Solution:

References:

• Law of Total Probability: https://en.wikipedia.org/wiki/Law_of_total_probability

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Link: Understanding Conditional Independence - Part 1

In our last newsletter, we covered the basics of conditional independence: what it is, why it matters, and how the math proves it. In this newsletter, we will explore how two events can be completely unrelated in one scenario, but deeply connected in another—and vice versa. Through simple, real-world examples, you'll see exactly how this twist works and why it changes how we look at probability.

The question has been refurnished again for your convenience.

Explain the concept of conditional independence with examples.

Solution:

Unconditional and conditional independence do not mean the same thing, and one does not guarantee the other. Two events may appear unrelated when we look at them on their own, but their relationship can change once we are given additional information. Likewise, two events that appear related overall can become unrelated once we account for a third event. This is one of the counterintuitive aspects of probability.

To see how they differ, we have to look at two distinct scenarios:

• Events can be both unconditionally independent and conditionally dependent.

• Events can be both unconditionally dependent and conditionally independent.

Now, let’s explore real world examples for each of the above mentioned scenarios.

1. Events can be both unconditionally independent and conditionally dependent.

• Example: You track two bonds alongside the overall economic state (Recession or Normal).

• Let us define the events as:

  • A: Bond 1 defaults.

  • B: Bond 2 defaults.

  • C: Recession occurs.

• Unconditional Independence: In a stable economy, the performance of Bond 1 does not affect Bond 2, hence P(A∩B) = P(A) × P(B).

• Conditional Dependence: When a recession (C) strikes, systemic economic stress links them together, meaning P(A∩B|C) ≠ P(A|C) × P(B|C).

• Below is a binary tree illustrating the scenario:

  

• Explanation of Table 1:

  • Stage 1 (Red Circles): Splits the world into two economic paths: a Recession economy or a Normal economy.

  • Stage 2 (Blue Circles): Tracks whether Bond 1 defaults, given the respective economic condition.

  • Stage 3 (Green Circles): Tracks whether Bond 2 defaults, based on Bond 1’s performance.

  • Analysis: The performance of the bonds is generally unconditionally independent. However, once we condition on the state of the economy, the performance of the bonds becomes conditionally dependent.

2. Events can be both unconditionally dependent and conditionally independent.

• Example: You observe a child’s height (Tall or Short), cognitive development (Vocabulary High or Vocabulary Low), and age (Age 5 or Age 12).

• Let us define the events as:

  • H: Height of the child.

  • V: Vocabulary size of the child.

  • A: Age of the child.

• Unconditional Dependence: Generally, taller children tend to have larger vocabularies because both factors naturally increase over time, i.e.,

  P(H∩V) ≠ P(H) × P(V).

• Conditional Independence: If you lock the age of the child (A), then a child’s height is no longer an indicator of the child’s vocabulary, i.e.,

  P(H∩V|A) = P(H|A) × P(V|A).

• Below is a binary tree illustrating the scenario:

• Explanation of Table 2:

  • Stage 1 (Purple Circles): Splits the children into two distinct age groups: Age 5 or Age 12.

  • Stage 2 (Blue Circles): Splits the children into two distinct height groups (Tall or Short), given the child’s age.

  • Stage 3 (Yellow Circles): Tracks the child’s vocabulary size (Vocab High or Vocab Low), given both their age and height.

  • Analysis: The height and vocabulary of a child are generally unconditionally dependent. However, once we condition on the child’s age, height and vocabulary become conditionally independent.

We hope you found the article both enlightening and valuable! For more insightful content delivered straight to your inbox, simply enter your email address below and hit the subscribe button. Stay tuned for future updates!

References:

• Conditional Independence: https://en.wikipedia.org/wiki/Conditional_independence

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Solution:

1. What is Conditional Independence?

   In probability, conditional independence means that two events may be related in general, but they become independent once we know a specific piece of information (a third event). Once we know that third event has occurred, knowing whether the first event happened gives us no additional information about whether the second event happened.

2. The Real-World Benefit of Conditional Independence.

In the real world, tracking how dozens of different things interact can be too overwhelming for computers. Conditional independence provides a shortcut by showing that, once we know certain information, some variables no longer provide information about each other. It simplifies complex computer models and machine learning algorithms by reducing the number of relationships the system needs to consider at the same time. By breaking a complex problem into smaller, more manageable parts, conditional independence makes it easier and faster for computers to calculate the probabilities of different possible outcomes.

3. Application of Conditional Independence.

Conditional independence is used in many real-world applications, especially when dealing with complex problems involving many variables. Some common applications include:

• Medical Diagnostic Tools: A computer can estimate the likelihood of a disease without having to consider every symptom and factor at the same time.

• Bayesian Networks: These are diagrams that show how different factors are connected. They use arrows to show how one factor can influence another, helping computers calculate probabilities by focusing only on the relationships that matter.

Now, let’s look at the mathematical foundation of conditional independence.

4. Mathematical Proof.

Two events A and B are conditionally independent if:

The conditional probability P(A∩B|C) is the probability of event A and B occurring given that event C has already occurred.

Recall the Bayes’ Theorem formula:

Using Equation (2), we can rearrange Equation (1) as follows:

Equation (3) can be rearranged as follows:

In Equation (4), P(A|B∩C) = P(A|C) means that once we know event C has occurred, knowing that event B also occurred provides no additional information about event A. In other words, event C is all the information we need to determine the probability of A; event B does not matter.

Another way Equation (3) can be rearranged is as follows:

In Equation (5), P(B|A∩C) = P(B|C) means that once we know event C has occurred, knowing that event A also occurred provides no additional information about event B. In other words, event C is all the information we need to determine the probability of B; event A does not matter.

Equation (4) and Equation (5) are equivalent formulations of Equation (1), rather than simplified versions.

• Equation (1) defines the conditional independence of A and B given C as a product of their probabilities: P(A∩B|C) = P(A|C) × P(B|C).

• Equation (4) and Equation (5) express this exact same relationship by showing that once event C is known, the occurrence of one event provides no new information about the probability of the other: P(A|B∩C) = P(A|C) and P(B|A∩C) = P(B|C).

In the next article, we’ll explore how two events can be unrelated in one situation but connected in another. Using simple, real-world examples, we’ll see how this works and why the information we have can change the way we understand probability.

We hope you found the article both enlightening and valuable! For more insightful content delivered straight to your inbox, simply enter your email address below and hit the subscribe button. Stay tuned for future updates!

