steveK

I will have to run that experiment.

wow... statistics and probability can be painful. This a tough punch in the gut.

@MishaLavrov sorry for the poor post. I revised it to remove the section on schemas as that should've never been there. I modified the section on Case2 and let readers know that this is my opinion and not approved by mathematicians!

@peterwhy Lockout by first draw or lockout probability ... I should've explained this better. Lets take the case for the 84 combinations that have all members divisible by 5. There are 9 numbers in a 6 of 49 lottery that are all divisible by 5. The means I have a probability of 40/49 to select a number that is not divisible by 5... and if I get on of those numbers which are not divisible by 5, the I have locked out all 84 combinations from being drawn.

Lockout by first draw or lockout probability ... I should've explained this better. Lets take the case for the 84 combinations that have all members divisible by 5. There are 9 numbers in a 6 of 49 lottery that are all divisible by 5. The means I have a probability of 40/49 to select a number that is not divisible by 5... and if I get on of those numbers which are not divisible by 5, the I have locked out all 84 combinations from being drawn.

signing off for now

hope that does not peeve people off

But since my schemas indicate high and low within the distributions... isn't that a form of bias? I think it's a form of bias -- I'll call it "ranking-bias".

I think I'm good here with this, I just seemed to take a different approach that got people peeved because I expressed it poorly. But I kinda think we're in agreement and maybe should have never made the claim: "not all lottery combinations have the same probability of being drawn".

@peterwhy yes I see that. I suspect in a backdoor-ish way Misha's analysis confirms my analysis... not directly, but indirectly. For example since there are only 9 numbers divisible by 5 equating to 84 combinations.... and since 40/49 > 9/49 then it stands to reason there is greater probability to get other numbers not divisible by 5. Thus the odds are stacked against combos with every member divisible by 5

@peterwhy ops a typo yes 7

@peterwhy This why I ran an analysis on the WA State lottery to see the distribution of combinations ... in schema form. Schema that has all members of its combination divisible by 5 have never been drawn. If there was true equity, then one of those combinations should have been drawn.

@peterwhy I'm not sure the lottery is unfair, but I believe there's bias that is against certain kinds of combinations who's members are in few supply... like all combinations where each member is divisible by 6, there are only 8 of those numbers. So every time a new lottery game is played those 8 numbers are competing against 41 other numbers to be selected.

I can't see how to calculate the bias, but it's there.

This bias I believe comes from the building of the combination one number at a time, which is the way all lotteries are done.

We are more likely to draw combinations where every member of the combination is all even or all odd because there a large sample to draw from. My goal is to show that there are combinations that will be difficult to draw because they fit in a certain schema.

There are 7 numbers divisible by 7, giving 6 combinations. This means there are 42 numbers not divisible by 7. So with that said, at the first draw attempt I have a 42/49 probability to lock out any of those 6 combinations. And this will happen every time we start a new lottery game. The odds are stacked against these 6 combinations because we have so many more options 42/49 to get something else other than a number divisible by 7.

@MishaLavrov That's good question because I'm not sure how to calculate it. If the balls had alphabetical characters on them, I would've never posted this question. But since the balls have numbers on them, they become self-identifying... prime, odd, even and therefore they have ranking. I'm really not trying to be a jerk here. We can do this for numbers divisible by 7.

@MishaLavrov to answer your question, if we take the pure mathematical probability then yes the probability of it being drawn is $\frac{1}{13,983,816}$ there is no disagreement. What I'm trying to point out, is that when we start the first draw attempt there are $43$ numbers that are not divisible by $8$ and so there is a slight (incalculable) bias because there is a $\frac{43}{49}$ probability to get something else... thus locking out that combination.

That's not the point I was making. I used primes, and numbers divisible by 5 and 8 to illustrate a point of locking-out. I know it's possible to draw a combination with all its members divisible by 5... but since there are 40 numbers not divisible by 5 to choose from, the probability of drawing that type of combination is less likely "because" we are constructing a combination one number at a time. I know mathematicians hate this question, but I think I'm correct.

If someone offered you a 1,000,000 dollars to reach in the pool of 49 numbers if you pull out any number divisible by $8$ or give you 100,000 dollars if you pull out any odd number. Which one offer will you take?

