ML Interview Q Series: Strictly Increasing Dice Rolls: Calculating Probability with Combinations.

Browse all the Probability Interview Questions here.

22. Say you roll three dice, one by one. What is the probability that you obtain 3 numbers in strictly increasing order?

Below is a comprehensive, in-depth explanation of how to approach and solve this problem, followed by potential tricky follow-up questions that might be asked in a rigorous interview setting. Each follow-up question is also in H2 format, and each answer is provided with as much clarity and thoroughness as possible.

Intuitive Understanding and Core Logic

A straightforward way to count the valid outcomes is to recognize that “strictly increasing” means each value must be unique and ordered from smallest to largest. Another perspective is:

• First, choose any 3 distinct faces out of the 6 (from {1, 2, 3, 4, 5, 6}). • Then, for strictly increasing order, there is exactly one way to arrange those 3 chosen numbers in strictly ascending order.

The number of ways to select 3 distinct faces out of 6 is given by the combination:

Deep Dive into the Reasoning

Total Number of Outcomes

Counting Strictly Increasing Triplets

Final Probability

Dividing by the total of 216 equally likely outcomes,

Edge Cases and Common Pitfalls

Strictly increasing means no two dice faces can match. A common mistake is to ignore the fact that dice can show the same number, but that would violate strict inequality. Another pitfall is to overcount by counting permutations instead of recognizing there’s only one strictly increasing arrangement per unique combination of three faces.

Potential Implementation in Python

Below is a very simple Python snippet that enumerates all possibilities and directly counts those that are in strictly ascending order, to confirm the analytical result.

count = 0
total = 0

for x1 in range(1, 7):
    for x2 in range(1, 7):
        for x3 in range(1, 7):
            total += 1
            if x1 < x2 < x3:
                count += 1

probability = count / total
print("Count of strictly increasing sequences:", count)
print("Total sequences:", total)
print("Probability:", probability)

Running this would print:

Count of strictly increasing sequences: 20
Total sequences: 216
Probability: 0.09259259259259259

which equals 5/54

Possible Follow-up Questions

Below are some additional questions an interviewer at a top tech company (like Uber, Google, Meta, etc.) might ask to probe depth of knowledge and reasoning. Each follow-up is posed in H2 format, followed by a thorough explanation.

How would the answer change if the question asked for non-decreasing order instead of strictly increasing order?

Why do we use combinations and not permutations to count strictly increasing sequences?

For strictly increasing triples, picking any 3 distinct values from {1, 2, 3, 4, 5, 6} and then ordering them in ascending order is unique. Hence:

Could this probability be computed using a more manual enumeration approach?

Yes. One could literally write down each possible triple and check the conditions. But this manual approach is typically tedious. Since we know that each outcome has a probability of 1/216, we only need to count valid triples. The combination argument or direct programmatic enumeration is much more straightforward.

How might this change if the dice were biased?

If the dice were biased, the probability of landing on any particular face would no longer be uniform (1/6). One would have to:

What if we rolled four dice and wanted them in strictly increasing order?

You can generalize the concept. For four dice in strictly increasing order (all distinct and strictly ascending):

How would you ensure your simulation code is correct in a real production environment?

In a production or large-scale environment, you would:

• Validate the code with smaller tests or unit tests where you know the exact expected outcome (like this example with a known closed-form solution). • Possibly run random sampling on a large scale, ensuring that the empirical frequency of strictly increasing rolls converges to the analytical result. • Keep track of edge cases: if the dice or random number generator is not truly fair or not truly random, you might see deviations. • Monitor for performance issues if you are enumerating or simulating a very large scenario.

Could Markov chains or other advanced probabilistic tools be relevant here?

Typically, for a small, finite scenario like rolling three dice, advanced tools such as Markov chains are not necessary. However, if the problem changed to a longer sequence of dice rolls (e.g., 10 dice in strictly increasing order) or the question involved conditional probabilities or more complex states, you might model transitions as states in a Markov chain. For three dice, the combinatorial approach is simpler and more direct.

What if the interviewer asks you to provide the reasoning behind the combination formula?

The logic is:

• Strictly increasing triple from a set of 6 faces means no duplicates are allowed. • Each triple can be viewed as a subset of size 3 from {1, 2, 3, 4, 5, 6}, because once selected, there’s exactly one way to arrange it in ascending order. • The number of ways to choose 3 distinct elements out of 6 is given by:

---

The number of ways to choose 3 distinct elements out of 6 is given by (6 choose 3) = 6! / (3! * 3!). This results in 20 such subsets, each corresponding uniquely to a strictly increasing triple. Hence you get 20 favorable outcomes among the 216 total.

How can we confirm there are no off-by-one or other miscounting errors?

A good way to confirm: • Double-check the sample space size: 6^3 = 216 is correct. • Cross-verify with direct computation: – Count any small portion by hand. For instance, if the first die is 1, then the second die can be 2 through 6, the third can be bigger than the second, etc. Summing up these small slices by hand can match the final count of 20. – Write a quick script to brute force all outcomes and check how many are strictly increasing (as shown in the Python snippet). Confirm it yields 20.

