ML Interview Q Series: Divide and Conquer: Finding the Heaviest Marble Using a Balance Scale

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3. How do you weight 9 marbles three times on a balance scale to select the heaviest one?

To solve this classic puzzle, you want to isolate the single heaviest marble out of nine marbles with at most three uses of a balance scale. A balance scale can only tell you if one side is heavier, lighter, or if both sides are balanced. The main insight is to divide the marbles into equal groups, compare them, and then continue narrowing down the candidates based on which group or marble must be heavier.

Below is a detailed explanation of how you can do this step-by-step, along with considerations for different outcomes and potential trick questions an interviewer might ask. The explanation focuses on the logic of dividing, comparing, and deducing which marble is heavier under each scenario. We’ll then explore common follow-up questions in further depth.

Strategy Overview

You have 9 marbles: label them M1, M2, M3, M4, M5, M6, M7, M8, M9. You can arrange them into three groups of 3:

• Group A: M1, M2, M3

• Group B: M4, M5, M6

• Group C: M7, M8, M9

The reason for having three groups of equal size is that each weighing on a balance scale can eliminate multiple candidates at once. One weighing tells us which of two groups is heavier (or if they are equal), and that knowledge informs the next step.

First Weighing: Compare Group A vs Group B

Place Group A on the left side of the scale and Group B on the right side of the scale:

• Case 1: A = B in weight If the scale balances exactly, then none of the marbles in A or B is heavier. This means the heaviest marble is in Group C (M7, M8, M9).

• Case 2: A > B If Group A is heavier than Group B, then you know the heaviest marble is in Group A (M1, M2, M3).

• Case 3: A < B If Group B is heavier than Group A, then you know the heaviest marble is in Group B (M4, M5, M6).

At this point, you will have reduced the problem from 9 marbles to just 3 marbles in the heavier group (or in Group C if A and B balanced).

Second Weighing: Compare Two Marbles from the Heavier Group

From the group of 3 marbles identified as “heavier” (let’s call this group H), select any two marbles. Suppose H = (M1, M2, M3) for illustration. We weigh M1 vs M2:

• Case 1: M1 = M2 in weight If they balance, both M1 and M2 are not heavier than each other, so the heaviest marble must be M3 (the one not weighed).

• Case 2: M1 > M2 If M1 is heavier than M2, then M1 is definitely the heaviest.

• Case 3: M1 < M2 If M2 is heavier than M1, then M2 is the heaviest.

Notice that by the second weighing, you already have your answer. You technically don’t even need a third weighing, because after the second weighing, you can identify the single heaviest marble.

However, the puzzle specifically mentions using the balance scale three times. In many versions of this riddle, it’s understood that you have “up to three weighings” to find the heavier marble. In practice, you only need two weighings if you divide them into groups of three:

1. A vs B to determine the heavier group (or C if A=B).

2. Within that heavier group, weigh two marbles to see which is heavier or if they are equal (in which case the third one is heavier).

In a real puzzle context, the third weighing is typically superfluous once you’ve narrowed it down. But if the requirement is strictly to “use the scale three times,” you can always do a redundant check. One way is:

• After identifying, for example, M3 as the heaviest from the second weighing, you could weigh M3 against any other known lighter marble (say M1) for the third weighing just to confirm M3 is indeed heavier. That’s purely optional and doesn’t alter the conclusion.

Subtleties and Pitfalls

1. What if you weighed A vs B and forgot to weigh or keep track of which side was heavier? This might sound silly, but losing track of which side was heavier can derail your entire strategy. Always remember the heavier side or if there was a balance.

2. What if the marbles are extremely close in weight so that the scale is not clearly tipping? The puzzle assumes a perfect balance scale with no measurement error. In real-world scenarios, small measurement inaccuracies can complicate the result.

3. Could you accidentally misplace a heavier marble if you re-group them incorrectly after the first weighing? Yes, that can happen if you mix the marbles from different groups without a well-defined scheme. That’s why labeling each marble and carefully tracking them after each weighing is critical.

