[quote=Flatlander;190586]What is the probability of finding a prime before n=500k?[/quote]
From 418k-500k, about 0.310 primes are expected, which gives a 26.6% (1 in 3.75) chance of finding at least one prime. (this is working from the file posted in the first post, which IIRC ignores Lennart's work on the other 3 k's, so this is only for the other 4)

Thanks. Good enough for me. :smile:

[quote=Mini-Geek;190587]From 418k-500k, about 0.310 primes are expected, which gives a 26.6% (1 in 3.75) chance of finding at least one prime. (this is working from the file posted in the first post, which IIRC ignores Lennart's work on the other 3 k's, so this is only for the other 4)[/quote]
Thinking that a more accurate figure could be obtained by calculating the odds of a prime for the entire 150K-1M range (rather than just the small slice of 418K-500K), I tried running the entire p=6T file (with all 8 k's in it) through Gary's odds of prime spreadsheet, and came up with a percentage chance of 98.613%, or 1 in 1.01, to find one prime. But that doesn't sound quite right. Only one prime in all of n=150K-1M? Gary, please tell me I did something wrong here! :smile:

Chris waits for Gary 'six-decimal-places' Barnes to take the bait. :motorhome:

[quote=mdettweiler;190593]Thinking that a more accurate figure could be obtained by calculating the odds of a prime for the entire 150K-1M range (rather than just the small slice of 418K-500K), I tried running the entire p=6T file (with all 8 k's in it) through Gary's odds of prime spreadsheet, and came up with a percentage chance of 98.613%, or 1 in 1.01, to find one prime. But that doesn't sound quite right. Only one prime in all of n=150K-1M? Gary, please tell me I did something wrong here! :smile:[/quote]
You're misreading it. If the chance is 98.613%, and there's a 1 in 1.01 chance of getting at least one prime, then the expected number of primes is far more than 1. What does the expected primes line say? I think it'll be around 4.28.
Remember that this doesn't consider what you remove as you find primes, so you might expect roughly more like 3 or 3.5 k's to be eliminated.

Also, it's really more inaccurate to consider the full file. You really ought to exclude any thought of ranges known to have no primes, and exclude the k that already has a prime. Flatlander's question was much more relevant, though you might want to include Lennart's k's.
While it's true that doing the full file would find what we were expecting at the start, we now have much more information and can make a [I]far[/I] more accurate estimation.

[quote=Mini-Geek;190597]You're misreading it. If the chance is 98.613%, and there's a 1 in 1.01 chance of getting at least one prime, then the expected number of primes is far more than 1. What does the expected primes line say? I think it'll be around 4.28.
Remember that this doesn't consider what you remove as you find primes, so you might expect roughly more like 3 or 3.5 k's to be eliminated.

Also, it's really more inaccurate to consider the full file. You really ought to exclude any thought of ranges known to have no primes, and exclude the k that already has a prime. Flatlander's question was much more relevant, though you might want to include Lennart's k's.
While it's true that doing the full file would find what we were expecting at the start, we now have much more information and can make a [I]far[/I] more accurate estimation.[/quote]
Ah, I see it now! Yes, it's about 4.28 expected primes. Okay, that makes sense. Seems I had it all wrong before...no wonder I always was coming up with overly pessimistic estimates for my individual conjecture efforts. :rolleyes:

As usual, Tim is spot-on on his statistical comments, both for the entire file and for 4 remaining k's for the n=418K-500K range. As he said, at the BEGINNING, our true expection would probably have been to find 3 to 3.5 k's (not 4.28) with primes for the entire n=150K-1M file. This is as a result of the stoppage of searching a k once a prime is found. Since it is now known information that we only have 1 prime up to n=418K, obviously that expection is much less.

Unfortunately we're probably only looking at an expectation of 1-2 primes the rest of the way to n=1M. (i.e. total of 2-3) The fact that we've found much less than expected so far has no bearing on what will happen the rest of the way so you can't use the entire file to determine what we should expect to find from here.

Max, if you want to take what is remaining to be searched from the sieve file, plug it into the spreadsheet, and see what the expectation is the rest of the way, go ahead. You'd have to break it up in 2 steps and add them together: (1) For the 4 k's remaining up to n=500K. (2) For all 7 k's for n=500K-1M.

A note about it: Sieving further does not increase the expected # of primes, it simply increases the chances of each candidate being prime but it reduces the total # of candidates, which evens things out. The spreadsheet "knows" this since the sieve depth is fed in. So the fact that we'll sieve more for n=500K-1M has no effect on the # of expected primes.


