You've $8$ identical-looking coins. All the coins weigh the same but $1$ coin is lighter than the rest. You're given $3$ double-pan balance scales. $2$ of the scales work, but the $3$rd is broken and gives random results (it is sometimes right and sometimes wrong). You do not know which scale is broken. How many least number of weighings will it take to find the light coin?
At first, I thought it this way:
Imagine the $8$ coins in $3$ stacks: $2$ stacks of $3$ coins & $1$ stack of $2$ coins.
Place the $3$ coins' stacks in the $2$ pans of any of the $3$ scales. Observe the outcome of the weighing. Do the same thing for the other $2$ scales too.
$2$ of the $3$ weighings outcomes must be same since $2$ of the scales work properly.
So by these $3$ weighings, we'll get whether the light coin exists in any of the two $3$ coins' stacks (if it is there, then we must see $2$ same unbalanced outcomes in our first $3$ weighings; otherwise, the light coin is in the $2$ coins' stack).
It then takes only $1$ more weighing to identify the light coin from within that lighter stack. However, in that way, it is only possible if we can find the broken scale in our first $3$ weighings. But all $3$ outcomes of our first $3$ weighings could be same. Then we're not sure which scale is broken. We need to do the same $3$ weighings again and again until we get the $1$ different outcome in a $3$ weighings. So our solution depends on whether we're getting such outcome.
If we get that outcome in $n$th $3$ weighings (for some positive integer $n$), then the total number of weighings it would take to find the light coin from the $8$ coin is $3n + 1$. For instance, if we get the different outcome in our first $3$ weighings, then $4$ weighings are sufficient (in total).
But that's not a least number of weighings process. After a bit of discussion with my friends, one of my friends, Sami, suggested that, without even detecting the broken scale, we can find the light coin in $6$ weighings:
First, by $3$ weighings, we figure out the light coin's stack (just like the previous).
Then take $2$ coins from the light coin's stack and compare them on the $3$ scales. Since at least $2$ of the scales will give the same outcome, we'll definitely get the light coin from that weighing.
But we're not sure whether there exists a least number of weighings less than $6$ for this problem. How can we prove that? And how can we generalize this?