I'm playing a computer game and I want to boost the high score. I assume that attempts are independent (say, you do not improve by playing), and in each game my achieved points are lognormally distributed with mean 500 and variance 1’000’000. How often do I have to play so that your high score exceeds 1000 with probability at least 0.7 ?
My first guess would be to compute the value of a r.v. X which is my high score and set that P(x>1000)=1-P(x<=1000)= 1-0.7. But I don't know how to handle that information
The "high score" you mentioned is the sample maximum of a sequence of random variables. We denote $$ X_{(n)} = \max\{X_1, X_2, \ldots, X_n\} $$
A key fact you need to know first is $$ \Pr\{X_{(n)} \leq x\} = \Pr\{X_1 \leq x, X_2 \leq x, \ldots, X_n \leq x\}$$
By assuming those $X_i$ are i.i.d., the RHS becomes $\Pr\{X_1 \leq x\}^n$
Now back to your question. I assume you are really quoting the moments but not the usual parameters of lognormal distribution, so we need to convert back first. From
https://en.wikipedia.org/wiki/Log-normal_distribution#Arithmetic_moments
We have
$$ \mu = \ln\left(\frac {500^2} {\sqrt{1000000 + 500^2}}\right) = \ln\left(\frac {500}{\sqrt{5}}\right) = \ln(100\sqrt{5}) \approx 5.40988914221 $$ $$ \sigma^2 = \ln\left(1 + \frac {1000000} {500^2}\right) = \ln5 \approx 1.60943791243$$
Therefore $$ \Pr\{X_1 \leq 1000\} = \Pr\{e^{\mu + \sigma Z} \leq 1000\} = \Pr\left\{Z \leq \frac {\ln1000 - \mu} {\sigma}\right\} = \Pr\left\{Z \leq \frac {\ln(2\sqrt{5})} {\sqrt{\ln5}}\right\}$$
where $Z$ is a standard normal random variable. The above value is approximately $0.8811371$ by R. Last we solve the inequality:
$$ \begin{align} 1-\Pr\{X_1 \leq 1000\}^n \geq 0.7 &\iff \Pr\{X_1 \leq 1000\}^n \leq 0.3\\ &\iff n\ln\Pr\{X_1 \leq 1000\} \leq \ln0.3 \\ &\iff n \geq \frac {\ln0.3} {\ln\Pr\{X_1 \leq 1000\}} \approx 9.514405 \end{align}$$
So the minimum required $n$ is $10$.