You toss the coin n times, and you have observed 60% of times, it is heads.

How large does n need to be in order to achieve 95% confidence that it is not a fair coin?


Attempt: Basically use Binomial distribution, but I have no idea how to take account of the 60% number into my calculation.

$\mu +/- Z_\alpha \frac{S}{\sqrt(n)}, \text{so I have, } 0.6 * \frac{\sqrt(n)}{0.6*\sqrt{0.6*0.4}} = Z_{0.05} =1.96$, so $n$ = 1.6


2 Answers 2


You want $n$ large enough that a confidence interval of the form $\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}},$ where $X$ is the number of heads and $\hat p = X/n,$ does not include $0.5.$

Roughly speaking the standard error is $\sqrt{.6(.4)/n}$ and the margin of error is about $2\sqrt{.6(.4)/n}\approx 0.98/\sqrt{n}.$ And you want the margin error to be less than $0.1,$ so something around $n = 96$ should suffice. I show examples with $n=100$ below.

n = 100;  x = 60;  z = qnorm(c(.025,.975))
CI = .6 + z*sqrt(.24/100);  CI
[1] 0.5039818 0.6960182

A superior kind of CI due to Agresti and Coull uses the point estimate $\tilde p = (x+2)/(n+4) = 62/104 = 0.5962$ and the endpoints are at $\tilde p \pm 1.96\sqrt{\tilde p(1-\tilde p)/104}.$ This interval also just misses covering $1/2.$

p.est + qnorm(c(.025,.975)) * sqrt( p.est*(1-p.est)/104 )
[1] 0.5018524 0.6904553

Finally, a Jeffries 95% CI uses quantiles $0.025$ and $0.975$ of the distribution $\mathsf{BETA}(60+0.5, 40+0.5),$ so that the interval is $(0.5023,0.6920).$

qbeta(c(.025,.975), 60.5, 40.5)
[1] 0.5022567 0.6920477

Depending on the kind of interval you are using and whether you want the smallest number just large enough so that the CI doesn't contain $1/2,$ I'll leave the rest to you.

  • $\begingroup$ I think it is possible to solve this problem using a Bayesian approach as I saw here, however, I don't really know why the y-axis of figure 4 is not in the 0 to 1 range. $\endgroup$
    – JMFS
    Jan 4 at 1:01
  • $\begingroup$ Link is to a whole book, not sure I see the relevance. $\endgroup$
    – BruceET
    Jan 4 at 8:19

I worked out similar math for the problem of seed germination, to estimate the population germination rate from a sample of n seeds of which k germinate.

The formula I got for the CDF is:

Computer-friendly Binomial CDF

where x is the germination probability. Substituting your problem variables in, letting n be the number of coin flips, k is 0.6*n, so the resulting formula would be:

Question-specific CDF

which you can solve for n to give you the number of total flips that will exclude a fair coin with 95% credibility. Note that this is a CREDIBLE interval instead of a CONFIDENCE interval. I don't know how this will work with an exact 60% instead of integer values for n and k, but the general CDF formula can get you there for any arbitrary credible interval. Let x equal the "fair coin probability" of 0.5, k be the number of heads, and n the total number of flips.

I actually made a Javascript calculator for the seed germination problem that you could co-opt for this use. There is also a link to a PDF of the worked-out math if you want to take a look or check it. Hope this helps!

UPDATE: I brute forced the math with my calculator and I calculate that the 95% credible interval excludes a fair coin (50% probability) at between 95 and 100 coin flips. A plot of the credible interval evolution with n is attached.

Coin flippz


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.