References:

• Conditional Independence: https://en.wikipedia.org/wiki/Conditional_independence

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Link 1: The Elbow Method: Finding the Optimal Number of K-Means Clusters - Part 1

Link 2: The Silhouette Method: Numerical Examples Using the Silhouette Score Formula - Finding the Optimal Number of Clusters (k) in K-Means Clustering - Part 2

This article shows a Python code example of implementing the silhouette method explained in the previous article.

The question has been refurnished again for your convenience.

What methods can be used to determine the optimal number of clusters (k) in a K-means clustering algorithm?

Solution:

The following Python example shows how the average silhouette score can be calculated for k = 2 to k = 10 using a synthetic dataset of 500 randomly generated data points around 3 different centers.

Explanation of the above code snippet:

• Average Silhouette Score Calculation & Values: The code calculates the average silhouette score for values of k ranging from 2 to 10. The highest average silhouette score is 0.9064, which occurs at k = 3. The calculation starts at k = 2 because the silhouette score requires at least two clusters to compare how well each data point fits within its assigned cluster relative to the nearest neighboring cluster.

• Optimal Choices: Since k = 3 produces the highest average silhouette score of 0.9064, the silhouette method suggests that 3 is the optimal number of clusters for this dataset. A score of 0.9064, being close to +1, indicates that the data points are generally well matched to their assigned clusters and well separated from neighboring clusters.

Explanation of the above code snippet:

• Visualizing the Maximum Silhouette Score Point: The code identifies the point corresponding to k = 3, where the average silhouette score reaches its maximum value of approximately 0.91, and highlights this point on the graph.

• Intersection Highlights: The graph uses red dotted horizontal and vertical lines that intersect at k = 3 and a silhouette score of approximately 0.91. This provides a clear visual indication of the optimal number of clusters suggested by the silhouette method.

We hope you found the article both enlightening and valuable! For more insightful content delivered straight to your inbox, simply enter your email address below and hit the subscribe button. Stay tuned for future updates!

References:

• K-means Clustering: https://en.wikipedia.org/wiki/K-means_clustering

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Link: The Elbow Method: Finding the Optimal Number of K-Means Clusters - Part 1

In the previous newsletter, we discussed the Elbow Method for selecting the optimal number of clusters (k) in the K-means clustering algorithm. In this newsletter, we'll explore another popular approach—the Silhouette Method—for determining the optimal value of k.

The question has been refurnished again for your convenience.

What methods can be used to determine the optimal number of clusters (k) in a K-means clustering algorithm?

Solution:

(b). The Silhouette Method.

The Silhouette Method is a clustering validation technique that evaluates how well data points fit into their assigned groups by comparing their internal similarity (similarity within their own data group) against neighboring groups of data. For any single point, a score between -1 and +1 is calculated using the following formula:

In Equation (1), a final silhouette score near +1 indicates that the point is well-placed within its own cluster group (close to its assigned cluster center) and far from neighboring clusters, whereas a score near -1 implies the point was likely assigned to the wrong cluster or group.

Understanding the Silhouette Score.

The Silhouette Score measures how well a data point fits within its assigned cluster and how clearly it is separated from other clusters. The score ranges from −1 to +1, with higher values generally indicating better-defined and more clearly separated clusters.

• Higher Silhouette Score: This indicates that a data point is close to other points in its own cluster while being far from points in the nearest neighboring cluster. This means the data point is well matched to its assigned cluster and clearly separated from other clusters.

• Lower Silhouette Score: This indicates that a data point is less clearly associated with its assigned cluster. It may be relatively close to a neighboring cluster, making its cluster assignment less certain.

• Negative Silhouette Score: This indicates that a data point is closer, on average, to a neighboring cluster than to its assigned cluster, suggesting that it may have been assigned to the wrong cluster.

In other words, a good clustering should have:

• Small y → The point is close to other points in its own cluster.

• Large x → The point is far from the nearest neighboring cluster.

Therefore, a small y and a large x results in a higher silhouette score.

Now, let's illustrate this with two examples.

Example 1: A Point Well Suited to Its Assigned Cluster.

Suppose point P belongs to cluster A.

First, we calculate y, the average distance from P to other points in its own cluster:

In Table 1, y is calculated as: (2 + 3 + 1) / 3 = 2.

Next, we calculate x, the average distance from P to the points in the nearest neighboring cluster, cluster B:

In Table 2, x is calculated as: (8 + 9 + 10) / 3 = 9.

Therefore, by plugging in x = 9 and y = 2 into Equation (1), we get:

Equation (2) Interpretation: The silhouette score of 0.78 indicates that point P is well suited to its assigned cluster. P is, on average, only 2 units away from points in its own cluster (cluster A), while it is 9 units away from points in the nearest neighboring cluster (cluster B). Since P is much closer to its own cluster than to the neighboring cluster, it is clearly separated from the neighboring cluster, resulting in a relatively high silhouette score of 0.78.

Example 2: A Point to its Neighboring Cluster.

Now consider point Q, which has been assigned to cluster A.

The distances from Q to other points in its own cluster are:

In Table 3, y is calculated as: (7 + 8 + 9) / 3 = 8.

Now consider the nearest neighboring cluster, cluster B:

In Table 4, x is calculated as: (2 + 3 + 4) / 3 = 3.

Therefore, by plugging in x = 3 and y = 8 into Equation (1), we get:

Equation (3) Interpretation: Q is, on average, 8 units away from points in its own cluster, while it is only 3 units away from points in the nearest neighboring cluster. Thus, Q is closer to the neighboring cluster than to its assigned cluster. The negative silhouette score of −0.625 indicates that Q may be better suited to the neighboring cluster and may have been assigned to the wrong cluster.

Key Takeaways.

The two examples above demonstrate the intuition behind the silhouette score:

The ideal situation is for a point to be close to members of its own cluster while remaining far from neighboring clusters. The silhouette score summarizes these two properties in a single measure.

Let us explore a Python code implementation example of the silhouette method in the next article!

We hope you found the article both enlightening and valuable! For more insightful content delivered straight to your inbox, simply enter your email address below and hit the subscribe button. Stay tuned for future updates!

References:

• K-means Clustering: https://en.wikipedia.org/wiki/K-means_clustering

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Solution:

Before we dive into the solution, let’s first explain what is k-means clustering.

K-means clustering is an unsupervised machine learning technique designed to group similar, unlabeled data points together into distinct clusters based on their inherent characteristics.

