I am bit puzzled on how we can interpret the posterior. Assume a coin which is 0.1 probable to be unfair. So our prior probability on the coin being unfair is 0.1, and being fair is 0.9. Also by unfair I mean, the probability of head is 2/3 instead of 1/2. Now, imagine a scenario where I toss this coin 10 times and I get 10 heads. So if I want to get the posterior, I do the following:

$$ P(\texttt{unfair}|\texttt{evidence}) = \frac{P(\texttt{evidence}|\texttt{unfair})*P(\texttt{unfair})} {P(\texttt{evidence})} = \frac{P(\texttt{evidence}|\texttt{unfair})*P(\texttt{unfair})} {P(\texttt{evidence}|\texttt{unfair})*P(\texttt{unfair})+P(\texttt{evidence}|\texttt{fair})*P(\texttt{fair})} =\\ \frac{(2/3)^{10}*(0.1)}{(2/3)^{10}*(0.1)+(1/2)^{10}*(0.9)} = \frac{0.0017}{0.0017+0.0009} = \frac{0.0017}{0.0026} = 0.653$$

Now how can I interpret this posterior? It's technically saying that the probability of the coin being unfair given the evidence is higher than the prior belief on the coin being unfair (0.653 > 0.1), but is still below the prior belief on the coin being fair (0.653 < 0.9). So, the only conclusion is, the coin is still more likely to be fair. So, in order to change our belief about the coin we have "being unfair", we need more evidence (i.e. 100 heads in a row). Is it correct?

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    $\begingroup$ Note that enclosing LATEX inside of dollar signs $ renders it as math. For example: \alpha is $\alpha$. $\endgroup$
    – Sycorax
    Commented Apr 27, 2015 at 2:51
  • $\begingroup$ For $n$ tosses of Heads to make your posterior suggest an unfair coin is now more likely than your initial prior probability for a fair coin of $0.9$, you need $\dfrac{(\frac23)^n}{(\frac12)^n} >(\frac{0.9}{0.1})^2$ i.e. $n > \dfrac{2\log(9)}{\log(\frac43)} \approx 15.3$ so $n \ge 16$. This is unlikely even if the coin is unfair in way described: $(\frac 23)^{16}\approx 0.0015$ $\endgroup$
    – Henry
    Commented Sep 5, 2023 at 12:49

1 Answer 1


Using the same approach, you can compute the posterior probability that the coin is fair. (Do the exercise!) What do you make of the result?

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    $\begingroup$ $$\frac{0.0009}{0.0026} = 0.34$$ which is lower than 0.653. Based on the updated posterior for fairness, we can conclude the coin is more likely to be unfair? $\endgroup$ Commented Apr 27, 2015 at 4:28
  • $\begingroup$ It should actually be 0.347, namely 1 - 0.653, and, yes, you would conclude that after seeing your experimental results, the probability of a fair coin is 0.347 and of an unfair coin is 0.653. Both priors (originally 0.1 and 0.9) must be updated. $\endgroup$ Commented Apr 27, 2015 at 4:38
  • $\begingroup$ @user3697176 Why must the priors be updated? Because of A difference, B significant difference or C other? $\endgroup$
    – BCLC
    Commented Aug 5, 2015 at 21:01
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    $\begingroup$ C. You have additional information. At the start you believe that the coin is very likely to be fair, but then you find evidence (10 heads in a row) that does not occur very often if the coin truly is fair. This should shift your believes towards thinking the coin might be biased. $\endgroup$ Commented Aug 5, 2015 at 21:46
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    $\begingroup$ @BCLC: The size of the difference has nothing to do with it. You update every-time there is new evidence. The updated probability P(unfair|evidence) is usually called the posterior probability, and it can, of course, be used as a prior probability for the next round. $\endgroup$ Commented Aug 9, 2015 at 12:26

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