Everything aligns, so the probability of 5/54 is correct.

If someone tries a purely arithmetic approach, how might they do it?

They might attempt: • Let X1, X2, X3 be random variables for the outcomes. • Probability(X1 < X2 < X3) = Probability of all permutations of (X1, X2, X3) being distinct times the fraction of permutations that are in ascending order. • For any three distinct faces, the probability to appear in one particular order is (1/6)^3 for the fair dice scenario. • Then note that if three values are distinct, there are 3! = 6 ways to order them, only 1 of which is strictly ascending. • Next, compute how often X1, X2, X3 are distinct: – The probability that we get three distinct rolls is: (6 * 5 * 4) / (6^3) = 120/216 = 5/9. – Among those distinct outcomes, 1 out of 6 permutations is strictly ascending. – So (5/9) * (1/6) = 5/54.

This matches the combination-based reasoning perfectly.

Below are additional follow-up questions

What if we only consider sequences that form a strictly increasing arithmetic progression?

A strictly increasing arithmetic progression of length 3 means the three outcomes (X_1, X_2, X_3) satisfy: [ \huge X_2 - X_1 = X_3 - X_2 \quad \text{and} \quad X_1 < X_2 < X_3. ] This implies there is a common positive difference (d = X_2 - X_1 = X_3 - X_2). For standard 6-sided dice (faces 1 through 6), the possible values for ((X_1, d)) are such that (X_1 + 2d \le 6) (because (X_3 = X_1 + 2d) must be at most 6). We must also have (d > 0) for the progression to be strictly increasing.

To enumerate systematically: • For (d = 1), (X_1) can be 1, 2, 3, or 4. That gives the triplets (1,2,3), (2,3,4), (3,4,5), (4,5,6). • For (d = 2), (X_1) can be 1 or 2. That gives (1,3,5), (2,4,6). • For (d = 3), (X_1) can only be 1. That gives (1,4,7), but 7 is outside the die range of 1 to 6, so no valid sequence here. • For (d > 3), you cannot fit three terms within 1 to 6 in strictly increasing order.

So the valid arithmetic progressions are: (1,2,3), (2,3,4), (3,4,5), (4,5,6), (1,3,5), (2,4,6).

There are 6 such strictly increasing arithmetic progressions in total. Among the 216 equally likely ordered outcomes for rolling three fair dice, only 6 yield an arithmetic progression of this form. Thus the probability is (6/216 = 1/36).

Potential pitfalls: • Forgetting that “strictly increasing arithmetic progression” imposes a positive difference (d). If (d) were allowed to be zero, you could have repeated values, which violates the strictly increasing condition. • Overlooking the upper boundary of 6 for the largest face, which invalidates progressions that extend beyond 6.

Edge scenarios: • If the die had more faces, say 10 or 20, you would do a similar enumeration but up to that new maximum face. • If negative or zero differences were allowed, it would not be “strictly increasing.”

How does the probability change if each die has a different number of faces (for example, 4-sided, 6-sided, and 8-sided dice) while we still want a strictly increasing result?

When the dice are not identical, we must carefully count the distinct ordered outcomes. Suppose the first die has (a) faces (numbered 1 to (a)), the second die has (b) faces (numbered 1 to (b)), and the third die has (c) faces (numbered 1 to (c)), with each face equally likely to appear on its respective die. The total number of possible outcomes is: [ \huge a \times b \times c. ] We want (X_1 < X_2 < X_3) where (X_1) is from ({1, \dots, a}), (X_2) from ({1, \dots, b}), and (X_3) from ({1, \dots, c}).

Approach for counting valid triples: • Consider each value (x_1) from 1 to (a). • For each (x_1), we can take (x_2) from ((x_1+1)) up to (b), provided (x_1 < x_2 \le b). • For each valid (x_2), (x_3) must be from ((x_2+1)) up to (c), provided (x_2 < x_3 \le c).

Hence the count of valid outcomes is: [ \huge \sum_{x_1=1}^{a} \sum_{x_2 = x_1+1}^{b} \sum_{x_3 = x_2+1}^{c} 1. ] In a real interview, you’d do explicit summation or a programmatic approach. For example, if (a=4, b=6, c=8):

1. Let (x_1) go from 1 to 4.

2. For each (x_1), choose (x_2) from ((x_1 + 1)) up to 6 if (x_1 + 1 \le 6).

3. For each (x_2), choose (x_3) from ((x_2 + 1)) up to 8 if (x_2 + 1 \le 8).

After enumerating, you divide by (4 \times 6 \times 8 = 192) total ordered outcomes. One could do a quick code snippet to do the counting automatically.

Potential pitfalls: • Mixing up the range: the first die’s maximum is (a), second’s is (b), etc. • If (a \ge b), there might be fewer or even no valid ways for (X_1 < X_2).

Edge cases: • If (a) is 1, it’s impossible to get (X_1 < X_2 < X_3). So the probability would be 0 if the first die has only a single face (or similarly if the second or third die has a range that forces no strictly increasing triple). • If one of the dice has fewer faces than the subsequent die, the enumeration might yield smaller or zero probabilities.

