ML Interview Q Series: Lotto Draw Non-Overlap Probability: A Combinatorics & Total Probability Approach.
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Question: In the lotto 6/49, six different numbers are drawn at random from {1,2,…,49}. What is the probability that the next drawing will have no numbers in common with the last two drawings?
Short Compact solution
Define Ak as the event that the last drawing has k numbers in common with the second last drawing. Then:
is the probability P(Ak). Let E be the event that the next drawing has no numbers in common with the last two drawings. By the law of total probability:
Hence the probability is approximately 0.1901.
Comprehensive Explanation
Overview of the Events
We have three consecutive lotto drawings. Each drawing selects 6 distinct numbers out of 49 possible numbers:
The second last drawing.
The last drawing (immediately after the second last).
The next drawing (after the last).
We want the probability that the next drawing shares no common numbers with either the second last or the last drawings.
Defining Ak: Let Ak be the event that the last drawing has exactly k numbers in common with the second last drawing.
The second last drawing picked 6 numbers.
We are looking at the probability that, out of those 6 positions in the last drawing, exactly k match the second last drawing's numbers, while the other 6 − k are chosen from the remaining 43 numbers not in the second last drawing.
Probability of Ak: The number of ways to choose k out of the 6 that match the previous draw is "6 choose k" in plain text. The number of ways to choose the remaining 6 − k numbers from the other 43 is "43 choose (6 − k)". The total number of ways to choose the last 6 from the entire set of 49 numbers is "49 choose 6". Therefore:
P(Ak) = ( (6 choose k)*(43 choose (6−k)) ) / (49 choose 6).
Defining E (no numbers in common with the last two draws):
If the last two drawings together have exactly k overlaps, their combined distinct numbers total 6 + 6 − k = 12 − k.
Therefore, to have no overlap in the next draw, those 6 new numbers must all be chosen from the remaining 49 − (12 − k) = 37 + k numbers.
Probability of E given Ak:
Once we fix that the last two draws share exactly k numbers, the union has 12 − k distinct numbers.
The next draw must exclude those 12 − k.
So P(E | Ak) = ( (37 + k choose 6) ) / (49 choose 6).
Combining with total probability: We use the law of total probability to get the overall probability of E:
P(E) = sum over k=0..6 of [ P(E | Ak) * P(Ak) ].
Substituting:
P(E | Ak) = ( (49 − (6 + 6 − k)) choose 6 ) / (49 choose 6 )
P(Ak) = ( (6 choose k)*(43 choose (6 − k)) ) / (49 choose 6 )
This summation evaluates to approximately 0.1901.
Further Discussion and Possible Follow-up Questions
How do we interpret the combinatorial terms in detail?
When computing P(Ak), we consider how the last draw picks its 6 numbers:
We want exactly k of those 6 to match numbers in the second last draw. That choice is (6 choose k) because we pick which k out of the second last draw’s 6 numbers will appear again.
We want the other 6 − k to come from the 43 numbers that were not in the second last draw. That choice is (43 choose (6 − k)).
The total possible ways to choose 6 from 49 is (49 choose 6). Thus:
P(Ak) = [ (6 choose k)*(43 choose (6 − k)) ] / (49 choose 6).
Why do we sum from k=0 to 6?
The two consecutive drawings could overlap by anywhere between 0 to 6 numbers (since each draw has exactly 6 distinct numbers). Hence k can be any integer from 0 up to 6.
Explain how the union of the last two draws yields 12 − k distinct numbers
If the last draw has k numbers in common with the second last draw, then the total unique numbers in those two drawings is 6 + 6 − k = 12 − k. This arises from basic set theory: if set A and set B each have size 6, and they share k elements in common, then the size of the union is |A| + |B| − |A ∩ B| = 6 + 6 − k = 12 − k.
What if the next drawing is chosen right after the last drawing?
