request for assistance for probability formula

Hi,

I tried to figure out the probability for this several times but was unsuccessful, please figure it out if you can:

Statistical distribution of likely next last digits of primes

The way that is easiest to see it is for primorial 30 equation sets that produce all primes (except 2,3,5) with the formula y=mx+b

y = 30x + 1 y = 30x + 7 y = 30x + 11 y = 30x + 13 y = 30x + 17 y = 30x + 19 y = 30x + 23 y = 30x + 29

y = 30x + 1 (produces half of primes ending in 1) y = 30x + 7 (produces half of primes ending in 7) y = 30x + 11 (produces half of primes ending in 1) y = 30x + 13 (produces half of primes ending in 3) y = 30x + 17 (produces half of primes ending in 7) y = 30x + 19 (produces half of primes ending in 9) y = 30x + 23 (produces half of primes ending in 3) y = 30x + 29 (produces half of primes ending in 9)

Each equation produces half of all primes ending in a give digit 1,3,7,9

Here are the prime1, prime2 consecutive last digit occurrence counts for the 4 permutations of b values 1,11 that can have consecutive primes both ending in 1: (full table at bottom of page)

p1,p2,count

1,11,2159 11,1,547 11,11,266 1,1,241

The sum of the four counts is much lower, than for example primes p1,p2 where p1 ends in 1, and p2 ends in 7:

p1,p2,count

1,7,2849 11,17,2193 1,17,1087 11,7,421

Right there that shows for a given primorial and a number range you can see the likelihood of consecutive primes ending digits, however I didn't do a full statistical formula considering all primorials and any number range, tried and gave up for now :D

Please figure out a generalized formula for any number range to show the probability of a two consecutive last primes last digit occurrence.

cheers, Jamie

full table:

p1,p2,count

7,11,2945 11,13,2916 17,19,2914 29,1,2904 23,29,2893 13,17,2892 19,23,2887 1,7,2849 23,1,2209 11,17,2193 17,23,2181 13,19,2163 1,11,2159 19,29,2154 29,7,2153 7,13,2087 23,7,1824 7,17,1624 13,23,1590 1,13,1585 19,1,1561 11,19,1551 17,29,1551 29,11,1526 29,13,1345 17,1,1278 13,29,1160 7,19,1138 11,23,1133 1,17,1087 19,7,1020 23,11,993 19,11,957 1,19,811 7,23,795 11,29,783 29,17,774 13,1,737 17,7,724 23,13,720 1,23,665 7,29,633 13,7,560 11,1,547 23,17,543 29,19,519 19,13,517 17,11,516 13,11,448 19,17,431 23,19,431 11,7,421 1,29,410 17,13,380 29,23,363 7,1,330 13,13,274 11,11,266 17,17,265 19,19,261 7,7,260 1,1,241 23,23,226 29,29,221
Reply to
Jamie M
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Shouldn't you find a mathematics newsgroup for all this prime-number stuff?

Speaking as a circuit designer, an embedded software engineer, AND a systems engineer, I don't give a fat rat's ass about the distribution of primes. For my purposes they can be considered completely random.

All these "prime number" posts are about as appealing to me as all of the political posts from the various left- and right-leaning freaks.

--
Tim Wescott 
Wescott Design Services 
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Reply to
Tim Wescott

Well they aren't random there's lots of patterns in prime number distribution. Conversely I don't care about any possible applications of primes related to engineering or computer science if they were shown to be non random so I guess we agree on something :)

cheers, Jamie

Reply to
Jamie M

On 8 Apr 2016, Jamie M wrote (in article ):

He posts to sci.math and crosses to s.e.d. You care about primes about as much as he cares about not cross-posting to inappropriate groups.

Jamie: why do you think electronics designers care about this? Probably just a numbers thing* (c;

  • s.e.d. oncebeing outed as one of the top most-posted-to USENET groups...
Reply to
DaveC

He already admitted he is trolling. just killfile him.

--
  \_(?)_
Reply to
Jasen Betts

I, for one, have done just that. The problem is that replies to his thread still show up.

Thanks.

Reply to
John S

Thats too bad you never get to try this:

30030 + 30030 in a calculator then press equals lots.

Reply to
Jamie M

To be honest I am mainly looking for free feedback to assist in my own learning. As long as anyone cares enough to reply that is great, engineers on here tend to whine a bit but I assumed they would have a tough time resisting a math problem. Of course with a well organized news client it isn't hard to organize into threads (usually that is default) so math based threads I post and reply to shouldn't take up any more room than other threads. As interesting as the math questions I have may or may not be, they pale in comparison to the question why I am posting here obviously, but I hope I answered as well as can be expected. I appreciate any math feedback, none of my questions are too difficult to answer I think, and I plan to keep making progress on whatever it is I am doing.

cheers, Jamie

Reply to
Jamie M

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