Assuming I play a Lottery where I can buy a slip with exactly one random number from $1$ to $10$. The Probability of winning this lottery is $\frac{1}{10}$, as only $1$ winning number will be drawn.
Now I add a second lottery game on this Lottery Slip (again $1$ random number from $1$ to $10$). But, when I draw a winning number, this counts for both lotteries.
Example $1$: My Lottery Slip says "Game $1$: $5$ and Game $2$: $8. \implies$ Winning Number is $8\implies$ I won 2nd Lottery.
Example $2$: My Lottery Slip says Game $1$: $8$ and Game $2$: $8. \implies$ Winning Number is $8$ $\implies$ I won both Lotteries.
Example $3$: My Lottery Slip says Game $1$: $1$ and Game $2$: $8. \implies$ Winning Number is $7$ $\implies$ I lost both Lotteries.
What is the probability of at least winning once with a Lottery Slip of $10$ such games? Why?
Your answer of ~65 is almost correct.
However, you have ten of these "double lottery" games. That means the not winning rate is $0.9\cdot 0.9 = 0.81$, since you have two independent picks in one game. This occurs $10$ times, so $0.81^{10}\approx 0.1215$. Subtracting from $1$ gets the answer of $$1-0.1215 = 0.8785 = \boxed{87.9\%}.$$