The draws in a lottery like 6/49 are typically independent from one draw to another (other than the numerical overlap we are analyzing). The event that exactly k matched previously does not affect how the next set of 6 numbers is picked from the 49, beyond telling us which numbers were possibly repeated. So each set of 6 is still a uniform random selection from the 49 numbers.
Why do we multiply ( (37 + k choose 6 ) )/(49 choose 6) by P(Ak)?
After conditioning on Ak, we have a scenario: exactly k overlaps in the previous two draws, so the union is 12 − k distinct numbers. We want none of these 12 − k to appear in the new draw, so we choose all 6 from the remaining 49 − (12 − k) = 37 + k. That probability is [ (37 + k choose 6 ) / (49 choose 6 ) ]. The law of total probability then tells us to weight this by how likely it was for the last two draws to have exactly k overlap, i.e., P(Ak).
Could there be any simplifications or alternative ways to approach the problem?
Yes. An alternative approach might involve enumerating the possible union sizes (12 − k) that the last two drawings can produce, then directly calculating the probability of that union size. But the law of total probability with Ak is typically the cleanest method to keep track of the overlap distribution.
How can we implement this computation in Python quickly?
Below is a simple Python snippet that uses math.comb (Python 3.8+) for binomial coefficients, then sums over the possible values of k:
import math
def comb(n, r):
return math.comb(n, r)
p_total = 0.0
for k in range(7): # k=0 to 6
p_Ak = comb(6, k)*comb(43, 6-k)/comb(49, 6)
p_E_given_Ak = comb(49-(6+6-k), 6)/comb(49, 6)
p_total += p_Ak * p_E_given_Ak
print(p_total) # Should be approximately 0.1901
This code uses the direct formula. The result should match the 0.1901 figure to a reasonable floating-point approximation.
Why is the final probability about 0.19?
Intuitively, having no numbers in common with the previous 12 unique numbers (when we do not know how much overlap those two draws had) is fairly restrictive. Yet it is not extremely rare: with 49 total numbers, there is still a substantial fraction of numbers left to choose from. Numerically, the final result is around 19%, indicating that in about 1 out of 5 draws, you will see zero overlap with the last two combined sets of numbers.
Could the probability be influenced by previous draws beyond the last two?
Under typical lottery rules and standard assumptions, each draw is from the same distribution with no memory beyond our combinatorial overlap calculation. We only considered the last two draws because we specifically want no overlap with exactly those two. Extending to more draws would require accounting for all previous sets, but the principle is the same, only more complicated combinatorics.
How might one approximate or simulate this probability?
A Monte Carlo simulation is a common approach in data science interviews for cross-checking combinatorial results. One could repeatedly:
Randomly draw 6 out of 49 for the second last draw.
Randomly draw 6 out of 49 for the last draw.
Randomly draw 6 out of 49 for the next draw.
Check if the next 6 is disjoint from the union of the second last and the last draws.
Aggregate the fraction of trials where there is no overlap.
Over many trials, the empirical fraction should converge near 0.19, aligning with the closed-form result.
Below are additional follow-up questions
What if the lottery had a bonus number drawn each time, and we wanted to exclude overlaps with that bonus number as well?
In many variants of lotteries, there is a main set of drawn numbers (e.g., 6 distinct numbers) plus a bonus or “power” number (sometimes chosen from a separate range, sometimes from the same 1–49 range). If we want to ensure that the new draw does not include any of the previous main numbers or the bonus number(s), we must adjust the count of excluded numbers accordingly.
Detailed Reasoning
Suppose each draw now includes 6 distinct “main” numbers plus 1 bonus number (making 7 total drawn numbers).
If the last two draws each have 7 distinct numbers (assuming they are from the same 1–49 range), then the union of those two draws could range from 7 + 7 − x distinct numbers, where x is the overlap between the sets of main+bonus from the two previous draws.
If we want absolutely no overlap with any of those main or bonus numbers, the next draw of 7 (6 main + 1 bonus) must be chosen from the remaining 49 − (14 − x).
We would again sum over all possible overlaps x, weighting by the probabilities of each overlap scenario.