Unsupervised machine learning is a branch of machine learning where algorithms are trained on unlabeled datasets with no predefined outcome or target variable (Y). Unlike supervised learning, the algorithm receives no correct answers or human guidance. Instead, it independently analyzes data to discover hidden patterns, relationships, clusters, and anomalies—unusual observations that differ from normal behavior. By uncovering the underlying structure of data, unsupervised learning helps organizations gain insights, detect unusual events, reduce complexity, and organize large datasets without knowing the desired outcome in advance.

We use K-means clustering to automatically group unlabeled data points into distinct clusters based on their similarities. This process helps uncover meaningful patterns and transform complex datasets into actionable insights through:

• Pattern Discovery: Identifying hidden groupings within data, such as segmenting customers based on purchasing behavior.

• Data Simplification: Reducing large, complex datasets into a smaller number of meaningful and easy-to-understand profiles.

• Actionable Insights: Supporting data-driven decisions in areas such as targeted marketing, fraud detection, and image compression.

K-means requires you to define the number of clusters (k) upfront. Choosing k blindly can ruin your model. Two data-driven techniques can help find the sweet spot.

• The Elbow Method

• The Silhouette Method

(a). The Elbow Method.

The Elbow Method finds the optimal number of clusters by plotting the model's total Within-Cluster Sum of Squared errors (WCSS) against different values of k. The WCSS measures the compactness of clusters by calculating the sum of the squared distances between each data point and its assigned cluster center (centroid). The general formula for WCSS is:

In Equation (1), a lower WCSS value indicates that the data points are closer to their respective centroids, meaning the clusters are tightly packed.

1. Why WCSS Always Drops?

As you add more cluster centers (centroids), each group covers a smaller neighborhood. Because the groups are smaller, every data point naturally sits much closer to its nearest center.

• Extreme Example: If you have 50 data points and create 50 clusters, every single point becomes its own centroid. The distance from each point to its center is exactly zero, making the total WCSS = 0.

2. The Law of Diminishing Returns.

Adding more clusters will always reduce the Within-Cluster Sum of Squares (WCSS) because each cluster contains fewer, more similar data points. However, after a certain point, the improvement becomes very small.

• From 1 to 2 clusters: WCSS may drop dramatically, indicating a significant improvement in clustering.

• From 7 to 8 clusters: WCSS may decrease only slightly, meaning the extra cluster provides little additional benefit.

3. Finding the “Elbow”.

When you plot WCSS against the number of clusters, the graph typically forms a shape that looks like a bent arm or an elbow. At first, the WCSS decreases sharply because adding more clusters greatly improves the quality of the grouping. However, after a certain point, the curve begins to flatten, indicating that each additional cluster provides only a small reduction in WCSS. This bending point, known as the elbow point, is considered the optimal number of clusters because it provides a good balance between keeping the model simple (using fewer clusters) and achieving high-quality clustering (low WCSS). Beyond the elbow, adding more clusters increases the model's complexity while offering only minimal improvement in clustering performance.

The following Python example shows how the elbow point can be identified using a synthetic dataset of 500 randomly generated data points around 4 different centers.

Explanation of the above code snippet:

• WCSS Calculation & Values: The code measures data grouping tightness by calculating Within-Cluster Sum of Squares (inertia) for 1 to 10 clusters, yielding a starting value of 33903.9042 for 1 cluster and dropping to 948.8903 at 4 clusters.

• Optimal Choices & Scale: The sharp drop slows down drastically after 4 clusters (confirming it as the ideal choice), and because this is synthetic, artificial data, the WCSS has no physical units.

Explanation of the above code snippet:

• Visualizing the Elbow Point: The code isolates the exact WCSS coordinate for k = 4 (WCSS ≈ 948.89) to clearly mark the point where the line’s steep drop suddenly levels off.

• Intersection Highlights: The resulting graph uses red dotted horizontal and vertical lines that intersect perfectly at this optimal point, providing a clear visual confirmation of the ideal cluster count.

Let us explore the Silhouette method in the next article!

We hope you found the article both enlightening and valuable! For more insightful content delivered straight to your inbox, simply enter your email address below and hit the subscribe button. Stay tuned for future updates!

References:

• K-means Clustering: https://en.wikipedia.org/wiki/K-means_clustering

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Solution:

Mean Absolute Error (MAE) and Mean Squared Error (MSE) are two of the most commonly used regression evaluation metrics. Both measure how far a model's predictions deviate from the actual target values, but they differ in how they penalize prediction errors.

1. Similarities.

Both MAE and MSE share several important characteristics:

• Both are used to measure the performance of regression models by comparing predicted values against the actual target values.

• Both summarize prediction accuracy by averaging errors across all observations.

• Both range from 0 to infinity, where 0 indicates a perfect model and lower values indicate better predictive performance.

• Both compare the model’s predictions against the true target values.

Additionally, both MAE and MSE can serve as either evaluation metrics or loss functions, depending on when they are used in the machine learning workflow.

• As evaluation metrics, they are calculated after the model has been trained to assess how accurately it performs on unseen validation or test data.

• As loss functions, they are calculated during model training. Optimization algorithms such as Gradient Descent continuously minimize the loss by adjusting the model’s parameters until prediction errors become as small as possible.

Despite these similarities, the two metrics differ significantly in how they measure and penalize prediction errors. Let us explore them one by one.

2. Differences.

(a). The Core Math: Distance, Ranges, Units, and Formulas.

At their core, both MAE and MSE quantify the distance between your model’s predictions and the true actual values, treating them like points in a multi- dimensional space.

• The Formulas:

  

• The Range: Both metrics range from 0 to ∞. A value of 0 represents perfect predictions, while smaller values always indicate better model performance.

• The Units:

  • MAE (Mean Absolute Error): It measures the average absolute distance between predictions and actual values. Because it uses an absolute value function instead of squaring, it keeps the exact same unit as your target variable.

  • MSE (Mean Squared Error): It measures the average variance of your errors. Squaring the differences means the unit becomes squared (e.g. $² or cm²), which removes it from a real-world context.

• Mathematical Penalty Visualization.

  

  In Figure 1, we observe the following:

  • The graph shows how each metric calculates its error penalty.

  • The linear MAE line stays constant, while the quadratic MSE curve shoots up aggressively for larger errors.

(b). Outliers: How Do They Penalize Prediction Errors?

The defining difference between MAE and MSE lies in how they treat large prediction errors.

• MSE is Highly Sensitive to Outliers.

