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I am not familiar with this kind of problem and I am trying to find what could be the general term or pattern used for the color assignment in the Mark Six lottery (This is the matrix of colors).

There are three sequences:

Red balls: [1, 2, 7, 8, 12, 13, 18, 19, 23, 24, 29, 30, 34, 35, 40, 45, 46];

Red(n) = ?

Blue balls: [3, 4, 9, 10, 14, 15, 20, 25, 26, 31, 36, 37, 41, 42, 47, 48];

Blue(n) = ?

Green balls: [5, 6, 11, 16, 17, 21, 22, 27, 28, 32, 33, 38, 39, 43, 44, 49];

Green(n) = ?

How do you think would be the best approach to find the pattern followed there? What would be the general term for those number sequences?

Thanks in advance!

JHH
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1 Answers1

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Here's a table of the ball colors in a format which makes the pattern a little more visible:

$$\begin{array}{|rr|rr|rr|} \hline \color{red}{1} & \color{red}{2} & \color{blue}{3} & \color{blue}{4} & \color{green}{5} & \color{green}{6}\\ \color{red}{7} & \color{red}{8} & \color{blue}{9} & \color{blue}{10} & ** & \color{green}{11}\\ \color{red}{12} & \color{red}{13} & \color{blue}{14} & \color{blue}{15} & \color{green}{16} & \color{green}{17}\\ \color{red}{18} & \color{red}{19} & \color{blue}{20} & ** & \color{green}{21} & \color{green}{22}\\ \color{red}{23} & \color{red}{24} & \color{blue}{25} & \color{blue}{26} & \color{green}{27} & \color{green}{28}\\ \color{red}{29} & \color{red}{30} & ** & \color{blue}{31} & \color{green}{32} & \color{green}{33}\\ \color{red}{34} & \color{red}{35} & \color{blue}{36} & \color{blue}{37} & \color{green}{38} & \color{green}{39}\\ \color{red}{40} & ** & \color{blue}{41} & \color{blue}{42} & \color{green}{43} & \color{green}{44}\\ \color{red}{45} & \color{red}{46} & \color{blue}{47} & \color{blue}{48} & \color{green}{49} & \\ \hline \end{array}$$

The pattern is easier to express mathematically if we subtract 1 from the ball number, so that the numbers start from zero.

We can calculate the color of a number $i$ using this formula:

Let $j = i - 1$
Then $$k = floor(((j + floor(j / 10)) \mod 6) / 2)$$ $k$ will be in ${{0, 1, 2}}$, with 0 = red, 1 = blue, 2 = green.

So subtract 1 from the ball number $i$ to get $j$. Then divide $j$ by 10, rounding down, and add that to $j$. Now find the remainder of that total, mod 6. Finally, divide that remainder by 2 (rounding down) to get the color number.

Here's a short Python script which tests that formula.

red = [1, 2, 7, 8, 12, 13, 18, 19, 23, 24, 29, 30, 34, 35, 40, 45, 46]
blue = [3, 4, 9, 10, 14, 15, 20, 25, 26, 31, 36, 37, 41, 42, 47, 48]
green = [5, 6, 11, 16, 17, 21, 22, 27, 28, 32, 33, 38, 39, 43, 44, 49]

groups = [[], [], []]
for i in range(1, 50):
    j = i - 1
    u = (j + j // 10) % 6 // 2
    groups[u].append(i)


print(red == groups[0])
print(blue == groups[1])
print(green == groups[2])

You can run it on the SageMathCell server here. The table at the start of this answer was also created in Sage / Python using the above formula. Here's the table script.

PM 2Ring
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