I've been told to assume $6$ coins are weighed on a chemical balance (two-pan) scale. We are told exactly two coins are fake and that fake coins are heavier than real ones. I've been given two designs of experiments that were used to determine the fake coins (I don't think the designs will be necessary to show for my question).

I'm asked to create output tables for the two designs as well as calculate their entropies.

I first calculated their partitions and obtained:

Design 1: $2+2+1+1+1+1+1+1+1+1+1+1+1=15$

Design 2: $3+1+1+1+1+1+1+1+1+1+1+1+1=15$

My partitions agree with those provided on the answer sheet so I'm fairly certain these are correct.

The issue I have however, is that I have different entropies to those provided by the solutions, however, my lecturer has a tendency to make mistakes so the solution sheet isn't always trustworthy.

We both use the same formulas:

Design 1: $-(2 \cdot \frac{2}{15} log(\frac{2}{15}))-(11\cdot \frac{1}{15}log(\frac{1}{15}))$

I obtain $3.64$, he obtains $2.52$. He doesn't specify what the base is on his $log$ but since for general $n$ the number of tests is $log_{[2]}(n)$ I assumed all the $log$'s were base $2$.

Design 2: $-(\frac{3}{15} log(\frac{3}{15}))-(12\cdot \frac{1}{15}log(\frac{1}{15}))$

I obtain $3.59$, he obtains $2.32$.

Overall we get the same answer that Design 1 is better than Design 2 but I'm not sure why our values are different?

Have I used the wrong base for my $log$'s or has he just miscalculated? (Or have I miscalculated?)

If anyone can see where either one of us went wrong please let me know.



1 Answer 1


Just plugging in the values you've provided into a calculator and using $log_{[2]}$ gave me the same values you have obtained so I believe you're answers should be correct. Although I did try a few different bases for $log$ and I never got your lecturers values so perhaps you can ask them how they obtained them? Either way they'll either show you your mistake or correct the solution sheet.

  • $\begingroup$ Yeah I'll probably go and ask them, even though they are error prone I doubt they would make a simple miscalculation error and upload it as the solution without checking at least once. $\endgroup$
    – Sean E.
    May 11, 2022 at 16:31

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