  MSE squares every prediction error before averaging. Because of this, larger errors receive quadratically larger penalties. For example:

  • Error = 3 → Penalty = 9

  • Error = 9 → Penalty = 81

  Although the error only triples, its contribution to MSE becomes nine times larger. Consequently, MSE strongly discourages large prediction errors and is preferred when large mistakes are particularly costly.

• MAE is Robust to Outliers.

  MAE simply takes the absolute value of each prediction error. Every error increases proportionally with its size, so extremely large errors do not dominate the metric. As a result, MAE is considerably more robust when datasets contain outliers or noisy observations.

• Outlier Penalty Breakdown.

  

  In Figure 2, we observe the following:

  • The visual illustrates how the two metrics handle a small error versus a large outlier.

  • The MSE penalty explodes exponentially as the error grows from size 2 to 10.

  • The MAE penalty is scaled linearly without being squared.

(c). Computational Speed: Training the Model.

When training a machine learning model, our main goal is to minimize L (which simply stands for Loss, or the score that measures the model's errors). Optimization algorithms like Gradient Descent use the derivative (slope) of this L curve to decide which direction to move and how large of a step to take.

The math behind your chosen loss function dictates how fast your model learns.

• MSE is Faster (The Smooth Bowl).

  Mean Squared Error maps out a perfectly smooth, round geometric bowl where your learning steps scale dynamically based on your distance to the target. Large initial errors create steep slopes that force the model to take massive predictive leaps forward. Because the variable x in Equation (3) stays active within the derivative math, the slope naturally flattens to zero as you get closer to your goal, allowing the algorithm to execute a gentle, precise parking job right at the minimum point according to Equation (3):

  

• MAE is Slower (The Sharp V-Shape).

  Mean Absolute Error maps out a rigid, sharp V-shaped trench with completely straight walls where your training steps remain entirely identical regardless of your distance to the target. Because the variable x in Equation (4) completely disappears from the derivative math, the slope has a constant magnitude of ±1 and never changes steepness or slows down, regardless of how far your predictions are from your goal. Because the gradient does not gradually approach zero, the model will aggressively overshoot the minimum point and bounce back and forth across the bottom in endless oscillation according to Equation (4):

  

• Visualizing the Loss Landscapes (L).

  

  In Figure 3, we observe the following:

  • The MSE Plot (Top Left): This curve visualizes the smooth bowl geometry mentioned above. As your prediction moves from 6 to 9.5, you can see the slope physically flatten out from an aggressive -8 to a gentle -1, proving mathematically why the model slows down to a perfect stop at the bottom target of 10.

  • The MAE Plot (Top Right): This curve illustrates the rigid, straight walls of the sharp V-shaped trench. Whether the prediction is far away at 6 or right next to the target value of 10 at 9.5, the slope remains completely unchanged at a constant -1, visually demonstrating why the model fails to decelerate and overshoots the minimum point.

(d). Statistical Interpretation: Mean vs. Median.

The two metrics also optimize different statistical quantities.

• Conditional Mean vs. Median.

  • MSE is Minimized by the Conditional Mean: Minimizing squared errors causes predictions to converge toward the conditional mean of the target distribution. Consequently, large outliers pull predictions toward themselves.

  • MAE is Minimized by the Conditional Median: Minimizing absolute errors leads predictions toward the conditional median. Since the median is resistant to extreme observations, MAE remains much more stable when outliers are present.

  • Statistical Distribution Shift Visualization.

    

    In Figure 4, we observe the following:

    • The graph shows the “Statistical Secret”: when a dataset is skewed (right skewed in this example) by an outlier, the distribution gets a long tail.

    • The Mean (MSE) gets dragged towards the right tail by the outlier, while the Median (MAE) stays near where the bulk of the actual data sits.

(e). A Quick Numerical Walkthrough.

Let’s walk through a simple, step-by-step example demonstrating how to calculate both MAE and MSE manually.

Imagine we are training a machine learning model to predict the weight in pounds (lbs) for 3 people:

Person 1: Actual = 150 lbs | Predicted = 152 lbs, Error = 150 - 152 = -2 lbs.

Person 2: Actual = 180 lbs | Predicted = 178 lbs, Error = 180 - 178 = 2 lbs.

Person 3 (Outlier): Actual = 120 lbs | Predicted = 130 lbs, Error = 120 - 130 = -10 lbs.

• The MAE Calculation.

  In Equation (1), the MAE removes negative signs by taking the absolute value of each error, then finds the simple average.

  • Take absolute values: |-2| = 2, |2| = 2, |-10| = 10.

  • Average the results: MAE = (2 + 2 + 10) / 3 = 4.67 lbs.

• The MSE Calculation.

  In Equation (2), the MSE removes negative signs by squaring each error, then averages those squares. (Note: this changes the final measurement unit to squared pounds)

  • Square the errors: (-2)² = 4, (2)² = 4, (-10)² = 100.

  • Average the results: MSE = (4 + 4 + 100) / 3 = 36 lbs².

• Step-by-Step Metric Breakdown.

  

  In Figure 5, we observe the following:

  • The visualization contrasts the calculation pipeline for both MAE and MSE metrics using the 3 people example discussed earlier.

  • Person 1 and Person 2 have small errors of 2, while the outlier is 10 (Person 3), keeping the gap between them normal.

  • Squaring the numbers turns the small errors into 4 but shoots the outlier up to 100, illustrating why MSE is much more sensitive to large errors.

(f). Summary of Differences between MAE and MSE.

Table 1 summarizes the differences between MAE and MSE.

We hope you found the article both enlightening and valuable! For more insightful content delivered straight to your inbox, simply enter your email address below and hit the subscribe button. Stay tuned for future updates!

References:

• MAE vs. MSE: https://www.reddit.com/r/learnmachinelearning/comments/15qusj3/mae_vs_mse/

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Solution:

The condition |x_i - x_i-1| = 1 means that each new number in the sequence must go up by 1 or down by 1 from the number before it. Let R be the number of times the sequence goes up (+1), and let L be the number of times it goes down (-1). We can find the exact values of R and L using two simple facts:

• Total Steps: There are 10 total changes from x₀ to x₁₀, so:

  

• Net Change: The sequence starts at 0 and ends at 4, so the net change is:

  

Solving Equation (1) and Equation (2) together by adding them gives 2R = 14, which means R = 7. Substituting back yields L = 3. This shows that out of the 10 total changes, exactly 7 must be increases (right steps) and 3 must be decreases (left steps).

