Re: question about probability
The law of big numbers states that the average $\nu_min$ is close to the $E{\nu_min}$.
$E\nu_min$ can be calculated directly for this experiment.
$P(\nu_min=0)$=0.623576
$P(\nu_min = 0.1)$ = 0.3764034
$P(\nu_min = 0.2)$ = 0.00002;
and $P(\nu_min>=0.3)=0$ for the purposes of calculating the mean.
Therefore, $E(\nu_min)$=0.037644, and the average proportion of heads for c_min should be close to this number.
