# Estimating the probability that a coin drawn at random will land on Heads

Suppose we have a large pool of biased coins with different Heads-probability; the Heads-probability is a random variable with some complicated distribution (e.g. truncated normal). We want to estimate the probability that, if we draw a coin at random and toss it, it will land on Heads. I consider the following three experiments:

• Draw $$n$$ coins; toss each coin $$m$$ times; count the number of Heads and divide by $$m n$$.
• Draw $$m n$$ coins; toss each coin one time; count the number of Heads and divide by $$m n$$.
• Draw $$1$$ coin; toss it $$m n$$ times; count the number of Heads and divide by $$m n$$.

Note: the pool is much larger than $$m n$$, so the fact that we draw $$m n$$ coins does not change the distribution (alternatively, we can draw a coin $$m n$$ times with replacement).

Are these experiments equivalent? If not, which of them would give a better estimate of the actual probability?

• Is this a homework?
– Tim
Jun 10 '20 at 17:12
• @Tim no... I am too old for homework. It is a question that arised in a research project in voting theory. Jun 10 '20 at 17:14
• I’m asking because the last option is so obviously bad, that it sounds like a homework problem.
– Tim
Jun 10 '20 at 17:15
• Why not $mn$ times draw with replacement a coin and toss it once?
– Tim
Jun 10 '20 at 17:16
• Are you at all interested in estimating a particular coins bias, or are you just interested in the probability of a head? Jun 10 '20 at 17:18

All three of your methods are unbiased estimators of the mean, but have different variances:

• Drawing $$m n$$ coins and tossing each coin one time has the smallest variance
• Drawing $$n$$ coins and tossing each coin $$m$$ times has a variance between the other two
• Drawing $$1$$ coin and tossing it $$m n$$ times has the greatest variance

so in a sense drawing $$m n$$ coins is the best of these options if you want to estimate the probability that, if we draw a coin at random and toss it, it will land on Heads.

The middle option might give you a better idea of the shape of the overall distribution, but that is not the question you asked.

(Another unbiased method would be drawing one coin and tossing it once. This would have an even higher variance.)

• It'd be great if you could actually prove how they differ in variance and that they are unbiased.
– Tim
Jun 10 '20 at 17:51
• @Tim - drawing one coin and tossing it once is unbiased - its expectation conditioned on being that coin is $p$ for that coin, and so its unconditional expectation is the mean of the distribution. The three bulleted cases are then unbiased by linearity of expectation. The variances can be found by the law of total variance Jun 10 '20 at 21:30