Let's think of each change in the sequence as a coin flip. Heads means we move one step right (+1) starting from x₀ = 0, and tails means we move one step left (-1) starting from x₀ = 0. In fact, finding the number of valid paths relies on the exact same logic and counting principles as answering the question: “If you flip a fair coin 10 times, what is the probability that you end up with 4 more heads than tails?”

Any valid path is simply an arrangement of 7 right steps and 3 left steps (similar to arranging the letters RRRRRRRLLL). Therefore, we can find the total number of paths by choosing 3 positions out of 10 for the left steps, or 7 positions out of 10 for the right steps. We can arrive at the solution using either of the methods shown below:

Therefore, according to Equations (3) and (4), the number of different paths to end up at x₁₀ = 4 is 120.

We hope you found the article both enlightening and valuable! For more insightful content delivered straight to your inbox, simply enter your email address below and hit the subscribe button. Stay tuned for future updates!

References:

• Counting Sequences with an Absolute Difference Condition: https://math.stackexchange.com/questions/4763451/number-of-sequences-that-satisfy-the-absolute-difference-condition

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Solution:

For this problem, we will evaluate the following methods to determine the best strategy for finding the highest safe floor while minimizing the absolute maximum number of drops:

• Brute-Force Logic

• Binary Search

• Divide and Conquer

• Balancing Linear Drops

1. Brute-Force Logic.

   Brute-force logic solves a problem by systematically testing every possible case until the correct solution is found. Instead of using shortcuts or deeper insights, it relies entirely on exhaustive trial and error.

   In this puzzle, the brute-force approach requires dropping the first egg from the 1st floor and moving upward one floor at a time until it breaks on floor X. This reveals that the highest safe floor is X-1.

   Example: If the highest safe floor is floor 42, the egg will survive drops on floors 1 through 42, and finally break on floor 43. This requires a total of 43 drops to find the answer.

   In the absolute maximum scenario, the egg only breaks when it reaches the 100^th floor. This requires a total of 100 drops to determine that the 99^th floor is the highest safe floor.

   This approach is highly inefficient. Because our objective is to minimize the absolute maximum number of drops, we must look for a strategy that avoids testing all 100 floors one by one.

2. Binary Search.

   Binary search is an efficient algorithm that finds a target value by repeatedly dividing a sorted search space in half. At each step, it eliminates half of the remaining possibilities, making it much faster than a standard linear search.

   In this puzzle, a binary search strategy begins by dropping the first egg from the 50th floor. If the egg breaks, the highest safe floor must be lower than floor 50. If it survives, the safe floor must be floor 50 or higher.

   While this works well in theory, it fails here due to the two-egg constraint. If the first egg breaks on floor 50, you are left with only one egg. You can no longer divide the remaining 49 floors in half because breaking your final egg would leave you with no answer. Instead, you are forced to switch to a brute-force approach, testing floors 1 through 49 sequentially from lowest to highest.

   Example: If the highest safe floor is floor 49, your first drop at floor 50 breaks Egg 1. With your single remaining egg, you must test floor 1, floor 2, floor 3, and so on, all the way up to floor 49. This requires 1 drop with the first egg and 49 drops with the second egg.

   Consequently, this strategy results in an absolute maximum of 50 drops, making standard binary search highly ineffective.

3. Divide and Conquer.

   Divide and conquer is a problem-solving strategy that breaks a large problem into smaller, more manageable subproblems, solves each subproblem separately, and combines the results to solve the original problem. This approach reduces overall complexity and increases efficiency.

   In this puzzle, this technique can be applied by dropping the first egg in fixed intervals, such as every 10 floors (the 10^th, 20^th, 30^th floors, etc.), as long as the egg survives. Once the first egg breaks on floor X, the second egg is used to brute-force the remaining subproblem, testing floors X-9 through X-1 sequentially from lowest to highest.

   Example: If the highest safe floor is floor 99, the first egg will survive drops from floors 10 through 90, and finally break on floor 100 (10 drops). The second egg must then test floors 91 through 99 sequentially. It will survive all of them, requiring 9 additional drops. This results in a total of 19 drops.

   This 19-drop total represents the absolute maximum scenario for this strategy. While this method is a significant improvement over brute-force logic and binary search, it is still not perfectly optimized. The fixed interval size means that as Egg 1 travels higher up the building, the total number of potential drops increases rather than stays balanced.

4. Balancing Linear Drops.

   This approach minimizes the absolute maximum number of drops by keeping the total drop count identical across all scenarios. To achieve this, the step size decreases by exactly 1 floor after every drop of the first egg. Every time you consume a drop with the first egg, you have 1 less drop available in your budget for the second egg. Shrinking the next floor interval by 1 perfectly counterbalances this spent drop.

   Let floor x be the first floor from which you drop the first egg.

   • If the first egg breaks on floor x, you have used 1 drop. The second egg must test the x - 1 floors below it sequentially from lowest to highest.

     • Total drops = 1 + (x - 1) = x drops.

   • If the first egg survives, you move up. Because you have already spent 1 drop, your next interval must shrink to x - 1 floors. Your second drop will be at floor x + (x - 1).

   • If the first egg breaks on this second drop, you have used 2 drops. The second egg must test the x - 2 floors in the remaining gap.

     • Total drops = 2 + (x - 2) = x drops.

This creates a consistent pattern: if the first egg breaks on the nth drop, the second egg has x - n floors left to test.

The total remains perfectly flat, i.e.,Total drops = n + (x - n) = x drops.

• The sequence of floors that we test the first egg until it breaks after the n^th drop is:

  • 1^st drop: x

  • 2^nd drop: x + (x - 1)

  • 3^rd drop: x + (x - 1) + (x - 2)

    .

    .

    .

  • n^th drop: x + (x - 1) + (x - 2) + … + (x - (n - 1))

To span across a 100-story building, the sum of this decreasing arithmetic series must equal or exceed 100:

Using the standard triangular number formula, the sum of integers from 1 to x in Equation (1) can be rewritten as:

By combining Equation (1) and Equation (2), we can solve for x:

Plugging x = 13 and x = 14 into Equation (3) yields the following results:

• For x = 13: (13 × (13 + 1)) / 2 = 91 floors, which is too low because it leaves 9 floors from 92 to 100 unaddressed.

• For x = 14: (14 × (14 + 1)) / 2 = 105 floors, which is sufficient to cover all 100 floors.

