# Identification of a fake coin. [duplicate]

I'm studing a bit of probability and I have a simple question about biased coins.

Is it possible, with an adequate numebr of tosses, to say that a coin is biased witha given probability?

In other words: how many toss are needed to say that the probability that a coin is unbiased is less that a given value?

I think that my question is different from How to determine the number of coin tosses to identify one biased coin from another? because I have only one coin and I don't know what is the probability of a face of my biased coin. I can suppose that the answer to my question is linked to this probability, but how?

## marked as duplicate by mathcounterexamples.net, Math1000, Ethan Bolker, amWhy, LeucippusJun 21 at 2:33

• @mathcounterexamples.net: That's not exactly the same problem; it's about two different coins. It may be possible to map the problems to each other, but then you should point out how. – joriki Jun 20 at 20:11
• This is typically done with hypothesis testing, which should arise in a statistics course. You assume the coin is unbiased, and show with tests that such an assumption is likely to be false. – Kaynex Jun 20 at 20:15

@Stan Tendijck gave a good answer in the Frequentist way of looking at things: You have no pre-conception of how likely the coin would be to be biased, but based on your results of (say) 10000 tosses and 5100 heads you can say that an unbiased coin would be closer than 50-50 than this result at least 19 times out of 20, so you suspect the coin is biased.

But an equally good approach is to assign some prior likelihood of bias (say you think that most coins are fair but one in a thousand would have a bias to come up heads at some amount uniformly distributed on $(0,1))$. Then when you get 5200 heads out of 10000 tosses, you can say that with this data the chance the coin is biased is only one in 50 or so! But if you continued flipping and got 510,000 heads out of 1,000,000 tosses the same calculation would say the coin is biased with probability (about) 99.9999%.

This second approach is called "Bayesian" statistics. It has the advantage of being more logically consistent but the disadvantage that your conclusion depends on your incoming "prior" assumption. That is mitigated by the fact that for large numbers of trials, the influence of your original "prior" gets smaller and smaller.

The standard approach here would be to take a to be determined sample size $n$.

Under the assumption of the coin being not biased, which should be your null hypothesis, the average times you see tails equals $1/2$. The variance of this average is given by $\frac{1}{2}\cdot\left(1-\frac{1}{2}\right)/n = 1/(4n)$. Based on the central limit theorem you can conclude that (approximately) with $95\%$ certainty $$\overline{X}_n \in \left[\frac{1}{2} \pm 2\cdot\frac{1}{2}\sqrt{\frac{1}{n}} \right]$$ holds, where I took $2$ as an approximation of $z_{0.975}$. So, if your average is outside of this confidence interval, it might be the case that your coin is biased.

You could formulate a similar result for a general $p$ as probability for the coin being tails.

Problem is that there is no general way of approaching your question because it depends on how rigged the coin may be. For example if the probability $p$ of getting tails equals $1/4$, a relatively small amount of trials is needed to conclude biasedness of the coin. However, if the probability $p$ of getting tails equals $0.49$, way more trials are needed. Suppose this is the case, then you need approximately $10.000 = 1/(0.5 - 0.49)^2$ trials. This amount increases exponentially as $p$ tends to $0.5$.

I hope this helps you a little bit and if you have any question don't hesitate to ask them!

There is no fixed number of tosses to decide this. You can make a test, for example how likely is it that the number of heads deviates at least this far from the mean. The confidence in the result will usually be higher if the number of tosses is higher, but it still simply depends on your tosses.