4. Is there a scenario where the puzzle changes if there are more marbles or a different number of weighings? Absolutely. The approach scales up, but you need a different group size and a more systematic way to handle more marbles within a fixed number of weighings. The general logic is to optimize each weighing so you eliminate the maximum number of possibilities each time.

5. How does this puzzle compare to the “Counterfeit Coin” or “12-coin problem”? The difference is that in the “Counterfeit Coin” puzzle, you might not know whether the counterfeit coin is heavier or lighter; plus, you can have more weighings or sometimes fewer. This changes the search strategy. In the 9-marble puzzle, you know exactly that one is heavier, making the approach simpler.

Potential Follow-Up Questions in an Interview

Below are some common follow-up questions that a FANG interviewer might ask to probe the depth of your understanding. After each follow-up question, there is a blank line before the in-depth answer to maintain clarity.

What if you had to confirm the answer exactly on the third weighing?

If the interviewer insists that you must use the third weighing to explicitly confirm the heaviest marble, you can still follow the main two-step solution but incorporate a third check. For example:

1. First weighing: A vs B to find the heavier group (or deduce it’s in C if they balance).

2. Second weighing: Pick any two marbles from that heavier group and weigh them.

3. Third weighing: If the second weighing didn’t already identify the heaviest (for example, in the rare scenario you suspect a measurement error or want an explicit final confirmation), weigh the presumed heaviest marble against a known lighter marble.

Since the puzzle states “How do you weight 9 marbles three times on a balance scale to select the heaviest one?” you can just note that the third weighing might serve as a verification step: weigh the identified heaviest marble (from the second step) against any other marble. If it’s heavier, that confirms your final answer.

Could there be a different grouping strategy that still works in exactly three weighings?

Yes, but the essence is the same. You want to split the marbles in such a way that after the first weighing, you’ve reduced the number of possible marbles that can be the heaviest from 9 to 3. Any grouping that still ensures this reduction will work. The most straightforward grouping, though, is indeed 3 marbles in each group.

One theoretical variation is to weigh a smaller subset first, but that can overcomplicate the second and third weighings. The standard grouping (3 vs 3) is the most elegant and direct.

What happens if there was a trick and one marble was lighter instead of heavier?

If the puzzle changes such that one marble is lighter (instead of heavier) and the rest are all identical heavier marbles, the methodology is similar except at each step you look for the lighter side or equality. You would still do group comparisons. However, you’d be searching for a lighter marble, so you pay attention to which side is lighter or whether the scale balances.

Can you show an example of incorrectly dividing the groups that would fail in three weighings?

A faulty approach might be:

• First weighing: Weigh 4 marbles vs 4 marbles.

• If they balance, the remaining single marble is heavier. That part sounds simple.

• But if one side is heavier, you have 4 marbles to test and only two weighings left. You can weigh 2 vs 2 in the second weighing:

  • If one side is heavier, you still have 2 marbles in that heavier side but only one weighing left, which might still let you identify the heavier one—so maybe it still works.

  • However, as soon as you do the second weighing, you can determine the heavier pair, then weigh 1 vs 1. Actually, in some sense, you can do it in three weighings too. So it’s not necessarily a failure, it’s just a less elegant approach.

In practice, the standard approach of “3 vs 3 in the first weighing” is more canonical, easier to remember, and straightforward to execute.

How could you write a quick Python check to simulate the weighings?

Below is a contrived, illustrative Python snippet that demonstrates how you might simulate weighings. It’s not strictly necessary for such a small puzzle, but it shows how an algorithmic approach could code the steps.

import random

def find_heaviest_marble(marbles):
    # marbles is a list of weights, each is a float or int
    # len(marbles) should be 9
    # The heaviest marble is the maximum by weight

    # Group them into 3 groups of 3
    groupA = marbles[:3]
    groupB = marbles[3:6]
    groupC = marbles[6:]

    # First weighing: compare sum of groupA vs sum of groupB
    weightA = sum(groupA)
    weightB = sum(groupB)

    if weightA > weightB:
        # The heavier marble is in groupA
        candidates = groupA
    elif weightB > weightA:
        # The heavier marble is in groupB
        candidates = groupB
    else:
        # If equal, the heavier marble is in groupC
        candidates = groupC