Note that x represents both the first floor from which you drop the first egg and the absolute maximum number of drops.

Therefore, you should drop the first egg starting from the 14th floor. Here, x = 14 represents both your starting floor and your absolute maximum number of drops. This balancing act makes it the optimal strategy.

Below is the sequence of drops to follow assuming the first egg doesn’t break:

• Drop 1: Floor 14

• Drop 2: Floor 27 (14 + 13)

• Drop 3: Floor 39 (27 + 12)

• Drop 4: Floor 50 (39 + 11)

• Drop 5: Floor 60 (50 + 10)

• Drop 6: Floor 69 (60 + 9)

• Drop 7: Floor 77 (69 + 8)

• Drop 8: Floor 84 (77 + 7)

• Drop 9: Floor 90 (84 + 6)

• Drop 10: Floor 95 (90 + 5)

• Drop 11: Floor 99 (95 + 4)

• Drop 12: Floor 100 (99 + 1)

Notice that the sequence naturally cuts short at Drop 12. Because the building has exactly 100 floors, Drop 12 tests Floor 100 instead of following the theoretical calculation of 99 + 3. This boundary creates a 12-drop exception where you do not need the full 14 drops. If the first egg survives up to Floor 99 on Drop 11 and breaks on Floor 100 on Drop 12, Floor 99 is immediately proven to be the highest safe floor. This specific scenario concludes in only 12 drops because Drop 11 already cleared every floor below it.

The true absolute maximum of 14 drops occurs if the first egg breaks one step earlier on Drop 11 at Floor 99. In that scenario, you have used 11 drops and are left with an unverified gap of three floors, which are floors 96, 97, and 98, because up to floor 95 had already been successfully tested in the first 10 drops. You must then use the second egg to test floors 96, 97, and 98 sequentially from lowest to highest. If the egg survives floors 96 and 97 but breaks on floor 98, it requires 3 additional drops, bringing the total calculation of drops to 11 + 3 = 14 drops.

Consequently, the Balancing Linear Drops strategy guarantees that you will find the highest safe floor within an absolute maximum of 14 drops. The only exception occurs at the very top of the building, where the 100-floor limit cuts off the theoretical 102-floor sequence, resolving those specific scenarios in just 12 drops instead of 14. Otherwise, every other path down the building requires a consistent maximum of 14 drops to identify the exact mystery floor from which an egg can be dropped without breaking.

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References:

• 2 Eggs and 100 Floors Puzzle: https://www.reddit.com/r/puzzles/comments/1kd100/medium_level_puzzle_2_eggs_and_100_floors/

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Solution:

To begin, let’s label each of the fox’s dens as 1, 2, 3, 4, and 5.

In order to solve this problem, we need to follow an even-odd rule. We will divide the five dens in Figure 1 into even and odd groups: 2 and 4 as even dens, and 1, 3, and 5 as odd dens. Because adjacent dens alternate between even and odd numbers, the fox must change its den group (even or odd numbered den) each night. If the fox is currently in an even den, it will be in an odd den the next day, and vice versa.

• Assumption 1: The fox starts in an even den on the first morning.

Under this assumption, the fox must be in either den 2 or den 4. The farmer begins by checking den 2. If the fox is there, the search ends immediately. If den 2 is empty, the initial assumption guarantees that the fox must be in den 4.

As shown in Figure 2, if the fox was in den 4 on the first morning, it must move to either den 3 or den 5 by the second morning. The farmer inspects den 3 on the second morning. If the fox is there, it is caught. If the fox is not in den 3, it means the fox was in den 5 during the inspection, which forces it to move to den 4 by the third morning. Therefore, the farmer inspecting den 4 on the third morning guarantees its capture.

As shown in Figure 3, if the fox was in den 5 on the second morning, it is forced to move to den 4 by the third morning. Therefore, the farmer inspecting den 4 on the third morning is guaranteed to capture the fox. If the fox is not found in den 4, it proves assumption 1 was incorrect.

• Assumption 2: The fox starts in an odd numbered den on the first morning.

  Based on assumption 2, the fox follows a strict pattern each morning:

  • Morning 1: Odd numbered den.

  • Morning 2: Even numbered den.

  • Morning 3: Odd numbered den.

  • Morning 4: Even numbered den.

  If the farmer follows assumption 1 and does not find anything for the first 3 days, the strategy moves to the day 4 scenario. Note that on the fourth morning, the fox is guaranteed to be in an even numbered den according to assumption 2. This scenario mirrors assumption 1. However, unlike the first morning, it is now a logical certainty that the fox is in either den 2 or den 4. Hence, the farmer can reapply the exact same checking sequence from assumption 1 starting from the fourth morning:

  • Morning 4: Inspect den 2. If the fox is not in den 2, it must be in den 4, meaning it will move to either den 3 or den 5 by the next morning.

  • Morning 5: Inspect den 3. If the fox is not in den 3, it must have been in den 5, which forces it to move to den 4 by the next morning since den 4 is the only adjacent option.

  • Morning 6: Inspect den 4. The farmer is guaranteed to capture the fox.

Therefore, to guarantee that the farmer captures the fox, he should inspect the dens over six days at most.

• Testing for an Even Start (Assumption 1):

  • Day 1: Inspect den 2.

  • Day 2: Inspect den 3.

  • Day 3: Inspect den 4.

Catching the Reset (The Reverted Setup):
If the fox is not caught by day 3, the first phase fails. However, according to assumption 2, a fox starting in an odd numbered den is forced into an even numbered den by the fourth morning. This means the entire setup completely reverts back to the conditions of assumption 1.

• The farmer can simply repeat the exact same loop to finish the job:

  • Day 4: Inspect den 2.

  • Day 5: Inspect den 3.

  • Day 6: Inspect den 4.

While the fox might evade the farmer during the first five days, it will inevitably be captured on the sixth day, ensuring the entire process takes at most six days.

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References:

• Fox in a Hole Puzzle: https://www.quora.com/Consider-five-holes-in-a-line-One-of-them-is-occupied-by-a-fox-Each-night-the-fox-moves-to-a-neighboring-hole-either-to-the-left-or-to-the-right-Each-morning-you-get-to-inspect-a-hole-of-your-choice-What-strategy-would-ensure-that-the-fox-is-eventually-caught

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Solution:

To begin, you first need to race all 25 horses in groups of five, so there will be five initial races. After conducting these first five races, you can record the results of all 25 horses in a table.

