Probability theory (heads and tails)

Sample size for a dichotomous variable with $p$ proportion of success:

$$N = p(1-p)bigg({Z over E}bigg)^2$$

where,

$Z$ is the $z$-value corresponding to your confidence interval (eg. $z$-value $1.96$ for $95%$ confidence interval),

$E$ is the margin of error (eg: $0.05$ for $5%$ MoE)

For the given example, let $p$ be the proportion of tails.

$$therefore 1 - p = 1.1 times p$$ $$implies 1 = 2.1p implies p = {10 over 21}$$

Substituting, we get

$$N = bigg({10 over 21}bigg)bigg(1 - {10 over 21}bigg)bigg({1.96 over 0.05}bigg)^2 = 383.28 approx 384$$

So, 384 trials will tell you with 5% margin of error in 95% Confidence Interval. This will change if you change your CI and margin of error parameters.