    # Second weighing: pick the first two marbles in candidates
    if candidates[0] > candidates[1]:
        heavier_marble = candidates[0]
    elif candidates[1] > candidates[0]:
        heavier_marble = candidates[1]
    else:
        # If they're equal, it must be the third one
        heavier_marble = candidates[2]

    # Third weighing (optional confirmation)
    # Weigh the identified heavier marble against a known lighter one
    # In practice, we already know the answer, but let's just do a final check
    # for demonstration.
    # We'll compare heavier_marble to the first marble of the entire list:
    if heavier_marble > marbles[0]:
        return heavier_marble
    else:
        return marbles[0]  # theoretically should never happen if we were right

# Example usage
if __name__ == "__main__":
    # Create marbles with random weights but keep track of index
    # Let's keep all marbles weight = 1 except one that is heavier
    marbles = [1]*9
    heavier_index = random.randint(0, 8)
    marbles[heavier_index] = 2  # heavier marble

    result = find_heaviest_marble(marbles)
    print("Heaviest weight found:", result, "Actual heavier index:", heavier_index)

The above code is just a demonstration. It embodies the same logic: group into three sets, weigh the sums of each set, pick the heavier set, weigh two marbles in that set, and so on.

Could we do it faster if we had more sophisticated tools than a balance scale?

If you had a scale that gave you the exact numerical weight, you could theoretically measure all 9 marbles at once or measure them one by one. But the puzzle is specifically about a balance scale that only compares two sets of marbles and tells you which side is heavier or if they are equal. Under that constraint, the minimal number of weighings to guarantee finding the heaviest among 9 marbles is indeed 2 weighings (3 if you count a verification step).

Conclusion of the Core Logic

1. Divide 9 marbles into three groups of 3.

2. Weigh the first two groups.

   • If one side is heavier, you know the heavier marble is among those 3.

   • If they balance, the heavier marble is in the third group.

3. Weigh any two marbles from that heavier group. The heavier side identifies the heavier marble, or if they balance, it’s the one not weighed.

4. (Optional) Use the third weighing to confirm your finding if the puzzle requires explicitly using the scale three times.

This approach is guaranteed to find the heaviest marble within three weighings, and often you only need two for the actual discovery. The third one can serve as a validation step if strictly needed.

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What if the interviewer asks about the reasoning behind each decision in more depth?

You might emphasize the concept of “information gain” in each weighing. Each weighing should eliminate as many possibilities as possible. By dividing the marbles into three equal groups, you ensure that the first weighing narrows down your candidates from 9 to just 3 (because if the scale balances, you know all 6 marbles on the scale are not the heavier ones; if it doesn’t balance, you immediately know the heavier group). This significant reduction from 9 possibilities to 3 possibilities is exactly what makes the puzzle solvable so efficiently.

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How does this puzzle relate to more general decision-tree or search problems?

In machine learning and algorithms, you often consider how to minimize the maximum depth of a decision tree (like in a 20 Questions game, or in binary search). Each weighing is akin to a branching in a decision tree. A well-designed grouping (3 vs 3 vs 3) ensures each test (weighing) splits the possibilities into roughly equal or manageable subsets, minimizing the depth needed to arrive at a single candidate. That general principle appears frequently in computational complexity and decision-making problems.

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Below are additional follow-up questions

If one of the marbles has a very slightly heavier weight that might not be detected by a non-ideal scale, how do we address potential real-world measurement errors?

In real-world scenarios, physical scales might have a small margin of error or limited sensitivity. This can complicate identifying the single heaviest marble if the difference in weight is very subtle. Even if you’re certain there is one heavier marble, you could end up with ambiguous outcomes, for example, the scale might appear balanced when in fact there’s a minuscule difference.