In Table 1, the five race groups are labeled A, B, C, D, and E, while the rankings within each group are labeled 1^st, 2^nd, 3^rd, 4^th, and 5^th.

From each race group in Table 1, we can eliminate the horses that finished in fourth and fifth place since we are only interested in the top three horses.

In Table 2, we can compare the speed of horses within the same race group (for example, B1 vs. B2), but we cannot compare the speed of horses across different race groups (for example, B2 vs. C3). Thus, we will need to conduct a sixth race among the fastest horses—the first-place winners in each group: A1, B1, C1, D1, and E1.

For simplicity, let’s assume the results of the sixth race are as follows:

• A1 finishes in 1^st place.

• B1 finishes in 2^nd place.

• C1 finishes in 3^rd place.

• D1 finishes in 4^th place.

• E1 finishes in 5^th place.

In Table 3, we can see that all horses in Groups D and E are slower than the top three horses from the sixth race (A1, B1, and C1). Therefore, we can eliminate all horses in Groups D and E.

In Table 4, we see that A1 is the fastest horse in Group A, as well as the overall fastest horse among all group winners. Thus, A1 secures first place in the competition. We also see from the sixth race that C1 finished in third place, meaning any horse slower than C1 can be eliminated. Hence, C2 and C3 are eliminated.

In Table 5, we see that B1 finished in second place among the group winners. This means B2 could potentially secure third place overall since B2 is the second-fastest horse in Group B. However, B3 can be eliminated because it is slower than B2, leaving it no chance to finish in the top three. Note that while A2 and A3 are slower than A1, we do not yet know how their speeds compare to B1, B2, and C1.

In Table 6, we already know that A1 is the overall first-place winner, so we do not need to race A1 again. Thus, we are left with five horses to compete for second and third place in the seventh race: A2, A3, B1, B2, and C1.

In the seventh race, we need to pick the second and third fastest horses from the group consisting of A2, A3, B1, B2, and C1. We can label these two horses as H2 and H3 to represent second and third place, respectively. Hence, A1, H2, and H3 are the three fastest horses in the competition.

Therefore, seven races are sufficient to identify the three fastest horses.

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References:

• 25 Horses Puzzle: https://www.reddit.com/r/learnmath/comments/1bv3r2h/there_are_25_horses_each_of_which_runs_at_a/

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Link: Binomial Distribution vs Negative Binomial Distribution

Link: Bear Gets Exactly What It Wants - Part 1

In the previous newsletter, we approached this problem through the lens of the Binomial Distribution and Bayes’ Theorem. In this edition, we revisit the same problem, but now tackle it using the framework of the Negative Binomial Distribution.

For ease of reference, the question is presented below.

A bear attempts to catch fish as they pass by in a river, each with a success probability of 1/2. Once the bear successfully catches 3 fish, it stops fishing and leaves. What is the probability that the 5^th fish to pass by is not caught by the bear?

Solution:

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References:

• Bear Catching Fish Problem: https://math.stackexchange.com/questions/2491326/probability-question-bear-catching-fish

• Binomial Distribution: https://en.wikipedia.org/wiki/Binomial_distribution

• Negative Binomial Distribution: https://en.wikipedia.org/wiki/Negative_binomial_distribution

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Solution:

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References:

• Bear Catching Fish Problem: https://math.stackexchange.com/questions/2491326/probability-question-bear-catching-fish

• Binomial Distribution: https://en.wikipedia.org/wiki/Binomial_distribution

• Negative Binomial Distribution: https://en.wikipedia.org/wiki/Negative_binomial_distribution

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Solution:

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References:

• Binomial Distribution: https://en.wikipedia.org/wiki/Binomial_distribution

• Negative Binomial Distribution: https://en.wikipedia.org/wiki/Negative_binomial_distribution

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Solution:

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Reference:

• The Elevator Puzzle: https://math.stackexchange.com/questions/214713/expected-value-problem-10-floors-and-12-people-get-on-an-elevator

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Note: Figure 1 is for logical reference and not drawn to scale.

In Figure 1, a large square is divided into four equal quadrants labeled A, B, C, and D. A smaller square is located at the center. In quadrants A, B, and C, the central square (light-blue area) is removed, creating L-shaped white regions, while quadrant D remains a complete white square.

Divide each region into equal-area pieces as follows:

• White part of quadrant A into 2 identical parts.

• White part of quadrant B into 3 identical parts.

• White part of quadrant C into 4 identical parts.

• Quadrant D into 5 identical parts.

Solution:

This problem tests two important problem-solving skills:

• Observational skills: The ability to notice, interpret, and remember details by using your senses and attention to detect patterns, changes, and subtle cues others may miss. These skills support better problem-solving, decision-making, and understanding of complex situations.

• Solving a problem without a conditioned mindset: Approaching problems without relying on habitual assumptions or familiar patterns, staying open to new perspectives. This leads to more creative, flexible, and unbiased thinking.

Now, let’s solve the problem.

(a). Divide the white part of quadrant A into 2 identical parts.

The white part of quadrant A can be divided into 2 identical parts as shown below in Figure 2.

Note: Figure 2 is for logical reference and not drawn to scale.

In Figure 2, the diagonal line cuts the white part of quadrant A into two equal sized parts labeled 1 and 2.

(b). Divide the white part of quadrant B into 3 identical parts.

The white part of quadrant B can be divided into 3 identical parts as shown below in Figure 3.

Note: Figure 3 is for logical reference and not drawn to scale.

In Figure 3, the 2 line cuts in the white part of quadrant B divides the region into three identical squares labeled 1, 2, and 3.

(c). Divide the white part of quadrant C into 4 identical parts.

This is where your observational skills really get tested. If you closely observe the shape of the white region of quadrant C in Figure 1, you can see that it is symmetrical to the shape of quadrants A, B, and C combined as shown in Figure 4.

Note: Figure 4 is for logical reference and not drawn to scale.

In Figure 4, you can see that the figure can be divided into four identical L-shaped regions: the three white L-shaped regions labeled L₁, L₂, L₃ and the light-blue region labeled L₄. Hence, you can also divide the white region of quadrant C, L₁, into four identical L-shaped pieces labeled 1, 2, 3, and 4 as shown in Figure 5.

Note: Figure 5 is for logical reference and not drawn to scale.

(d). Divide quadrant D into 5 identical parts.

