Depends on what you mean by 'truly random'. A distribution can be skewed, in which case the bias would itself be biased and then clearly different from a perfec

Depends on what you mean by 'truly random'. A distribution can be skewed, in which case the bias would itself be biased and then clearly different from a perfectly fair coin. Assuming you mean a uniform distribution, we can calculate this. Been a while since I've done stats, so bear with me... We can first argue from symmetry that we should expect as many heads as tails. But this doesn't mean that this is necessarily the same distribution as a perfectly fair coin (Bernoulli with p = 0.5). Anyway, the distributions we have here are a Bernoulli distribution and a uniform distribution X ~ {0, 1}; 0 with probability 1-q, 1 with probability q q ~ [0, 1], uniformly The expected value of X is E[X] = 0 * p(0) + 1 * p(1) = p(1) I think we could prove p(1) = E[q], but I'm honestly not sure, and don't know if that holds up in general. But from the above, we know it's 0.5 anyway. Now, we need higher moments: E[X^n] = 0^n * p(0) + 1^n * p(1) = p(1) = 0.5 The moment generating function of a probability distribution is defined as M_X(t) = E[e^(tX)]. There's a theorem in probability that two distributions with the same MGF are the same. You can see from the above that this has the same moments as a Bernoulli distribution with p = 0.5. Hence, a biased coin with a random uniform bias should be identical to a Bernoulli distribution. In fact, I believe the above argument works for not just uniform bias, but any random bias that is symmetric about p = 0.5.