To handle this:

• You could repeat each weighing multiple times to average out small random errors. If a slight difference is present, over several trials, the heavier side should tip more frequently.

• Another approach is to ensure that the weight difference is large enough relative to the measurement error. For instance, if you suspect that the difference between normal marbles and the heavier marble is extremely small, you might re-weigh subsets in different configurations to confirm consistency.

• As a further step, you can try rotating marbles between the two pans to confirm that the same marble always tips the scale in its favor. If M1 and M2 appear the same in the first weighing but you suspect measurement fluctuations, swap M1 with M2’s position or weigh them alone.

A real-world pitfall is underestimating how repeated small-scale calibration issues might mislead you into thinking both sides are balanced when they’re not. This underscores how puzzles like these assume an ideal, perfectly calibrated scale. When the margin of error is not negligible, you must develop additional validation steps and replicate weighings to be fully certain which marble is heavier.

How would you modify the strategy if you had to ensure no marble is ever weighed more than once?

This constraint significantly changes the puzzle. In the classic approach, the heaviest group from the first weighing is narrowed down, and you weigh two of those marbles again. That means at least some marbles are weighed twice. If you are not allowed to re-weigh any marble:

1. You must design a schedule of weighings where each marble appears at most once on the scale.

2. A naive way is to weigh three distinct groups of three marbles each in three weighings, but the puzzle requires each weighing to produce useful information. If you weigh group A vs group B in the first weighing, you can’t reuse those marbles in subsequent weighings. This means your second weighing might involve group C vs group D, but you only have nine marbles, so constructing brand-new groups from leftover marbles is tricky.

3. One possible approach is a “tournament bracket” style, but with each marble weighed only once, you get little direct comparison data. You’d end up not having enough data to guarantee isolating the heaviest in just three weighings because you wouldn’t have a direct way to compare winners across weighings without re-weighing them.

In practice, this constraint is so limiting that, with nine marbles, you cannot ensure identification of the single heaviest in only three weighings if no marble is weighed more than once. You’d need either a different puzzle structure (fewer marbles or more weighings) or accept that some marbles must be weighed more than once.

The pitfall here is that you might try to replicate the standard three-step approach without noticing you’ve reintroduced a marble onto the scale. Another subtle issue is that any error in labeling or tracking these marbles becomes even more detrimental because you cannot recheck a previous measurement.

What if you need to prove rigorously that two weighings suffice (or do not suffice) without the optional third check?

To prove that two weighings suffice in the ideal puzzle scenario:

1. Show that the first weighing (3 marbles vs 3 marbles) narrows the search down to exactly 3 candidate marbles (either the heavier group or the group left out when the scale balances). This step alone is crucial for ensuring you can handle 9 possibilities in a single weighing.

2. Demonstrate that with one additional weighing among the 3 candidates, you can identify the single heaviest marble definitively. By weighing two of them: if they balance, the third is heaviest; if not, the heavier side is immediately clear.

A key subtlety:

• You must confirm that none of the possible outcomes leaves ambiguity. For instance, if the first weighing is balanced, you confidently dismiss six marbles. If it’s not balanced, you confidently dismiss six marbles that are known lighter.

• If you consider alternative groupings, you must verify that no scenario leaves more than 3 candidates after the first weighing; otherwise, the second weighing would not be guaranteed to find a unique heaviest.

The pitfall is forgetting to prove you can do it for all outcomes. Many puzzle solutions show you how to handle the “best-case” scenario. But a rigorous proof must show that even in the worst-case path (like if the first weighing was balanced and you move to the third group), the second weighing still resolves the heavier marble.

How can you adapt the puzzle if the heaviest marble could possibly weigh the same as the others?

This variant is trickier because you’re no longer guaranteed one marble is strictly heavier. The logic changes:

• After you weigh two groups of three, if they balance, you either suspect the heavier marble is in the third group or it doesn’t exist at all (i.e., all marbles are the same).

• You would then need an additional step to verify whether the potential “special” marble is truly heavier or actually the same as the others. You might weigh one candidate from the third group against a known standard marble (one from the balanced group) to check.

