\[ \begin{align} 11 &\rightarrow 1 \\ 12 &\rightarrow 2 \\ 13 &\rightarrow 3 \\ 14 &\rightarrow 4 \\ 15 &\rightarrow 5 \\ 16 &\rightarrow 6 \\ 17 &\rightarrow 7 \\ 21 &\rightarrow 8 \\ 22 &\rightarrow 9 \\ 23 &\rightarrow 10 \\ \end{align} \]

Then we can ignore every other combination and we have our random generator for 1-10. Any problem with this approach?

Obviously, this approach wastes the majority of draws. Can we do better?

There are 49 combinations (\(7 * 7\)) that we can get from two consecutive draws. We combine them as follows:

\[ f(\text{draw1, draw2}) = (\text{draw1}-1) * 7 + \text{draw2} - 1 \]

where we shifted the minimum combination to be zero:

\[ 0 * 7 + 0 = 0\\ 0 * 7 + 1 = 1\\ ...\\ 6 * 7 + 6 = 48 \] Now we need to map the numbers 0-48 to 1-10 such that we can sample uniformly. The maximum numbers (0-48) that we can use for each mapped number 1-10 without biasing the results is:

\[ 48 // 10 = 4 \]

This means that number 1 can be mapped by at most 4 unique numbers between 0,48, number 2 can be mapped by at most 4 unique numbers between 0,48, etc. This also suggests that we will have to sacrifice 9 numbers (i.e, 49 unique numbers - 4 * 10 = 9 numbers that will have no mapping).

We can achieve this goal by using modulo arithmetic as follows:

\[ map(draw1, draw2) =f(\text{draw1, draw2}\big) \% 10 + 1 \]

which will give us the following mapping:

\[ \begin{align} 0\%10 + 1 &\rightarrow 1 \\ 1\%10 + 1 &\rightarrow 2 \\ ...\\ 9\%10 + 1 &\rightarrow 10 \\ ...\\ 39\%10 + 1 &\rightarrow 10 \\ 40-48 &\rightarrow \text{Ignore} \end{align} \]

One way to code this up is as follows:

def random_1_to_7():
    return np.random.randint(1, 7)

def random_1_to_10():
    while True:
        a = random_1_to_7()
        b = random_1_to_7()
        combined = (a - 1) * 7 + b - 1
        if combined < 40:
            return (combined) % 10 + 1

To run it:

import numpy as np
np.random.seed(1234)
from collections import Counter
# Test the function
samples = 1000
results = Counter([random_1_to_10() for _ in range(samples)])
print([f"({i}:{results[i]/samples:.3f})" for i in range(1,11)])

## ['(1:0.091)', '(2:0.121)', '(3:0.104)', '(4:0.092)', '(5:0.085)', '(6:0.105)', '(7:0.089)', '(8:0.086)', '(9:0.124)', '(10:0.103)']

## True

## Counter({10: 132, 6: 118, 2: 114, 9: 109, 3: 105, 5: 92, 4: 87, 7: 83, 8: 82, 1: 78})

And as a result in this case we only wast 8/49 draws.