4
$\begingroup$

A king has rounded up 1000 people suspected of counterfeiting coins, which look and feel the same as the official coin. However, only the official coin is truly fair (Pr(heads)=0.5), while all fake coins produce strongly skewed coin flip results (biased towards heads or tails).

The king decides to identify the counterfeiters by flipping a coin taken from each of the suspects. He leaves the actual flipping to 6 of his guards, handing them a suspected fake coin AND one official coin. The latter is to assure the king that his guards carried out the coin flips faithfully. Unfortunately, the king forgets to tell the guards how often to flip each coin, only that both coins (fake and official) must be flipped an equal number of times. All 1000 suspected fake coins tested in this manner go through the hands of all 6 guards.

Because heads of proven counterfeiters will roll, the king cannot afford to be wrong and commit too many innocent people to die. But he's comfortable executing up to 50 innocent people.

How should the king analyze the coin flip data from his 6 guards to identify the counterfeiters among the 1000 suspects? Can he also find a way to keep the number of innocently convicted under 50?

To get started, here's a sample from the first suspect (heads/tails):

Guard 1: suspect coin (88/11), official coin (42/47)
Guard 2: suspect coin (38/5), official coin (22/21)
Guard 3: suspect coin (115/15), official coin (70/60)
Guard 4: suspect coin (39/33), official coin (35/37)
Guard 5: suspect coin (70/13), official coin (43/40)
Guard 6: suspect coin (22/18), official coin (19/21)

Thank you!

$\endgroup$
1
  • 1
    $\begingroup$ This question is weird and I love it. $\endgroup$ – David Marx Nov 13 '13 at 3:36
4
$\begingroup$

If the guards are independent of each other and the tosses are fair then it doesn't matter which guard tossed which coin or how many times each guard tossed each coin. The results for each coin can be grouped together. Thus for the coin that you give data the grand result is 372 heads from 467 tosses (a fairly convincingly biassed coin).

Rank the coins in order of the ratio of the likelihood of the maximally likely Pr(heads) divided by the likelihood of Pr(heads)=0.5 and the owners of the coins with the 50 highest ratios are your 50 best choices of culprits.

The likelihood function you need is:

$$ L(\theta) \propto \binom{n}{h}p^h(1-p)^{n-h} $$

where $\theta$ is the set of all possible values of $p$, Pr(heads), $n$ is the total number of tosses and $h$ is the number of heads observed. Plug in $p=\frac{372}{467}$ to get the likelihood of the most likely value of $p$ for the coin in your question and $p=0.5$ and divide the two values to get the likelihood ratio that represents the maximal strength of the evidence for that coin being biassed.

There is no need to do a significance test for this problem and so you do not need to combine P-values.

You can set criteria for how strong the evidence needs to be before you sentence a coin owner to death, or you can just kill the 50 against whom the evidence is strongest.

$\endgroup$
4
  • $\begingroup$ Thank you for your answer, Michael. I upvoted & implemented this solution as well as Glen_b's on the original question and see similar results (his uses the simulated data while yours manages without it). I will likely accept this answer as well. My only remaining reservation is that in my actual data I want to avoid lumping together the 6 "guards" as that will result in giving more weight to the those that flipped the coin more often, perhaps disproportionately (flips can vary over 2 orders of magnitude). $\endgroup$ – reviewer3 Nov 13 '13 at 5:43
  • $\begingroup$ Also, the limit of 50/1000 was specific to the innocent. That's why I suggested flipping a fair coin as well to determine an FDR. There is no limit on actual counterfeiters (theoretically all 1000 could be guilty). But I very much like the idea of ranking the coins based on their likelihood ratios. Thank you! $\endgroup$ – reviewer3 Nov 13 '13 at 5:48
  • $\begingroup$ @Stefan The likelihood function weights the guards in the proper proportion to how many times they flipped the coins. More flips is more evidence and so should be more heavily weighted. The function I give weights each flip equivalently as there is no information in the fact that more than one guard was involved (under the assumptions I state). $\endgroup$ – Michael Lew Nov 13 '13 at 5:56
  • $\begingroup$ You're absolutely right of course. In this case there is no benefit to knowing that more than one guard was involved, as all 6 flip the same coin. The data that inspired this question consisted of 6 biological replicates, however, which may introduce some confounding factors, hence the caution against weighing one replicate over the other. But in the context of this question here, I completely agree. $\endgroup$ – reviewer3 Nov 13 '13 at 6:10

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.