• If that second weighing also balances, you conclude all marbles are of equal weight. If it tips, you find the heavier one.

A pitfall arises if you incorrectly assume there must be a heavier marble. You could end up misattributing a normal marble as heavier due to scale or measurement errors. Also, if you have multiple marbles that could be heavier but all weigh the same, you need a strategy to confirm uniqueness or to identify that there’s no single heavier marble at all.

How do you handle the situation if you suspect multiple marbles might be heavier, but only one is the “heaviest of all”?

If more than one marble can be heavier than normal but you only need the absolute heaviest:

• You could end up with multiple candidates that are heavier than the baseline. One approach is to weigh in a way that eliminates the obviously lighter or baseline-weight marbles quickly. However, if some marbles have slightly varying degrees of heaviness, you might need more comparisons.

• In the standard puzzle, it’s assumed all marbles but one are identical. The moment you drop that assumption, you can’t reliably isolate just one heaviest marble in only three weighings without additional structure. The puzzle’s simplicity dissolves because each weighing might only tell you partial information about a subset.

A subtle pitfall is discovering that even if two marbles are heavier, they might differ by a minuscule margin. If your scale only shows which side is heavier, you might never know which of the two heavy marbles is heavier unless they go head-to-head on the scale. That might require extra weighings.

Can we assign numeric values to the marbles and use a weighting code approach (like a ternary search) to find the heavier one more systematically?

Yes, there is a concept of coding your marbles with labels or weights in a systematic way—this is seen in some variants of the “coin-weighing” problems:

• You might place marbles in certain combinations on each side of the scale, effectively creating a ternary-based system of reasoning. The outcome of each weighing (left heavier, right heavier, or balanced) can be mapped to digits in a ternary representation.

• However, with only three weighings and nine marbles, a straightforward grouping is still the simplest approach. Ternary coding is more relevant when the puzzle is extended to more marbles while still keeping the number of weighings small.

A subtle pitfall is that while a code-based approach is elegant for large expansions of the problem (like 27 marbles or 81 marbles in four weighings), it can be overkill for the simpler 9 marble scenario. Furthermore, any slip in setting up those coded weighings or misreading the scale’s outcome can ruin the entire scheme.

What if the marbles are of different colors or shapes, and you have extra information suggesting some might be more or less likely to be the heaviest?

When you have prior probabilities (e.g., you suspect a certain marble is more likely to be heavier), you could adapt your weighing strategy to reduce the expected number of weighings. For instance, if you strongly suspect marble M9 is heavier, you might place it directly on one side in the first weighing.

Yet for a worst-case guarantee (the puzzle’s typical requirement), you still need a systematic way to ensure you isolate the heavier marble in three weighings no matter what. In an interview context, highlight that “expected-case optimization” vs “worst-case guarantee” is an important distinction:

• Expected Case: You put your high-priority suspects on the scale first, hoping to resolve the puzzle faster in typical scenarios.

• Worst Case: You still ensure that if your suspicion was wrong, you can pivot your strategy to examine the other marbles thoroughly in the remaining weighings.

Pitfall: leaning too heavily on prior information might produce a strategy that fails if the assumptions are incorrect. Realistically, you want an adaptive approach that can incorporate new scale outcomes efficiently.

Does this method extend cleanly to 27 marbles using three weighings?

If you scale up to 27 marbles, one might recall a classic approach: you can still do three weighings if you break them into three groups of nine. However, that method only works if you’re certain one of the 27 marbles is heavier. After the first weighing (9 vs 9), you reduce the problem to 9 marbles, and the second weighing (3 vs 3) reduces it to 3. The third weighing identifies the single heaviest from that final trio.

Yes, you can handle 27 marbles with three weighings by a 9-9-9 grouping. The puzzle with 9 marbles is essentially a smaller instance of this 3^3 arrangement. Each weighing gives you three possible outcomes, so in theory, 3 weighings can uniquely identify one heavier item among 3^3 = 27 possibilities.

