In an urn, there are 101 balls, among which exactly 3 are red. The balls are drawn one by one without replacement. At which position is it most likely that the second red ball will be drawn?
The color of the other balls does not play a role in the solution, so we can consider them to be of the same color. The red balls can then be drawn from the urn in $\binom{101}{3}$ different ways, and these cases can occur with equal probability.

If the second red ball is drawn at the $k$-th position $(1<k<100)$, then among the $k-1$ balls drawn before it and the $101-k$ balls drawn after it, there is one red ball each. This means that the positions of the first and third red balls can be determined in $(k-1) \cdot(101-k)$ ways, which is exactly the number of cases in which the second red ball can end up in the $k$-th position.

The probability in question will therefore be maximal for the value of $k$ where the product $(k-1) \cdot(101-k)$ is the largest. Transforming the product:

$$
(k-1) \cdot(101-k)=-k^{2}+102 k-101=-(k-51)^{2}+2500
$$

Since $-(k-51)^{2} \leqq 0$, the above product reaches its maximum when $-(k-51)^{2}=0$, i.e., when $k=51$. Therefore, the answer to the question is: the second red ball is most likely to be drawn at the 51st position.

Remarks. The second red ball can appear at the $k=2,3, \ldots, 51, \ldots, 100$ positions; the probability that it is drawn at the 51st position is 2500/( $\left.\begin{array}{c}101 \\ 3\end{array}\right) \approx 0.015$, which, although not very large, is still the largest among the probabilities of the individual positions. For comparison, the probability in the cases of $k=2$ (and $k=100$) is $99 /\binom{101}{3} \approx 0.0006$, which is $1 / 25$ of the previous value.

Let's consider that if we observe the first red ball, its drawing at the first position is the most likely, with a probability of

$$
\frac{\binom{100}{2}}{\binom{101}{3}}=\frac{3}{101} \approx 0.03
$$

If, therefore, two people are playing, one betting on the position of the first red ball and the other on the position of the second red ball, and both bet on the most likely outcome for them, then the first has about twice the chance of winning compared to the second.