The calculations become more cumbersome, as we now track the possible ways the bonus numbers might match or differ. But the same law of total probability approach remains valid.
A potential pitfall here is forgetting to account for the scenario in which the bonus number in one draw might be the same as a main number in another draw. This complicates how we count the overlap. Thus, we must be very explicit about how many distinct numbers were used in each previous draw and which sets we are excluding in the next draw.
How does this calculation change if each draw can contain repeated numbers (i.e., not necessarily distinct)?
Some lotteries or random selections might (in theory) allow repeats within a single draw—though this is rare in real-world lotto games. If repetition within a single draw were allowed, the combinatorial structure changes drastically.
Detailed Reasoning
In the classic 6/49 game, each draw is strictly a combination of 6 distinct numbers out of 49, so there are C(49, 6) possible outcomes for each draw.
If repetition is allowed, each draw is effectively a sample of 6 (not necessarily distinct) numbers from 49. There would be 49^6 possible outcomes (if order matters) or a different count if we only count multisets.
Overlap computation becomes more complex: the concept of “k matches” needs to factor in how many times a number appears. For example, if the second last draw had the number 10 three times and the number 11 one time, etc.
The probability that the next draw has no overlap at all would become even more nuanced, because we would exclude any instance of numbers that appeared in either of the previous draws, but each draw might feature those numbers multiple times.
A key pitfall is incorrectly applying “choose” formulas that assume no repetition. One must switch to the appropriate counting methods (multiset combinations) or simply re-derive from first principles how many total ways each draw can occur, and how to define “k matches” in that setting.
What if we suspect the draws are not truly uniform, but we do not know the exact bias?
In real lotteries, we typically trust that each number is equally likely to be drawn. However, if we suspect that the draw mechanism might be slightly biased toward certain numbers (e.g., mechanical bias in the draw machine), the exact combinatorial model would no longer be completely accurate.
Detailed Reasoning
Under uniform randomness, every 6-combination from 49 is equally likely. If the random mechanism is biased, certain combinations might be more likely than others.
We can still define events like Ak (“k numbers overlap”), but computing P(Ak) is no longer just [ (6 choose k)*(43 choose (6−k)) ] / (49 choose 6 ). Instead, we need to integrate (or sum) over the probabilities of each specific combination that yields exactly k overlap.
Without knowledge of the bias distribution, we can at best estimate or bound probabilities. For instance, we might attempt a maximum entropy approach or assume a small perturbation from uniform.
If the bias is severe or we have partial knowledge (e.g., certain “hot” numbers appear more frequently), we would incorporate a probability weight w(n) for each number n, normalizing so the sum of w(n) across all 49 numbers is 49. Then the probability of a particular 6-subset would be proportional to the product of the weights of the chosen numbers.
A pitfall is continuing to rely on uniform combinatorial formulas (like binomial coefficients over total combinations) when the underlying distribution is not uniform. This can yield misleading or completely inaccurate results.
What if the draw size is changed from 6 to n and the total pool is changed from 49 to N?
We might want to generalize from a “6 out of 49” lottery to drawing n out of N. The same question about “what is the probability that the next draw has no overlap with the previous two draws?” can be posed in a more general sense.
Detailed Reasoning
Let each draw select n distinct numbers from a set of size N.
Define Ak = “the last draw has k numbers in common with the second last draw.”
By an analogous argument to the 6/49 case, P(Ak) = [ (n choose k)*(N−n choose n−k ) ] / (N choose n ).
The union of these two draws would have 2n − k distinct numbers.
The probability that the next draw is disjoint from those 2n − k numbers is [ (N − (2n − k) choose n ) ] / (N choose n ).
Summing over k = 0..n yields the final probability: sum over k of P(Ak) * P(E | Ak).
A potential pitfall is forgetting that k cannot exceed n (nor be negative), or incorrectly substituting the parameters. Another subtlety is that as n grows larger relative to N, the chance of zero overlap becomes very small, so a purely combinatorial approach can yield extremely small probabilities that might cause floating-point underflow in a naive numerical implementation.