Pitfall: you must ensure you don’t attempt a misguided grouping that doesn’t neatly funnel you down to 3 or fewer marbles after the second weighing. The 27 scenario is a direct extension only if you keep dividing by three at each step.

Is there a risk of mixing marbles between weighings and losing track of which group was heavier?

Practically speaking, if you don’t label or track your marbles systematically (M1, M2, M3, etc.), you can easily forget which ones were on which side. Especially after the first weighing, you might accidentally swap marbles from the heavier side with others in a second weighing and end up with ambiguous results.

To mitigate this:

• Always mark or list the marbles in each group.

• Physically separate groups after each weighing: place the heavier group in a “candidate” area, and keep the other marbles aside so they don’t get confusingly mixed.

• Double-check your notes before proceeding to the next weighing.

This real-world housekeeping issue is often overlooked in the puzzle’s theoretical solution. In an interview, emphasizing accurate labeling and systematic note-taking can show attention to detail, which is critical in large-scale, real-world testing or data experiments.

How would you verify or test your strategy is robust if the puzzle was scaled to many more marbles, but you still only have three weighings?

When scaling up the problem, the essential factor is the maximum number of distinct outcomes three weighings can produce. Since each weighing has three possible results (left heavier, right heavier, or balanced), three weighings can produce at most 3^3 = 27 distinct outcome paths.

To systematically verify if your strategy works for N marbles in 3 weighings:

• Ensure N ≤ 27. You need the number of possible marbles to be no more than the total unique outcomes you can get from your weighings.

• If N > 27, you cannot guarantee identifying a single heavier marble in only three weighings because the decision tree doesn’t have enough leaves.

• Map each possible scenario (which marble is heavier) to a unique sequence of weighings outcomes. This ensures that any real outcome path points to exactly one heavier marble.

A subtle pitfall is forgetting that each outcome path must be uniquely associated with only one of the N marbles. If you have 28 marbles, there’s no way to design weighings that produce 28 distinct outcome sequences in just three weighings. This ties directly to the concept of information theory and decision-tree complexity: each weighing must partition the possibilities into up to three sets, and you need enough partitions overall to distinguish all marbles.

Could you use a trick like combining marbles in one pan to match the suspected heavier marble?

Sometimes in puzzles, a clever step is to put known baseline marbles on one side to equalize the suspected heavier marble on the other side, but that usually requires knowledge of each individual marble’s weight or at least the combined weight. In the 9 marble puzzle, you don’t know the exact heavier weight, just that one marble is heavier:

• You can’t “equalize” the scale reliably without a reference weight. The puzzle scenario doesn’t provide a standard reference marble of known weight.

• If you try to guess how many normal-weight marbles match the heavier marble’s weight, you’re back to speculation. So it typically doesn’t help in systematically identifying the single heavier marble.

This is a pitfall in many puzzle variations: overcomplicating the approach by trying to guess or approximate weights in a single pan. The standard “divide-and-weigh” method remains the most foolproof for a puzzle that only reveals heavier/equal/lighter outcomes.

If the marbles were extremely large and heavy in a real-world experiment, how does the practicality of the puzzle change?

Physical constraints come into play:

• A giant scale might be needed, and physically placing large marbles on the scale is more cumbersome. The time and labor of setting up each weighing can be non-trivial.

• You must also consider safety aspects if these are truly large or dense objects. Additionally, environmental factors like friction on the scale or slight differences in gravity distribution could matter if the marbles are significantly larger.

• The puzzle logic remains sound, but the real-world implementation might require forklift-scale operations or mechanical assistance. Each weighing could be error-prone due to mechanical friction or the angle at which each marble rests.

One subtle real-world pitfall: if the marbles are so large that the difference in weight is overshadowed by measurement noise, your scale readout might be effectively useless. For instance, if each marble weighs hundreds of kilograms and the difference is only 0.1 kg, you need a scale with very high accuracy to detect that difference.

How do you ensure that the puzzle logic remains the same if you do the weighings in a different order or sequence?

