The probability that the serial number, which was drawn 3 times, does not win is $\frac{700}{1000}=\frac{7}{10}$, because out of 1000 possible cases, 700 do not win. Accordingly, the probability that the other serial numbers do not win is $\frac{800}{1000}=\frac{4}{5}$, and $\frac{900}{1000}=\frac{9}{10}$. Therefore, the probability that the owner of the ticket does not win with five tickets, despite 10 attempts, according to the multiplication rule is

$$
v=\left(\frac{7}{10}\right)^{2} \cdot \frac{8}{10} \cdot\left(\frac{9}{10}\right)^{2}=\frac{49 \cdot 8 \cdot 81}{10^{5}}=\frac{31752}{100000}=0.31752
$$

thus, the probability of winning is $(1-0.31752=0.68248)$ more than twice as large as the probability of not winning.

Uray László (Bp., V. Piarista g. II. o. t.)
A certain lottery ticket owner, from among the various serial numbers of their tickets, had five serial numbers drawn in the first draw, specifically two numbers 3 times each, one number 2 times, and two numbers 1 time each. What is the probability that the ticket owner did not win despite having 10 attempts? (It is known that each series consists of $1000$ tickets, of which only one 100-ticket group wins in each draw of serial numbers.)