What happens if we only care about having no overlap with the last two draws in a single dimension, such as parity (odd/even)?
Sometimes, rather than caring about the specific integer values drawn, one might only look at a particular characteristic like “no overlap in terms of being all odd or all even.” This is a more abstract question but can illustrate how the principle generalizes to partial information about the numbers.
Detailed Reasoning
Suppose each draw is still from {1,..,49} with 6 numbers selected. But we decide we only care about whether the next draw shares any odd or even numbers with the set of previously drawn odd or even numbers.
For instance, if the second last draw had 4 even and 2 odd, and the last draw had 3 even and 3 odd, the union of their “even numbers” might be 5 distinct even integers total, and the union of their “odd numbers” might be 4 distinct odd integers total.
Ensuring no overlap in parity space might mean the next draw does not include any of those specific “even” or “odd” integers.
We would similarly partition the 49 numbers into odd/even, see how many odd/even were used in the union, and exclude them from selection.
The main pitfall is that once you reduce the problem to parity, you might lose the ability to precisely track which actual numbers were used. This approach only works if the question is specifically about parity or some other property. For the exact question of numeric overlap, you must use the original combinatorial approach.
How do we deal with computational challenges when implementing this for very large N?
If the lottery had, for example, n=10 out of N=1,000, or even bigger numbers, direct computation of binomial coefficients might become very large, risking overflow or precision errors.
Detailed Reasoning
Binomial coefficients like (N choose n) can be extremely large. Even (49 choose 6) is in the tens of millions, which is usually still safe in 64-bit integer arithmetic but may approach floating-point precision issues when dealing with the division of large numbers.
For bigger N (e.g., N in the thousands or millions), it might be necessary to use logarithms to compute these probabilities in a numerically stable fashion.
Alternatively, we can rely on specialized libraries for exact rational arithmetic, or do a prime factorization approach to handle binomial coefficients exactly.
Monte Carlo methods could also be used as an approximate approach for very large N.
A key pitfall is simply plugging large N and n values into direct binomial functions, encountering either run-time errors or floating-point inaccuracies that can yield results like 0.0 or NaN. Proper numerical techniques (e.g., log-sum-exp) are critical in large-scale problems.
Could we have a scenario where the two previous draws were known to be identical?
Sometimes, by a freak occurrence (albeit very unlikely), the last two draws might have drawn exactly the same 6 numbers. In that extreme case, the union of the last two draws is just those same 6 numbers, so if k=6, the union has 6 distinct numbers total.
Detailed Reasoning
If we consider the event that the two previous draws matched completely, that’s the k=6 case.
Then the question becomes: what is the probability that the next draw is disjoint from that repeated set of 6 numbers?
This is simply ( (49−6) choose 6 ) / (49 choose 6 ) = (43 choose 6) / (49 choose 6 ).
Because the probability of k=6 is quite small, in an actual scenario, it contributes only a small amount to the total sum.
However, it still must be included if we want an exact probability, which is precisely why the summation goes from k=0 to k=6.
One subtlety: if we know from real news or from some anomaly that the same 6 numbers appeared in the last two draws, we can condition on that fact and simply compute the one term (the probability of no overlap with those 6). This yields a simpler expression but obviously changes the final answer from 0.1901 to just that single conditional scenario’s probability.
In practice, do lotteries typically publish these probabilities to the public?
Lotteries often publish general odds of winning or partial matches, but less commonly publicize the probability of no overlap with previous draws. They might publish raw statistics (like how many times a particular number has been drawn historically). In principle, the same combinatorial logic could be adapted to produce “fun facts” about repeated or disjoint draws.
Detailed Reasoning
Because each draw is typically considered independent in the sense that each set of 6 from 49 is equally likely, the odds of any specific event (like no overlap with the last two draws) are constant over time.
The lottery commission might not find it necessary to highlight that the probability is about 0.19 for no overlap, but a mathematically inclined audience may find it interesting.
