ML Interview Q Series: Probability Calculation: One Card Number Doubling Another with Replacement

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Two cards are taken from a set of 100 numbered cards (1 to 100), with replacement. What is the probability that one card’s number is exactly twice the other’s number?

Comprehensive Explanation

To address this, consider two random variables X and Y. Each is uniformly sampled from 1, 2, …, 100 (because we are drawing with replacement, both draws come from the full set independently). The total number of equally likely outcomes is 100 * 100 = 10,000.

We want the probability that one card is double the other, meaning either X = 2Y or Y = 2X.

• For X = 2Y to hold, Y must be in {1, 2, …, 50}, because 2Y must be at most 100. This gives 50 valid (X, Y) pairs.

• For Y = 2X to hold, X must be in {1, 2, …, 50}, for the same reasoning. This also gives 50 valid pairs.

• The pairs where X = 2Y and Y = 2X cannot both be true unless X = 0 (or Y = 0), but 0 is not in the set {1, 2, …, 100}. Hence there is no overlap between these two sets of pairs.

Hence, the total number of favorable outcomes is 50 + 50 = 100.

Below is the core formula for the probability:

Where:

• Number of favorable pairs is 100.

• Total number of pairs is 10,000 (because X and Y can each be one of 100 values independently).

Therefore, the probability is 1/100.

How to Count the Favorable Outcomes

We systematically count:

• The set of pairs satisfying X = 2Y: (2,1), (4,2), (6,3), …, (100,50).

• The set of pairs satisfying Y = 2X: (1,2), (2,4), (3,6), …, (50,100).

Each set has 50 unique pairs, and no pair appears in both sets. Hence, 100 total satisfying pairs.

Potential Follow-up Questions

If the Cards Were Drawn Without Replacement, What Would Change?

When drawing without replacement, X and Y would not be independent. The number of total outcomes would become 100 * 99, since the second draw would have only 99 remaining cards (assuming order matters: first draw is X, second is Y).

The counting of favorable pairs would also change. For example, if we pick a card with value x, the second card can only be 2x if 2x is still in the remaining deck. You would sum over all valid x for which 2x is still in the deck, plus all valid x for which x/2 is in the deck (if x/2 is an integer). The probability would be slightly different from 1/100.

Does the Order in Which We Draw the Cards Matter?

Often, when we talk about probability in drawing two cards, we need to decide if (X, Y) is considered distinct from (Y, X). In our derivation, we allowed order to matter (first draw is X, second draw is Y). That is why the total sample space was 10,000 for a deck of 100 with replacement. If, in a different interpretation, order did not matter, we would adjust the total number of outcomes accordingly. However, the question explicitly or implicitly uses the approach where each draw is an independent event, so (X, Y) is indeed different from (Y, X).

How Would You Verify This Probability Experimentally?

A quick way to verify is through a simple simulation in Python. Below is an example code snippet:

import random

trials = 10_000_00
count = 0

for _ in range(trials):
    card1 = random.randint(1, 100)
    card2 = random.randint(1, 100)
    if card1 == 2*card2 or card2 == 2*card1:
        count += 1

empirical_probability = count / trials
print(empirical_probability)

Over many trials, the empirical probability should hover around 0.01 (i.e., 1/100).

How Can This Be Generalized to Cards Numbered from 1 to N?

If there are N cards labeled 1 to N, and you draw two with replacement, the total number of outcomes would be N * N = N². The number of favorable pairs is:

• Count of pairs (x, 2x) for x in {1, …, floor(N/2)}, which is floor(N/2).

• Count of pairs (2x, x) for x in {1, …, floor(N/2)}, also floor(N/2).

• There is no overlap if N > 1.

Hence, total favorable outcomes = 2 * floor(N/2). Thus, the probability = 2 * floor(N/2) / N². For N = 100, floor(N/2) = 50, so we get 2*50 / 100² = 100/10,000 = 1/100.

Below are additional follow-up questions

What If the Distribution of Card Values Is Not Uniform?

In some scenarios, each card might not have an equal probability of being drawn. For example, certain card values might appear more frequently or less frequently due to biases in the deck or in the selection process. In that case, we can no longer treat each card draw as having probability 1/100 for each label. Instead, we would have to assign a probability p(x) to each label x in {1, …, 100}, ensuring that the sum of p(x) over all x equals 1. Then, the probability of drawing cards X = x and Y = y (with replacement) would be p(x) * p(y), and we would sum over all pairs satisfying x = 2y or y = 2x.

A key pitfall: People might assume that the probability remains 1/100 for each label if they do not realize the deck or the process has biases. That oversight can significantly alter the results. Hence, if the deck or draws are not uniform, one should carefully account for each possible probability p(x).

How Would You Extend the Same Logic to a “Triple” Condition?

Instead of looking for two cards where one is double the other, consider a scenario where one card is triple the other. The count of favorable outcomes changes because now we need x and 3x to be within the range {1, …, 100}. If we let N = 100, then x must be in {1, …, floor(N/3)} for 3x to remain valid.

We can derive a more general formula for one card being k times the other. Denoting N as the largest card, the probability that one card is exactly k times the other is:

Where 2 appears because we can have X = kY or Y = kX, and there is no overlap if k > 1 and N > 1. For k = 3 and N = 100, we get 2 * floor(100/3) / 100^2 = 2 * 33 / 10,000 = 66/10,000 = 0.0066. A subtle point is that floor(100/3) = 33, and so the largest x for 3x to stay ≤ 100 is 33.

A pitfall: People might forget to handle floor(N/k) when k does not evenly divide N. Simply using N/k without flooring can lead to off-by-one errors or fractional counts of valid pairs.

How Can We Adapt the Method for a Continuous Numeric Range?

