Solving Coin-Weighing Problem (81 Coins, 1 Fake) Using Information Theory So I have a coin-weighing puzzle under these situations:

*

*There are 80 real coins and 1 fake coin (total of 81 coins).


*The real coins are all the same weight, and the weight of the fake coin is different from the real coin.


*The real and fake coins cannot be distinguished, other than weight.


*A balance is used to identify the fake coin.


*When using the balance, the same number of coins is placed on either plate. The result is either "the left plate is heavier," "the right plate is heavier," or "the two plates are the same weight."
According to prior research, this problem cannot be solved by using the balance 4 times or less. I tried to show this by using information theory as follows, but I feel like I am missing something here.

Label the coins 1,...,81, and let $C$ be the number of the fake coin. Also, let $W$ be a random variable defined so that $W=1$ when the fake coin is heavier than the real coin, and $W=0$ when lighter. At the initial state, $C$ and $W$ can be regarded as independent random variables following a uniform distribution.


Then, define random variable $R_k$ so that $R_k = 0$, $R_k = 1$, $R_k = 2$, when, the result after using the balance for the $k$th time is, respectively, "the left plate is heavier," "the right plate is heavier," or "the two plates are the same weight."


Then, $R_k$ can be regarded as uniquely determined by $R_1, ..., R_k,$ and the real values of $C$ and $W$. (Do I have to add additional proof that this is true?)


Assume that the fake coin can be surely identified by four measurements using the balance, then this means that the value of $C$ is determined when $R_1,...,R_4$ have been determined. Thus, using the entropy function and its chain rule, \begin{eqnarray*}
H(C, W, R_1, ..., R_4) &=& H(C) + H(R_1 | C) + H(R_2 | R_1, C) + H(R_3 | R_2, R_1, C) + H(R_4|R_3, R_2, R_1, C) + H (W| R_4, R_3, R_2, R_1, C)
\end{eqnarray*}


And, we also have\begin{eqnarray*}
H(C, W, R_1, ..., R_4) &=& H(W) + H(C|W) + H(R_1 | C, W) + H(R_2 | R_1, C, W) + H(R_3 | R_2, R_1, C, W) + H(R_4|R_3, R_2, R_1, C, W)
\end{eqnarray*}


Then we notice that $H(C|W) = H(C)$, as $C$ and $W$ are independent variables. Also, $H(R_1 | W, C) < H(R_1 | C), H(R_2 | R_1, C, W)  < H(R_2 | R_1, C) $, and so on. Thus, from the two equations, we get
$H(W) < H(W | R_1, ..., R_4, C)$


But we know that $H(W | R_1, ..., R_4, C) < H(W)$, so there is a contradiction.

OK. So I somehow arrived at a contradiction, showing that the problem cannot be solved by using the balance for times. But then I realized that the same logic applies when using the balance for any number of times, so apparently there is something wrong...
What sort of logic am I missing? I would appreciate any help.

The conclusion in this statement is wrong:

Then we notice that $H(C|W) = H(C)$, as $C$ and $W$ are independent variables. Also, $H(R_1 | W, C) < H(R_1 | C), H(R_2 | R_1, C, W)  < H(R_2 | R_1, C) $, and so on. Thus, from the two equations, we get ${\color{red}{H(W) < H(W | R_1, ..., R_4, C)}}$

The conclusion would be $H(W) > H(W|R_1,\ldots, R_4,C)$.

Your logic would be like saying "If $a+b=c+d$ and $b < c$, then $a < d$"  The inequality is backwards.