The puzzle typically follows a sequence:

1. Compare Group A vs Group B.

2. Select the heavier group or, if they balance, select the remaining group.

3. Compare two marbles from that group.

You could reorder these steps, but the principle must remain: the first weighing must reduce the pool from 9 to 3 marbles. If you start in a less optimal configuration, such as weighing only 2 marbles vs 2 marbles first, you might not reduce the candidate pool as quickly, potentially forcing extra weighings.

As soon as you deviate from the optimal partition (3 vs 3 vs 3), you have to double-check that your subsequent weighings still conclusively narrow down the possibilities. An interviewer might ask you to prove your alternative sequence yields the same worst-case resolution. The main pitfall is inadvertently creating a scenario where you can’t isolate the heavier marble in just three weighings because your first comparison was too small or not informative enough.

Could this puzzle strategy be applied if we had some marbles heavier and some lighter, but one is still the absolute heaviest?

If multiple marbles differ from the standard weight in both directions (lighter and heavier), the puzzle transforms into a more complicated version:

• Each weighing can have multiple reasons for an imbalance: you could have one heavier marble on one side and one lighter marble on the other side. Parsing the results can be extremely complex.

• Traditional “divide into three groups” might not directly solve the puzzle unless you have a well-defined approach for tracking how many heavier or lighter marbles are in each group.

A major pitfall is that the single measurement from the first weighing may not be sufficient to parse which marble caused the tilt. You might end up needing a more advanced strategy or more weighings. Indeed, the puzzle’s guarantee that exactly one marble is heavier simplifies the logic drastically. When that guarantee is removed or replaced, the complexity ramps up significantly.

If you discover after two weighings that you already know the heaviest marble, why might a puzzle insist on a third weighing?

Some puzzle wordings say “use the scale three times,” which might lead you to think all three weighings are mandatory. In many standard solutions, you only need two weighings:

1. Identify the heavier group (or the group not on the scale if they balance).

2. Identify the single heavier marble from the final group of three.

You might still perform a third “confirmation” weighing for completeness if the puzzle text is strict that you must “perform three weighings.” For instance, the puzzle setter might want to ensure that each step of the logic is physically confirmed rather than inferred.

The pitfall is misunderstanding the puzzle requirements. If it literally demands three weighings, you can’t claim you solved it in two weighings unless you add an explicit final check. From a purely logical standpoint, you don’t need the third weighing. But puzzle constraints can sometimes be about procedure rather than minimal logic.

How can overconfidence in your initial assumption lead to a wrong conclusion in an interview setting?

In an interview, a candidate might jump straight to the statement, “We only need two weighings. Done!” This is correct logically, but the question states “How do you weight 9 marbles three times on a balance scale to select the heaviest one?” Some interviewers might then probe, “Are you sure you’re following instructions about using the scale three times?” or “What if the third weighing is explicitly required?”

Being too quick to say “two weighings suffice” might appear as not listening to instructions or missing a puzzle nuance. Also, you might fail to consider verifying the result in the third weighing. The real pitfall is not clarifying whether the puzzle demands a strict usage of all three weighings or if it merely states an upper limit of three weighings. Always confirm the puzzle’s exact constraints in an interview setting.

What is a typical real-world lesson from the 9 marble puzzle regarding data collection and experiment design?

A key lesson is that each “experiment” or “measurement” should maximize information gain. In the puzzle, the “experiment” is placing marbles on the scale. By dividing the marbles into equal groups, you ensure each measurement splits the possibilities in the most balanced way possible, enabling a quick narrowing of candidates.

In practical data science or machine learning:

• You want to design experiments (like feature selection or A/B tests) that reduce uncertainty quickly.

• Inefficient testing or partial data might force more experiments (or weighings) to gain the same level of certainty.

A subtlety is that in real-world experimentation, an unbalanced setup might still be chosen if it’s easier to implement or if certain specialized knowledge justifies it. But from a purely information-theoretic standpoint, a balanced split is usually optimal, echoing the puzzle’s logic.