It’s also worth noting that players might exhibit behavioral patterns (e.g., picking the same “lucky numbers” repeatedly), but that doesn’t affect the official random draw’s probability distribution.
One potential pitfall is to assume that if a certain combination appeared recently, it’s less likely to appear again. True, it might feel improbable psychologically, but the official random mechanism does not have memory, so each combination remains equally likely each time.
How would we adjust if we had partial knowledge about the last two draws but not full?
Imagine a situation where we only know that the second last draw and the last draw overlapped by at least 1 but we do not know exactly how many. Or we only know that the union was at most 10 distinct numbers. We do not have the exact distribution across k=0..6, only partial constraints.
Detailed Reasoning
If we know the union had at most 10 distinct numbers, then we know 12−k ≤ 10, meaning k ≥ 2. So we restrict the sum to k=2..6.
The probability then becomes a conditional sum of the terms for k=2..6, normalized by the probability that k≥2.
More generally, if we have partial information about k, we incorporate it in the law of total probability by restricting the range of k to those consistent with our knowledge and renormalizing accordingly.
Alternatively, if we only know that “k≥1,” then we exclude the k=0 portion and rescale to ensure the probabilities sum to 1 within the feasible range.
A pitfall is forgetting to renormalize after restricting to partial information. For instance, if we only sum the probabilities for k≥2, we get a partial probability that must be divided by P(k≥2) to get a conditional probability.
Are there any real-world implications if a gambler tries to avoid previous draws?
A common question in lottery discussions is whether a gambler should avoid numbers that appeared recently, hoping to “improve” their chances of picking a unique combination. From a strict probability standpoint, it makes no difference—each combination of 6 from 49 is equally likely on every new draw.
Detailed Reasoning
Some players believe in “hot” or “cold” numbers. The no-overlap scenario is relevant if the gambler aims to pick sets that do not share numbers with recent draws, possibly to avoid sharing a jackpot with other players who might pick the same repeated numbers.
However, from the official draw perspective, this does not affect the underlying probabilities. The draw is still uniform, so the chance of winning is not inherently improved by avoiding or including previously drawn numbers.
The only real difference might come if many other players also avoid repeated numbers, slightly changing the distribution of chosen tickets among the player base. If you happen to pick repeated numbers and those repeated numbers come up again, you might have a bigger share of the jackpot (since fewer people would have chosen them). But this is more about game-theory and player behavior than about the fundamental drawing mechanism.
A pitfall is conflating player strategy with the actual probabilities of the draw itself. The random mechanical process (or RNG-based) is still uniform, so skipping recent numbers doesn’t change the probability of them being drawn next.
Are these results still valid if each drawing is done from 1..49 with replacement (i.e., between draws), but each 6-number draw remains without replacement?
Yes. Generally, standard lotteries do not remove numbers permanently from the pool across multiple drawings. After each drawing, all 49 numbers are “reset” for the next draw. So each draw is from the full 49 without replacement within that draw, but there is replacement between draws.
Detailed Reasoning
Each drawing is from the same set {1..49} with no knowledge of previous draws; that is exactly the usual assumption.
The question about overlap is purely a combinatorial count after the fact. The draws remain independent in the sense that each is a random combination of 6 from 49.
The formula for P(Ak) remains the same under this assumption, because for each new draw, the chance of picking any particular combination of 6 from 49 is 1 / (49 choose 6).
A subtle pitfall is mixing up “without replacement within a single draw” with “without replacement across draws.” The latter would be a scenario where once numbers are drawn in a previous drawing, they are no longer in the pool for the next draw. That is generally not how standard lotteries operate. If it were, the entire logic would have to be reworked because the sample space for the second draw would not be all 49 numbers.
If we wanted to calculate the probability that exactly one number is shared with the last two draws, could we adapt the formula?
Sometimes interviewers ask about partial overlap beyond the “no overlap” scenario. For example, “what is the probability the next draw shares exactly one number with the last two draws?”
