Given a bag of 10 coins, 9 are ordinary coins and one is a double headed coin. You select one coin at random and toss it three times. It comes up heads each time what is the probability its the double header?
This can be solved using bayes rule the answer is $\frac{8}{17}$.
However the follow up question asks how many tosses would you need to be 95% sure that the coin is double headed?
Based on Mark Galek's answer here
The probability of getting a head n times is:
$(\frac{1}{2})^n$
and I need to find the value of n that gives us 95% confidence that the coin is not fair.
$(\frac{1}{2})^n=0.05$
$n=ln(0.05)/ln(\frac{1}{2})$
n=5
Is this correct my gut instinct is that it is a little low?
Here I have used the CI for the fair coin to try and find how many heads would be required for us to be 95% sure that it isn't fair. Given that the only other choice here is for the coin to be double headed I believe this is correct.
Is there any other way this calculation could be done without using the CI for the fair coin?
Best Answer
As pointed out at Number of tosses to ensure $95\%$ that the coin selected is double-headed, the existing answer to this question is incorrect, as it doesn't take into account the given information on how the coin was chosen.
The existing answer answers the question how many times you'd have to throw heads in a row so that the probability for a fair coin to produce that result would be less than $5\%$. I understand the question to ask instead how many times you'd have to throw heads in a row so that the conditional probability for the coin to be fair, given how it was chosen, would be less than $5\%$.
As calculated by @Idonknow in the question linked to above, the answer is $8$. The conditional probability for the coin to be double-headed after $n$ heads in a row is
\begin{eqnarray*} P(\text{double}\mid\text{$n$ heads}) &=& \frac{P(\text{double}\land\text{$n$ heads})}{P(\text{$n$ heads})} \\ &=& \frac{P(\text{$n$ heads}\mid\text{double})P(\text{double})}{P(\text{$n$ heads})}\;. \\ &=& \frac{1\cdot\frac1{10}}{1\cdot\frac1{10}+\left(\frac12\right)^n\cdot\frac9{10}} \\ &=& \frac1{1+9\cdot2^{-n}}\;. \end{eqnarray*}
For this to be $\ge95\%$, we need $1+9\cdot2^{-n}\le\frac1{0.95}$ and thus
$$ n\ge\log_2\frac9{\frac1{0.95}-1}=\log_2171\approx7.4\;, $$
so we need $8$ consecutive heads to be $95\%$ sure that the selected coin is the double-headed coin.
Note that the number of fair coins appears in the numerator of the argument of the logarithm. That means that every time we double the number of fair coins, we need one more throw of heads to reach the same level of certainty. For instance, if we had $4\cdot9=36$ fair coins and one double-headed one to choose from, we'd need $10$ consecutive throws of heads to reach $95\%$ certainty that we chose the double-headed one.