Deriving Random10() function from Random7() in JavaScript

Suppose we have a random7() function that generates random numbers from 1 to 7. We need to create a random10() function that generates random numbers from 1 to 10, using only the random7() function.

Problem

const random7 = () => Math.ceil(Math.random() * 7);

This function yields a random number between 1 and 7 (inclusive) every time we call it. We need to write a random10() function that returns random numbers between 1 and 10 (inclusive) using only this random7() function.

Using Rejection Sampling (Optimal Solution)

The most efficient approach uses rejection sampling. We generate numbers in a range that's evenly divisible by 10, then map the results.

const random7 = () => Math.ceil(Math.random() * 7);

const random10 = () => {
    let result;
    do {
        // Generate a number from 1-49 (7×7 grid)
        result = (random7() - 1) * 7 + (random7() - 1);
    } while (result >= 40); // Reject 40-48 to keep uniform distribution
    
    return (result % 10) + 1; // Map 0-39 to 1-10
};

// Test the function
for (let i = 0; i 

3
7
1
9
4


How It Works

The algorithm works by:

• Creating a 7×7 grid (49 possible outcomes) using two random7() calls
• Rejecting results 40-48 to maintain uniform distribution
• Mapping the remaining 40 outcomes (0-39) to 1-10 using modulo

Alternative: Summation Approach (Less Optimal)

Here's a corrected version of the summation approach, though it's less mathematically sound:

const random7 = () => Math.ceil(Math.random() * 7);

const random10Summation = () => {
    let sum = 0; // Initialize sum to 0
    for (let i = 0; i 

8
3
6
2
9


Comparison

Conclusion

The rejection sampling method provides perfectly uniform distribution and is the optimal solution. The summation approach, while simpler, doesn't guarantee uniform randomness across all outcomes.