Overall, this scenario highlights that each die’s range of faces must be respected individually.

What if we need the three outcomes to be in strictly increasing order and also all be even numbers?

This follow-up question adds the condition that each of the rolled values must be both strictly increasing and even. For a standard 6-sided die, the even faces are 2, 4, and 6. We want an ordered triple ((X_1, X_2, X_3)) such that: [ \huge X_1, X_2, X_3 \in {2, 4, 6}, \quad X_1 < X_2 < X_3. ] We can enumerate quickly: • The set of even faces is ({2, 4, 6}). • If we pick all three, the only way to put them in strictly increasing order is ((2, 4, 6)). So there is exactly 1 favorable outcome. Out of 216 total outcomes, the probability is (1/216).

Potential pitfalls: • Forgetting that “strictly increasing” means we cannot skip or reorder them any other way. With only three distinct even faces, there is exactly one valid ascending triple if we insist all be even. • If the die had more even numbers (e.g., an 8-sided die with faces 2, 4, 6, 8), the count of strictly increasing even triples might be different.

Edge cases: • If we also allowed odd numbers or any partial condition, the counting changes drastically. • If the question asked for “non-decreasing” instead of “strictly increasing,” then duplicates could appear, and we’d have more scenarios to consider.

How can we handle the case where the dice are rolled repeatedly until a strictly increasing sequence appears, and we want the expected number of rolls?

In this scenario, each “roll” means rolling all three dice once. We keep repeating this 3-dice roll until ((X_1, X_2, X_3)) is strictly increasing for the first time. We want the expected (average) number of trials (each trial is one triple-roll) to see the first success.

If the probability of success on any single triple-roll is (p = \frac{5}{54}), and each trial is independent of previous attempts, then the expected number of trials until the first success in a Bernoulli process follows a geometric distribution with parameter (p). The expected value (E[\text{Trials}]) is: [ \huge \frac{1}{p}. ] Given (p = 5/54), we have [ \huge E[\text{Trials}] = \frac{54}{5} = 10.8. ]

Interpretation: • On average, you expect to roll the three dice about 10.8 times before seeing a strictly increasing triple for the first time.

Potential pitfalls: • Mixing up “number of rolls of a single die” vs. “number of triple-rolls.” We must clarify that each trial consists of three dice rolled at once. • If dice were not fair, (p) would differ, and the result would change accordingly.

How do we handle partial knowledge of outcomes or missing data in real settings?

Imagine a scenario where you rolled three dice, but you only observed two of them, or one result was hidden. You might have incomplete information on ((X_1, X_2, X_3)). A follow-up question is: “Given partial observations, what is the probability that the actual triple was strictly increasing?”

For instance, suppose you see (X_1 = 2) and (X_2 = 4), but you do not know (X_3). If the die is fair, then (X_3) could be any of ({1,2,3,4,5,6}) with equal probability (1/6). The triple is strictly increasing if (2 < 4 < X_3), so (X_3) must be in ({5,6}). Thus the conditional probability, given you know the first two are 2 and 4, is (2/6 = 1/3).

Potential pitfalls: • You must carefully condition on the observed values. If the partial data already breaks the strictly increasing chain (e.g., if you observe (X_1 = 4) and (X_2 = 2)), the probability is immediately 0. • In real-world settings, partial data is common, so thinking through conditional probabilities is key.

What if the dice are colored, and we only care about the set of values, not the order in which they appear?

Sometimes, a follow-up question might invert the problem: if we do not care about the order of the three dice, but only the distinct set of values that appear, does that affect our strictly increasing question? Strictly speaking, “strictly increasing order” is inherently an ordered property. If you only track the set of values, you lose the sequence information.

To adapt the question: • If the dice are physically distinguishable (e.g., red, blue, green), we keep the concept of an ordered triple. • If the dice are indistinguishable, we can’t talk about a strictly increasing sequence in the same sense; we’d only know that the set of numbers has 3 distinct elements. But the question about “strictly increasing” doesn’t quite apply if order is not observed.

Pitfalls: • Confusing “multiset of results” vs. “ordered triple” can lead to incorrect probabilities. • Some interview problems might ask how to handle permutations vs. combinations for dice that are identical or differently colored.

How can real-world factors like dice imperfections or rolling methodology alter the theoretical probability?

In a purely theoretical model, each face is equally likely with probability (1/6). Real dice might have imperfections; the table or rolling method might bias certain faces. In practice: • If the dice are physically unbalanced (heavier on one side), the probability of each face is not uniform. • You would then need an empirical approach: roll the dice many times, estimate each face’s probability, then compute the probability of (X_1 < X_2 < X_3) accordingly. • Even how a person tosses the dice can introduce subtle biases.

Potential pitfalls: • Ignoring the possibility of slight biases could lead to systematically wrong probability estimates. • Over a large number of trials, these biases might show up in the data (e.g., certain faces appear more frequently than 1/6).

Real-world usage often requires verifying assumptions or measuring them experimentally before relying on the “fair dice” model.