Hi Misha. Would you agree that during the first number drawn, there is a higher probability to draw a prime numbered ball over a numbered ball divisible by $8$? There are $15$ prime numbers between $1$ and $49$ and only $6$ numbers that are divisible by $8$ ... so basic probability says yes we have a higher probability to draw a prime number ball... thus locking out that combination where all of its members are divisible by $8$. So it appears I'm correct. That's the bias... simple basic probability.

@TonyK Why was I downvoted? Is there something wrong with my proof? Let me know if I made a mistake please.

Let us continue this discussion in chat.

@MishaLavrov I have just a Bachelors degree and I could NEVER compete at level you do. You have a PhD. So, with that said, if you would be kind enough to provide another proof, I would like it very much. Sincerely!! My comment about DEI is because I've been assimilated into the Borg.

@MishaLavrov if you have another proof, please submit it in the answer section. Being a supporter of DEI, I would like to see other proofs. As I said sorry for the miscommunication.

@MishaLavrov [if we were to randomly choose pairs of $P$ and $Q$ many times over and calculate $N$, we would discover that the furthest $P$ can ever be from $\sqrt{N}$ is $P$] . So I made this claim. My apology to confuse the situation. Looks clear to me.

@MishaLavrov then why didn't you shoot off a proof then?

@MishaLavrov you'll soon find out that to factor semiprimes quickly, you'll need to stop think that $Q$ cannot be even. You'll need to be brave and step into the world of The Reals. As you can see, the partial derivative gives us the slope which is in The Reals. Once $4P$ is substituted back into $d_1$... it all works out... even though it makes you sick to see $Q=4P$. Cheers! There will new surprises to come, not yet revealed. Cool stuff coming. Most all people working with Primes think they need to stay inside the integer box and that's wrong thinking.

@Peter I removed the confusing parts, and provided a proof.

@TonyK I have provided a proof.

@MishaLavrov I provided a proof I believe is solid.

Okay so I removed the "confusing part". I updated it once again.

I updated the question to be more clear

@MishaLavrov so $4P^{2}=N$

@MishaLavrov but the other condition is that we must maintain $PQ=N$ along with $Q=4P$. I’m not trying to be annoying or a jerk.

@MishaLavrov but mathematically it appears to be a true statement. It’s okay to step into the domain of the Real Numbers to see this is a true statement.

@KeithBackman the condition appears to only occur when $Q=4P$.

@Peter I snuck in the condition that this can only happen if $Q=4P$.

@MishaLavrov I snuck in the condition that this can only happen if $Q=4P$

I sneaked in the condition at the end that this can only happen if $Q$ is be equal to $4P$

Never mind I fixed it

Misha is there something wrong with my answer? It got downvoted but nobody provided feedback on why they downvoted it.

@MishaLavrov I have just a Bachelors degree and I could NEVER compete at level you do. You have a PhD. So, with that said, if you would be kind enough to provide another proof, I would like it very much. Sincerely!! My comment about DEI is because I've been assimilated into the Borg.

@MishaLavrov if you have another proof, please submit it in the answer section. Being a supporter of DEI, I would like to see other proofs. As I said sorry for the miscommunication.

@MishaLavrov [if we were to randomly choose pairs of $P$ and $Q$ many times over and calculate $N$, we would discover that the furthest $P$ can ever be from $\sqrt{N}$ is $P$] . So I made this claim. My apology to confuse the situation. Looks clear to me.

@MishaLavrov then why didn't you shoot off a proof then?

@MishaLavrov you'll soon find out that to factor semiprimes quickly, you'll need to stop think that $Q$ cannot be even. You'll need to be brave and step into the world of The Reals. As you can see, the partial derivative gives us the slope which is in The Reals. Once $4P$ is substituted back into $d_1$... it all works out... even though it makes you sick to see $Q=4P$. Cheers! There will new surprises to come, not yet revealed. Cool stuff coming. Most all people working with Primes think they need to stay inside the integer box and that's wrong thinking.

@Peter I removed the confusing parts, and provided a proof.

@TonyK I have provided a proof.