This is a riddle… and I need your help to solve it.
Let me explain...
I have 12 marbles, 11 of which I know are identical, and one which looks the same, but
weighs a slightly different amount.
I need to figure out which marble is different.
I have a scale I can use to compare groups of marbles, but I can only use it three times.
And to make matters worse, I don’t know whether the odd-marble-out weighs more or
less than the other marbles.
How can I find the different marble?
To try figuring it out for yourself, pause the video on the next screen.
This is known as the 12-marble problem.
And to solve this, let’s start by weighing 4 marbles against 4 marbles, leaving 4 off
of the scale.
This is our first weigh, and it can go one of two ways: either the scale is balanced,
or it’s not.
If the scale is balanced, we know that the odd-marble-out is in the other 4, and that
the 8 on the scale are all normal.
Next, we can weigh 3 verified normal marbles, against 3 marbles from the questionable group.
If this is balanced, we know the fourth marble is the Odd-marble-out, and we can weigh it
against a normal one to find out if it weighs more, or less, than the others.
If the three and three aren’t balanced, however, then we know if the Odd-Marble-Out
weighs more or less than the others, and that it’s in that group of three.
Next, weigh two of those marbles against each other.
If they’re the same, you know the third marble is different.
If they’re different, you know which one is the Odd-Marble-Out.
But, what if the first time you weigh four marbles against four marbles they’re different?
You know that THE Marble is one of eight, and you still don’t know if it’s heavier
or lighter than the rest.
If it’s heavier, it’s one of the 4 on the heavy side, and if it’s lighter, it’s
one of the four on the lighter side.
You also have 4 marbles you know are normal and only two weighs left.
Here’s the trick:
Weigh three “heavy-side” marbles and one “light-side” marble against three normal
marbles and one “heavy-side” marble.
If it’s balanced, you know the odd-marble-out is lighter than the others, and that it’s
one of the 3 “light-side” marble not on the scale.
Weigh one of those marbles against another.
If it’s not balanced, the lighter one is the winner.
If it is balanced, then the third marble is it.
What if your second weigh isn’t balanced?
If it’s heavier on the 3-normal and 1-heavy-side side, then you know either the 1 heavy-side
marble is the odd-marble-out, and is heavier than the others, or the 1 light-side marble
on the other side is the odd-marble-out, and is light than the others.
Weigh the heavy-side marble against a normal marble to find out.
If the second weigh is heavier on the 3-heavy-side and 1-light-side marbles side, then you know
the odd-marble-out is heavier than the others, and is one of the 3-heavy-side marbles.
Weigh one of those marbles against another.
If it’s balanced, the third is your heavy-marble-out.
If it’s not balanced, the heavier one is your guy.
Getting to this solution can take a number of different paths.
You could brute-force a solution, by trying any and every possible combination of weighs
as a thought problem until you found a sequence that works, but this doesn’t sound very
elegant, or efficient.
More likely, you’d want to use both inductive and deductive reasoning to design your solution.
Deductive reasoning is “top-down” logic.
For example: if I weigh four marbles against four marbles, what does that tell me?
Inductive reasoning, on the other hand, is bottom-up.
In this problem , that’s a question like: If I have 3 marbles left, and I know if the
odd-marble-out is heavier or lighter than the rest, then I can solve it in 1 weigh,
so how do I get from 4 possibly heavy, 4 possibly light, and 4 normal to there?
We use deductive and inductive reasoning all the time in real life.
“If I leave at 5:30, what time will I get there?” and, “if I have to arrive by 6,
what time do I have to leave?” are just two simple examples.
Practicing this form of logic in marble-finding thought puzzles can help us get better, faster,
and more efficient at using that logic in
other situations.