What is the probability that the smallest number in a lottery draw will be odd?
In the task, we naturally thought of the so-called Genoese lottery (5 numbers are drawn from 90); in Hungary (even at the time of the problem's appearance), this is the better-known game form.

The smallest number in a draw can be 1, 2, ..., 86. We group these numbers into pairs of the form $(2k-1, 2k)$, where $k=1,2, \ldots, 43$. In each such pair, the odd number appears more frequently as the smallest drawn number than the even number. The excess for the pair $(2k-1, 2k)$ is:

$$
T_{k}=\binom{91-2k}{4}-\binom{90-2k}{4}=\binom{90-2k}{3}
$$

(we used the well-known identity for binomial coefficients

$$
\binom{n}{t-1}+\binom{n}{t}=\binom{n+1}{t}
$$

).

Therefore, the total excess of the odd numbers appearing is:

$$
T=\sum_{k=1}^{43} T_{k}=\sum_{k=1}^{43}\binom{90-2k}{3}
$$

Given the value of $T$, the number of results starting with an odd number is

$$
T+\frac{1}{2}\left[\binom{90}{5}-T\right]=\frac{1}{2}\left[\binom{90}{5}+T\right]
$$

and the desired probability is

$$
\left.\frac{1}{2}\left[\binom{90}{5}\right]+T\right] /\binom{90}{5}=\frac{1}{2}+\frac{T}{2\binom{90}{5}}
$$

To fully solve the problem, we still need to calculate the value of $T$.

For simpler calculation, we introduce a new summation index; let:

$$
k=44-m, \text { i.e., } m=44-k
$$

Thus, we get the terms of the sum in reverse order:

$$
T=\sum_{m=1}^{43}\binom{2m+2}{3}
$$

Expanding the running term of the sum,

$$
\frac{2}{3}(m+1)(2m+1)m=\frac{4}{3}m^3+2m^2+\frac{2}{3}m
$$

and using known sum formulas, the value of $T$ is given by the following polynomial at $r=43$:

$$
\begin{gathered}
\frac{4}{3} \frac{r^2(r+1)^2}{4}+2 \frac{r(r+1)(2r+1)}{6}+\frac{2}{3} \frac{r(r+1)}{2}= \\
=\frac{1}{3} r(r+1)(r(r+1)+(2r+1)+1)= \\
=\frac{1}{3} r(r+1)^2(r+2)=(2r+2)\binom{r+2}{3}
\end{gathered}
$$

Thus, the probability that the smallest drawn number is odd is

$$
\frac{1}{2}+\frac{44 \cdot\binom{45}{3}}{\binom{90}{5}} \approx 0.5142
$$

Remarks. 1. In the lottery draw, however, the even numbers have the advantage in the largest number - at least theoretically. For every draw pattern, the "arithmetic mirror image" can be written with respect to the center of the complete set of numbers, i.e., $(1+90)$ : $2=45.5$. Thus, the place of the $E$ first, $M$ second, ..., Ö fifth number is $(91-E), \ldots,(91-\ddot{O})$. This makes the sequence decreasing, and $91-E$ has the opposite parity of $E$.

2. The origin of our question was the following: we saw tables from the first 1000 draws of the Hungarian lottery series starting in 1957 (up to early 1976) about the "distribution" of the smallest and largest numbers, i.e., the "frequency" of the 1-65 and 26-90 values that occurred. Summing these in odd and even categories, the smallest place showed a 523:477 advantage for the odd numbers, and the largest place showed a 501:499 advantage. The expected 51.4% is instead 52.3%, and 49.9%.

The 1000-trial "sample" is very small compared to the nearly 44 million possible draw outcomes; on the one hand, this makes the deviations of the "actual figures" from the expected understandable. Furthermore, in the first 1000 draws, the drawn 5000 numbers had a significant advantage for the odd numbers, 2560:2440, in a 1.049 ratio. (Even today, this ratio is around 1.02.)

We can try to "filter out" the latter fact from the results of the first and last place. We use the 523 and 501 "actual figures" for the odd numbers in a 2560:2440 ratio: 498.5 and 477.5, respectively, and calculate the ratios as follows:

$$
\begin{aligned}
& 498.5:(498.5+477)=498.5: 975.5=0.5110 \\
& 499:(499+477.5)=\quad 0.5110 .
\end{aligned}
$$

Surprisingly, a 51.1% advantage appears from two different data sets, compared to the expected 51.4%.

However, it must be added that such "I will explain my evidence" type calculations can only be used with great caution!

3. An interesting detail is that the largest value that occurred for the smallest number, and the smallest value that occurred for the largest number, are exactly arithmetic mirrors: $65+26=91$.