Detailed Reasoning
We could define an event Fm = “the next draw has exactly m numbers in common with the union of the last two draws.”
We would again sum over k, the overlap between the second last and the last draws. Condition on Ak, then count how many ways the next draw can share exactly m out of the 12−k distinct numbers from the union.
So P(Fm|Ak) = [ (12−k choose m)*(49−(12−k) choose (6−m)) ] / (49 choose 6 ).
We would multiply by P(Ak) and sum for k=0..6.
The method is the same—just a more general counting approach.
A pitfall here is forgetting that the union of the two previous draws has size 12−k, not 12. Another common mistake is mixing up exactly m numbers shared with the entire union vs. exactly m shared with each draw separately.
How do we ensure we haven’t double-counted any configurations?
Double-counting is a common worry in combinatorial reasoning. In this particular approach, we avoid double-counting by carefully structuring the events based on k, the overlap between the second last and last drawings:
Detailed Reasoning
For each k in 0..6, we identify exactly those configurations of the second last and last draws that share k numbers, which form a disjoint event from those that share a different k’.
Within each such event, we count the ways the next draw can be disjoint from the union of the previous two draws. Because these counts are conditioned on that specific k-overlap event, we are partitioning the space of (second last draw, last draw) outcomes.
Summing over k is a partition-based summation, ensuring each scenario is counted exactly once.
A subtle pitfall might arise if we tried to define the overlap sets or union incorrectly. But as long as we fix k exactly for each scenario, the events are mutually exclusive, so no double-counting occurs.
Could random rounding or approximations affect an interviewer’s acceptance of the final numeric answer?
Interviewers will typically allow for a small rounding difference (e.g., 0.1900 vs. 0.1901 vs. 0.19). However, giving an answer like 0.19 is usually acceptable as an approximate decimal.
Detailed Reasoning
We might do the computation by hand or with a typical calculator and get 0.1901 (like in the official snippet).
Using a different floating-point approach might yield 0.19007, etc. Depending on the precision displayed, some might show 0.190.
Interviewers generally care about whether the candidate can derive the correct combinatorial expression and understand the reasoning. The final decimal is typically secondary, as long as it is close.
A pitfall is if you claim an exact fraction but do it incorrectly—some candidates might try to reduce the binomial expressions to a fraction of small integers. If done improperly, that can raise doubts. It’s safer to say 0.19 or 0.1901 to four decimal places unless you have confirmed an exact fraction.
Does the result change if we only want no overlap with the second last drawing (but overlap with the last drawing is allowed)?
This is a different question: “What is the probability that the new draw has no overlap with the second last drawing only?” We ignore whether it overlaps with the last drawing. The logic and the result are simpler.
Detailed Reasoning
In that scenario, the second last draw had 6 numbers. We just want the new draw to exclude those 6.
The probability of picking 6 distinct numbers from the remaining 43 is ( (43 choose 6 ) ) / ( (49 choose 6 ) ).
This is straightforward and does not require summation over k. We do not care how many numbers the last draw repeated from the second last draw.
A pitfall is accidentally mixing in the last drawing’s numbers if we only want to exclude the second last. Each scenario is a separate question and must be addressed with the correct set of excluded numbers.
How would you present your reasoning and code to a non-technical stakeholder?
Sometimes interviewers want to see if you can communicate these ideas to non-engineers. You might focus on a conceptual explanation: “We look at how much the previous draws overlapped. Then we see how many distinct numbers they used in total. We compute the chance that the next draw avoids all those numbers. We sum over all possible degrees of overlap.” You can also show a quick simulation or a small fraction–type example.
Detailed Reasoning
Use simpler examples with smaller sets (like drawing 2 numbers from 5) to illustrate. Show how the overlap might be 0, 1, or 2, then how we compute each probability.
Provide short, clear code or short tabular enumeration for small cases. For a real 6/49 lottery, a direct table would be huge, but the approach remains the same.
For a less technical person, highlight that about 19% of the time, the next draw will have no overlap with the last two draws combined.