This is where solving a problem without a conditioned mindset gets tested. In parts (a), (b), and (c), we have undergone a process called conditional thinking where our brain is trained to think in a certain way based on prior assumptions. As the level of difficulty was increasing in the previous 3 tasks in parts (a), (b), and (c), we would assume that this final task will be the most difficult. However, by looking at Figure 1, quadrant D is the easiest to divide into 5 identical parts. Quadrants A, B, and C have white regions which are irregularly shaped, so it requires some thinking to divide those white regions equally. Quadrant D is simply a white square which can be easily divided into equal parts without much thinking. We can divide quadrant D into 5 identical rectangles by drawing 4 evenly spaced lines inside quadrant D as shown in Figure 6.

Note: Figure 6 is for logical reference and not drawn to scale.

In Figure 6, quadrant D is divided into 5 identical rectangles labeled 1, 2, 3, 4, and 5.

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References:

• Dividing Square Regions into Equal Parts: https://puzzling.stackexchange.com/questions/137722/dividing-square-regions-into-equal-parts

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Solution:

The game is guaranteed to include a 4 because that is the stopping condition. Since a 4 is always present in the results, any 1, 2, or 3 rolled will automatically be smaller and can never be the maximum value. These smaller numbers are essentially “mathematical noise” that does not affect the outcome for the highest roll. The only numbers capable of being the highest are 4, 5, and 6. Therefore, we focus only on the order of first appearance of these three specific numbers.

There are 3! = 6 possible ways in which the numbers 4, 5, and 6 can first appear. The process stops the moment a 4 is rolled, so any number that would have appeared after the first occurrence of 4 is never observed.

By counting the outcomes from the table above, where k is the maximum value observed (among 4, 5, 6, with the stopping rule at 4), we find:

• P(k = 4),i.e., 4 is the maximum number in 2 out of 6 cases = 2/6 =1/3.

• P(k = 5),i.e., 5 is the maximum number in 1 out of 6 cases = 1/6.

• P(k = 6), i.e., 6 is the maximum number in 3 out of 6 cases = 3/6 = 1/2.

Therefore, by summing the weighted probabilities, the expected value of the highest number rolled—i.e., E[k] = 31/6 ≈ 5.17.

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Reference:

• Expected Value of Highest Die Roll: https://math.stackexchange.com/questions/4576114/expected-value-of-highest-die-roll

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Solution:

Reference:

• 30-Day Pond Puzzle: https://math.stackexchange.com/questions/3067148/one-lily-pad-doubling-in-size-every-day-covers-a-pond-in-30-days-how-long-wou

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• May 15, May 16, May 19

• June 17, June 18

• July 14, July 16

• August 14, August 15, August 17

Shiva then privately tells Krishna the month of his birthday and Ram the day.

After Shiva leaves, the following conversation takes place:

• Krishna: I don’t know Shiva’s birthday, but I am certain that Ram doesn’t know either.

• Ram: At first I didn’t know Shiva’s birthday, but now I do.

• Krishna: Then I also know Shiva’s birthday.

When is Shiva’s birthday?

Solution:

(a). Krishna’s first claim:
“I don’t know the birthday, but I’m sure Ram doesn’t know either.”

The key idea here is uniqueness. A person can determine the birthday immediately only if the information they receive points to a single possible date in the list.

Krishna knows the month, and Ram knows the day. Krishna’s statement means that, based on his month, there is no possible day Ram could have been given that would allow him to identify the birthday immediately.

Looking at the list, the days 18 and 19 each occur only once:

→ June 18.

→ May 19.

If Ram had been told either of these days, he would instantly know the birthday.

Now suppose Krishna had been told May or June. In those cases, it would be possible that Ram received 19 or 18, respectively—and in such a situation, Ram would already know the answer. Since this possibility cannot be ruled out, Krishna would not be able to say with certainty that Ram does not know the birthday.

However, Krishna is certain. Therefore, his month cannot be May or June. It must be a month in which all days appear more than once, so that Ram would remain unsure regardless of which day he was given.

(b). Ram’s statement:
“At first I didn’t know Shiva’s birthday, but now I do.”

From Krishna’s earlier statement, both now know that the birthday must be in July or August.

Let’s look at the remaining possibilities:

→ July: 14, 16.

→ August: 14, 15, 17.

Ram knows the day, and initially he did not know the birthday. This means his day must not have been unique in the original list (so it wasn’t 18 or 19, which we already eliminated).

Now, after Krishna’s statement, Ram re-evaluates using the reduced list.

If Ram’s day were 14, there would still be two possibilities:

→ July 14.

→ August 14.

So he would still be unsure.

However:

→ If Ram’s day is 16, the only remaining option is July 16.

→ If Ram’s day is 15, the only remaining option is August 15.

→ If Ram’s day is 17, the only remaining option is August 17.

In these cases, Ram can now uniquely identify the birthday.

Since Ram says he now knows the birthday, his day cannot be 14.

(c). Krishna’s second claim:
“Then I also know Shiva’s birthday.”

After Ram’s statement, Krishna updates his reasoning using the reduced set of possibilities:

→ July: 14, 16.

→ August: 15, 17.

We already know Ram’s day cannot be 14, since that would leave ambiguity (July 14 vs August 14), and Ram said he now knows the birthday. So 14 is eliminated entirely.

Now only four possibilities remain:

→ July 16.

→ August 15.

→ August 17.

At this stage, Ram’s certainty implies that his day must uniquely identify the date within this reduced set.

Now check each case:

→ If Ram had 16, the only possible match is July 16. → unique ✔

→ If Ram had 15, the only possible match is August 15. → unique ✔

→ If Ram had 17, the only possible match is August 17. → unique ✔

So Ram’s statement alone does not yet distinguish between the three remaining candidates.

Now Krishna uses his knowledge of the month:

• If Krishna had been told July, the only remaining consistent possibility is July 16, which would immediately make the answer certain.

• If Krishna had been told August, there are still multiple consistent possibilities (August 15 or August 17), and nothing in Ram’s statement would allow further elimination.

Since Krishna says he now also knows the birthday, the only consistent resolution is that his information must collapse to a single definite date, i.e. July 16.

Hence, Shiva’s birthday is July 16^th.

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References:

• Cheryl’s Birthday Puzzle: https://www.reddit.com/r/explainlikeimfive/comments/7fnvf7/eli5_the_cheryls_birthday_logic_puzzle/

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(a) Sampled with replacement.
(b) Sampled without replacement.

Solution:

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References:

• Probability Drawings Problem: https://math.stackexchange.com/questions/3743113/intuitive-understanding-of-this-probability-drawings-question

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