In some real-world problems, “cards” might represent continuous values rather than integers (for example, measuring random real-valued events). If we want the probability that one continuous draw is exactly twice the other, the probability of hitting an exact real value would be zero in a continuous distribution. This is because the set of points where x = 2y in a continuous space has measure zero. Hence, the probability that x is exactly twice y would be 0 for continuous uniform distributions.

This underscores a conceptual pitfall: Extending discrete logic directly to continuous scenarios can be misleading, because discrete probabilities for exact matches do not translate straightforwardly to continuous spaces.

What if We Consider an Ongoing Stream of Draws and Look for the First Time This Event Occurs?

You might find yourself in a situation where you are repeatedly drawing pairs of cards (with replacement) and you want to compute the expected number of trials until one card is double the other. For each draw, the event “one card is double the other” has probability 1/100 (in the uniform scenario from 1 to 100). If each pair of draws is independent from the others, the expected number of draws to get a “double” match can be approached by a geometric distribution with success probability p = 1/100. The expected number of draws (each draw is one pair) is 1/p = 100.

A tricky edge case: If the draws are somehow dependent (for example, the deck composition changes in a more complicated manner, or the sampling method depends on history), the geometric distribution assumption breaks down. One needs to carefully revisit the probability of success on each draw to see if it is still constant.

Could Floating-Point Precision Affect a Simulation?

When implementing a Monte Carlo or simulation approach in Python (or another language), the random generation of integers 1 through 100 is not at much risk for floating-point imprecision. However, in other numerical problems where card values might be replaced with continuous or large floating numbers, floating-point inaccuracies can affect whether we detect that one value is “double” the other. For instance, if we were checking if x == 2.0*y in a floating-point context, subtle rounding might cause false negatives or false positives.

A pitfall: Relying on direct equality checks with floating-point numbers can lead to incorrect results. In practice, one might introduce a small tolerance for equality checks in such continuous or large floating contexts.

How Would the Result Change If Each Draw Included an Additional Probability of Discarding the Card?

Imagine a scenario where each time you draw a card, there is some chance you discard it from the deck entirely, altering the available pool for subsequent draws. This is neither exactly “with replacement” nor exactly “without replacement.” The distribution of subsequent draws becomes more complicated, and one needs to keep track of the evolving deck composition. As cards get discarded, the probability of drawing certain values changes over time.

A subtlety here is that even though we start with a deck of 100 labeled cards, if some labels are more frequently discarded early on, the deck distribution shifts. The probability calculation of “one card is double the other” would need to consider the dynamic deck state at each draw. Consequently, a naive approach that assumes independence among all draws or a fixed probability 1/100 for each label would be incorrect.

What If We Label Cards Differently, Such as Starting from 0 Instead of 1?

If the deck is labeled from 0 to 99, we still have 100 cards, but zero can pose an interesting edge condition. For instance, the statement “one card is double the other” cannot hold if one of the cards is zero and the other is nonzero, because 0 * 2 = 0 does not match any nonzero number. And if both cards are zero, the condition “one card is double the other” trivially holds, but only because 0 = 20. In fact, (0, 0) would satisfy 0 = 20. That gives us one additional pair to consider. So for a deck from 0 to 99, you would do a careful count of pairs (x, 2x) for x in {1..49}, plus the pair (0,0), plus the symmetrical set (2x, x). This subtle shift can change the probability slightly. People often forget to check boundary cases like zero or negative values.

How Does One Verify That the Two Conditions (X=2Y or Y=2X) Never Overlap?

It might seem obvious that if X=2Y, you cannot simultaneously have Y=2X for positive integer values X and Y. But occasionally, interviewers want to confirm that the candidate has explicitly verified that there is no overlap. A quick check:

If X=2Y and Y=2X, then X=2(2X) => X=4X => 3X=0 => X=0. But our deck has strictly positive integers 1..100 in a typical setup. So there is no overlap. For non-positive or alternate deck definitions (including zero or negatives), carefully re-check that logic.

A subtle pitfall: If you had negative integers or zero in your deck, you might find overlaps. So it always helps to do a quick direct check for overlap rather than assuming it blindly.

How Could We Adjust the Counting for a Partial Deck?

Sometimes, we do not have the full deck from 1 to 100. For example, maybe some specific values are missing or we only have a subset of the deck. The core idea is still to count all valid pairs (x, 2x) and (2x, x) that exist in the subset. However, the total number of possible outcomes changes because we do not have 100 unique labels anymore, and the distribution might not be uniform. The probability then becomes the ratio of the count of valid pairs to the total count of all pairs in that partial deck scenario. One must be very explicit about the deck's exact composition to avoid miscounting.

A potential edge case: If the partial deck has very few low or high numbers, it can drastically change the probability of “one card is double the other.” For instance, if the deck is only {25, 50, 75, 100}, then (50, 25) and (100, 50) are possible double combinations, and (25, 50) or (50, 100) also. With fewer total cards, these pairs significantly shift the probability relative to the full deck.

How Do We Count “Close to Double” Instead of “Exactly Double”?

There may be a variation where we ask the probability that one card is within 1 or 2 units of twice the other (for instance, x ≈ 2y within some small tolerance). With discrete integers 1..100, we might define conditions like |x - 2y| <= δ, for some δ in {1, 2, …}. This approach changes the counting method, since now multiple pairs around the exact doubling condition also qualify. One would systematically go through all (x, y) in {1..100} x {1..100} and check if |x - 2y| <= δ. Because this is a more relaxed criterion, we would see more favorable pairs than in the strict double case, thus yielding a higher probability.

A pitfall: If the interviewer wants the probability for “approximately double,” do not assume it is just the same 50 + 50 pairs. Carefully define the approximation bound, then count systematically or derive an analytic approach.