A pitfall here is diving into heavy notation or advanced combinatorial formulas that can confuse a business stakeholder. Instead, aim to convey the essence: each draw is random, and we are calculating the chance that 6 new random picks avoid up to 12 previously used numbers.
Is there a generalized principle behind using the law of total probability in problems like this?
Yes. Any time you can partition the sample space into disjoint events (like A0, A1, …, A6), you can express the probability of an event of interest (E) as the sum over those disjoint events of P(E | Ak) × P(Ak).
Detailed Reasoning
In these problems, it is often easier to compute the probability of E in pieces, by conditioning on a simpler or more direct event Ak that slices the sample space.
The formula is typically: P(E) = ∑ P(E ∩ Ak) = ∑ P(E | Ak)P(Ak).
This approach is especially handy for overlap questions, because “the last two draws have k overlap” is a discrete partition: k=0,1,2,...,n.
Carefully enumerating each partition ensures we do not miss any scenario or double-count.
A pitfall is sometimes to attempt to handle the entire scenario in one step, which may become very complicated or lead to mistakes. Partitioning simplifies the logic, even if it requires an additional summation.
How might the results change if the number of draws we are excluding overlaps with is larger than 2?
If we wanted the next draw to have no numbers in common with the last three or four draws, we would have to track the union (and overlap) of more draws. The principle is the same, but the complexity increases quickly.
Detailed Reasoning
If we had 3 previous draws, each of 6 numbers, the size of their union could be anywhere from 6 to 18, depending on overlaps among all three draws.
We would define multiple overlap variables: k1 (overlap between the first and second draws), k2 (overlap between the second and third draws), and so on, plus the triple overlap among all three.
We then find the union’s size via set-inclusion-exclusion. This becomes complicated: we have to do something like |A ∪ B ∪ C| = |A| + |B| + |C| − |A∩B| − |B∩C| − |A∩C| + |A∩B∩C|.
Finally, the next draw has to avoid that union.
A key pitfall is failing to systematically apply set-inclusion-exclusion when more than two sets are involved. The combinatorial sums can get large, and it is easier to make errors. Hence, many prefer simulation for more than two or three sets.
Could we incorporate data on how many tickets are sold per draw and how many winners there were?
This question might arise if the interviewer wants you to think about the difference between the official random draw vs. the distribution of players’ ticket selections. For instance, “What is the probability that no winning ticket in the next draw shares any number with previous winning tickets?”
Detailed Reasoning
You would have to incorporate not only the official random draw distribution but also the distribution of how players pick their numbers.
If many tickets share some historical pattern (like “favorite numbers”), the chance that a new random draw matches any of those tickets depends on how big that portion of the ticket pool is, how many total tickets, etc.
This is more about expected value of winners or overlap among winners rather than overlap among drawn numbers.
A major pitfall is confusing the probability that a drawn set is disjoint from previously drawn sets with the probability that no purchased ticket is a winner. They are related but require careful analysis of how tickets are distributed among potential number choices.
If an interviewer asked you to estimate the probability with a quick approximate numerical technique without doing the full summation, how might you respond?
In a time-constrained interview, one might try a rough approximation:
Detailed Reasoning
First, note that on average, the last two draws produce about 12 − (expected overlap) distinct numbers. The expected overlap k between two consecutive draws is 6*(6/49) = 36/49 ≈ 0.7347. So the expected union is about 12 − 0.7347 ≈ 11.2653 distinct numbers.
That means, on average, the next draw of 6 must come from the remaining 49 − 11.2653 ≈ 37.7347 numbers.
One might say the approximate probability is ( (37.7347 choose 6 ) ) / ( (49 choose 6 ) ), plugging in the expected union size.
This is only an approximation, ignoring the distribution of overlap. But it can yield a value close to the more accurate 0.19. The pitfall is that you might lose some accuracy by ignoring the variance in overlap. An interviewer might accept a rough approach if you justify it properly and mention that the exact value requires summation or simulation.