Finding the Heavy Coin by weighing twice Suppose you have $100$ coins. $96$ of them are heavy and $4$ of them are light.  Nothing is known regarding the proportion of their weights. You want to find at least one genuine (heavy) coin. You are allowed to use a weight balance twice. How do you find it?
Assumptions: 
Heavy coins all have the same weight; same for the light coins.
The weight balance compares the weight of two sides on the balance instead of giving numerical measurement of weights.

I think this works.
Divide the coins into three groups: $A$ with $33$ coins, $B$ with $33$ coins and  $C$ with $34$ coins.
Weigh $A$ and $B$ against each other.
Now if $A$ is heavier than $B$, then $A$ cannot have two or more light coins, as in that case, $A$ would be lighter (or equal to $B$). Now split $A$ into groups of $16$ plus one odd coin. Weigh the groups of $16$ against each other. If they are the same, then any of those coins is heavy. If not, then any of the heavier 16 coins is heavy.
Consider the case when $A$ and $B$ are equal.
The possibilities for $A$, $B$ and $C$ are:


+-----------+------------+-----------+
|  A        |   B        |  C        |
+-----------+------------+-----------+
|  33H      |  33H       |  30H + 4L |
|           |            |           |
|  32H + L  |  32H + L   |  32H + 2L |
|           |            |           |
|  31H + 2L |  31H + 2L  |  34H      |
+-----------+------------+-----------+


Now move one coin from $A$ to $B$ (call the resulting set $B'$) and weigh it against $C$.
If $B' > C$, then the coin you moved from $A$ is a heavy coin.
If $B' = C$, then the coin you moved from $A$ is a light coin and the remaining coins in $A$ are heavy.
If $B' < C$, then all the coins in $C$ are heavy.
