Analysis: Would constructing lottery combinations by drawing one number at a time give combinations different probabilities of being drawn?

Question

I’m not sure that all lottery combinations have the same probability of being drawn. Numbers are drawn one at a time to construct the $6$ numbered combination we see in a straight pick $6$ lottery (without extra power-ball structure). I’ll go through two cases I’ve examined to demonstrate my claim. I’ll work with the $6$ of $49$ Washington State Lottery which has $13,983,816$ combinations.

Do you agree?

Case1: This case will show the only true way to guarantee all combinations have the same probability of being drawn. This case requires that all combinations of $6$ possible numbers to be generated pre-drawing, and that each combination is assigned a random index number $1$ through $13,983,816$ before each draw. If index numbers were assigned directly to each combination in an ascending order 1-to-1 assignment, there would be bias that would come from a Geometric (1/13,983,816) distribution of the index numbers. This case requires the generation of only $1$ random index number which will be the address into the lookup table containing the pre-generated combinations. This guarantees equal probability to all combinations.

$P = \frac{1}{13,983,816}$

Case2: (Posters Opinion, not approved by Mathematicians)

This case will show there is bias when combinations are constructed one number at a time, rather than just using the method in Case1. Those combinations come prepackaged.

To show this, we need only to analyze the probability of the first number drawn from the pool of $49$ numbers. It will be the first draw only that shows not all combinations have the same probability of being drawn… because they're constructed one number at a time due to a condition I call the lockout by first draw.

Let’s examine three combinations, a combination where every number in the combination is divisible by $8$, a combination where every number in the combination is prime, and a combination where every number in the combination is divisible by $5$.

There are $6$ numbers that are evenly divisible by 8, which yields only $1$ combination out of $13,983,816$.

There are $15$ numbers that are prime which yield $5005$ combinations out of $13,983,816$.

There are $9$ numbers that are divisible by $5$, which yield $84$ combinations out of $13,983,816$.

The Process of Building Combinations

To build any combination we need to select our first number out of the pool of $49$. Using basic probability here, the probability of selecting a number that is divisible by $8$ is $P_8 = \frac{6}{49}$ and the probability of selecting a number that is not divisible by $8$ is $P_{!8} = \frac{43}{49}$.

The probability to select a prime number is $P_P = \frac{15}{49}$ and the probability to select a number that is not prime is $P_{!P} = \frac{34}{49}$.

The probability to select a number divisible by $5$ is $P_5 = \frac{9}{49}$ and the probability to select a number that is not divisible by $5$ is $P_{!5} = \frac{40}{49}$.

So right away there’s a greater probability to lockout by first draw numbers that are divisible by $8$ because $\frac{43}{49} > \frac{6}{49}$.

This says clearly that the combination $8$ $16$ $24$ $32$ $40$ $48$ has a lower probability to be constructed through the process of building combinations one number at a time. Looking at the first draw probability shows this.

$P_P > P_5 > P_8$

$\frac{15}{49} > \frac{9}{49} > \frac{6}{49}$

$P_{!8} > P_{!5} > P_{!P}$

$\frac{43}{49} > \frac{40}{49} > \frac{34}{49}$

But no matter how you do the math, all combinations do have the same probability to be drawn... even if the combination is constructed one number at a time. Below for the only combination where all numbers are divisible by $8$.

$P = (\frac{6}{49})(\frac{5}{48})(\frac{4}{47})(\frac{3}{46})(\frac{2}{45})(\frac{1}{44}) = 13,983,816$

Conclusion

Case2 tried to show that as combinations are constructed one number at a time, there may be bias in the draw. But this is false.

I took a beatdown on this post and had to revise it. I took a different approach on this topic and stepped on my own toes because I did not see the forest for the trees. I had to remove the section discussing "schemas" as this should've never been added here.

Answer 1

This is not evidence of bias. If the $28$ different "schemas" were equally likely, that would be a biased drawing.

Why?

• Only $\binom93 = 84$ combinations out of $\binom{49}{6} = 13\,983\,816$ fall under schema $28$.
• Meanwhile, schema $7$ (all six numbers are from category B) corresponds to $\binom{20}{3} = 1\,140$ combinations out of $13\,983\,816$, so of course it should be more common.
• In your sample, the most common schema is schema $11$, with a $132$ pattern; it corresponds to $\binom 91 \binom{20}{3} \binom{20}{2} = 1\,949\,400$ different combinations. So of course it should be much more common than a schema with only $1\,140$ combinations!

However, it is silly to say that schema $11$ is somehow advantageous in the lottery. Sure, it is more likely that some combination of this type is chosen. But if you picked a schema-$11$ combination, and a schema-$11$ combination did come up, you still have only a $\frac{1}{1\,949\,400}$ chance of winning! On the other hand, if you picked a schema-$28$ combination, and a schema-$28$ combination comes up, you have a $\frac1{84}$ chance of winning. The overall probability is $\frac1{13\,983\,816}$ both ways, it's just that you've broken it up into stages differently.