely: that the number of white ravens on the birch became the same as it was initially, or that it changed?
Let there be $a$ white and $b$ black ravens on the birch, and $-c$ white and $d$ black ravens on the oak. The probabilities of black and white ravens flying from the birch to the oak and vice versa can be illustrated using a tree. Point $S$ is the start, event $B$ is the flight of a black raven, and event $W$ is the flight of a white raven.

![](https://cdn.mathpix.com/cropped/2024_05_08_22585010b2534c7ab9ebg-21.jpg?height=460&width=649&top_left_y=199&top_left_x=722)

Event $A$ "The number of white ravens is restored" is marked by the shaded area, to which the chains $SBB$ and $S W W$ lead. The probability of this is $\mathrm{P}(A)=\frac{b}{50} \cdot \frac{d+1}{51}+\frac{a}{50} \cdot \frac{c+1}{51}=\frac{b d+a c+a+b}{50 \cdot 51}$.

The probability of the opposite event "The number has changed" is obtained by summing the probabilities along the chains $S B W$ and $S W B: \quad \mathrm{P}(\bar{A})=\frac{b}{50} \cdot \frac{c}{51}+\frac{a}{50} \cdot \frac{d}{51}=\frac{b c+a d}{50 \cdot 51}$.

We have $50 \cdot 51(\mathrm{P}(A)-\mathrm{P}(\bar{A}))=(b-a)(d-c)+a+b \geq a-b+a+b=2 a>0$ (by the condition $b-a \geq 0, d-c \geq-1)$.

## Answer

The number of white ravens will become the same as before.