12. (3b, 8-11) In belly-pushing competitions, the chances of opponents winning are proportional to their body masses. Jumbo weighs more than Dumbo, and Pink weighs less than Bonk. There can be no draws in a match. Jumbo and Dumbo must take turns pushing Pink and Bonk. Which of the following events is more likely:

$$
\begin{gathered}
A=\{\text { Jumbo wins only against Pink, and Dumbo — only against Bonk }\} \\
\text { or } \\
B=\{\text { Jumbo wins only against Bonk, and Dumbo — only against Pink }\} ?
\end{gathered}
$$
# Solution.

Let's denote the masses of the opponents as follows: $u, j, p, b$ corresponding to the first letters of their names. Then $u>j, b>p$. The probability that Yumbo defeats only Pink is the product of the probabilities that he defeats Pink and loses to Bonk, i.e., $\frac{u}{u+p} \cdot \frac{b}{u+b}$. Similarly, the probability that Jumbo defeats only Bonk is $\frac{j}{j+b} \cdot \frac{p}{j+p}$. Therefore, the probability of event $A$ is

$$
\mathrm{P}(A)=\frac{u b j p}{(u+p)(u+b)(j+p)(j+b)}
$$

Similarly,

$$
\mathrm{P}(B)=\frac{u p j b}{(u+b)(u+p)(j+b)(j+p)}
$$

It is clear that $\mathrm{P}(A)=\mathrm{P}(B)$.

Answer: the probabilities are equal.