Gary

[quote=gd_barnes;190702]Max, if you want to take what is remaining to be searched from the sieve file, plug it into the spreadsheet, and see what the expectation is the rest of the way, go ahead. You'd have to break it up in 2 steps and add them together: (1) For the 4 k's remaining up to n=500K. (2) For all 7 k's for n=500K-1M.[/quote]
Okay, here's what we've got for n=418K-500K for the 4 k's in the main drive:
[I]odds of 1 6.7341187E-05[/I]
[I]1 in 14,849.75[/I]
[I]odds of all candidates 26.684%[/I]
[I]1 in 3.75[/I]
[I]expected # of primes [B]0.310[/B][/I]

And then for 500K-1M on all 7 k's:
[I]odds of 1 4.1213031E-05[/I]
[I]1 in 24,264.17[/I]
[I]odds of all candidates 81.692%[/I]
[I]1 in 1.22[/I]
[I]expected # of primes [B]1.698[/B][/I]

Adding the expected #'s of primes, we get [B]2.008[/B] expected primes before n=1M. Definitely not a pretty prospect, though it does make the fact that we've only found one prime yet sound a bit more realistic after all. :sad: Gary, I see now why you were guessing 10M to prove the whole conjecture assuming we can't knock out k=1597 early on.

Taking 428-438.

[quote=mdettweiler;190708]Okay, here's what we've got for n=418K-500K for the 4 k's in the main drive:
[I]odds of 1 6.7341187E-05[/I]
[I]1 in 14,849.75[/I]
[I]odds of all candidates 26.684%[/I]
[I]1 in 3.75[/I]
[I]expected # of primes [B]0.310[/B][/I]

And then for 500K-1M on all 7 k's:
[I]odds of 1 4.1213031E-05[/I]
[I]1 in 24,264.17[/I]
[I]odds of all candidates 81.692%[/I]
[I]1 in 1.22[/I]
[I]expected # of primes [B]1.698[/B][/I]

Adding the expected #'s of primes, we get [B]2.008[/B] expected primes before n=1M. Definitely not a pretty prospect, though it does make the fact that we've only found one prime yet sound a bit more realistic after all. :sad: Gary, I see now why you were guessing 10M to prove the whole conjecture assuming we can't knock out k=1597 early on.[/quote]


What n-value did you use for the above calculations? Using the average n-value for each range, i.e. n=459K and n=750K respectively will understate the estimate. The spreadsheet is not designed for computing the expected # of primes over a wide n-range. Sorry...I forgot to mention that. It would take calculus that I am unfamiliar with to determine it exactly. As Chris said, Gary "6 decimal place" Barnes talking here. lmao Good one Chris!

A closer estimate without getting into calculus would be as follows:

Use the avg. n-range for the following and determine the expected # of primes for each range for the # of candidates in each range:
n=418K-500K, i.e. 459K
n=500K-600K, i.e. 550K
n=600K-700K, etc.
n=700K-800K
n=800K-900K
n=900K-1M

Now add all of the expected # of primes together and you should end up with a value somewhat higher than 2.008. Even more accurate would be to do it in n=10K ranges but that would take too long. The most accurate (actually exact way) would be to use infinitely small increments, which is what calculus would help you do in a reasonable time frame.

The reason why using the average doesn't work is because the reduction in expected # of primes as the n-value increases is not linear.

Regardless, whatever expectation you come up with, likely to be around 2.1 to 2.4 somewhere, would need to be reduced due to stopping the searching of a k once a prime is found. Therefore, we're likely to end up with a current expectation slightly under 2 primes from this point forward.

My initial estimate of n=10M was undoubtedly too low anyway. I was throwing out a SWAG and should have done a little more analysis to start with. It likely should have been more like n=25M-30M. But now with just 1 prime for n=150K-418K, I think it will be more like n=60M-100M, possibly higher.

If we can get 2 more primes up to n=1M, then I think there's a chance we could still prove this thing in the next 10-25 years. If not, the proof may extend beyond many of our lives here and that's with "only" 7 k's remaining! Believe it or not! :smile:

Tim, if you're familiar with how you would properly compute the EXACT expected # of primes over a wide n-range since it is non-linear, feel free to offer up some calculus here for doing so. Likely you'll need to look at the formulas in the spreadsheet. 1-2 of which were designed by axn1 in a thread at RPS and most of the rest were designed by yours truly. I've been meaning to take a college-level calculus class at some point. All I've had is very basic high school calculus and that was over 25 years ago and so I don't remember much. :smile:

Max, you compared this expectation of ~2 the rest of the way to the fact that we've only found 1 prime so far to draw a conclusion that finding 1 prime so far is not unusual. You concluded correctly but based on the incorrect reason. To conclude it correctly, you need the original expection for n=150K-418K, which was likely ~2 to ~2.5 or so. True, it's not terribly unusual to have only found 1 prime with an expectation of 2-2.5 but drawing such a conclusion based on the expectation of the # of primes for n=418K-1M could have easily caused you to draw an incorrect conclusion had that expectation been < 1 prime or > 5 primes.


Gary

Ah, I see. Yeah, you're right, it would be rather time-consuming to have to do all of those range individually...but, you know, I just had a thought. Maybe I could write up some kind of Perl script to automatically read in a sieve file and do the same calculations on it that your spreadsheet does! Even though I wouldn't even begin to understand the calculations, as long as I copy them verbatim from the spreadsheet, I should be OK, and I can of course use the spreadsheet to verify that I've gotten everything set up right. But, the key is, with a computer at the helm, it wouldn't be so time-consuming to break things down into little n-ranges; it could, say, automatically split everything up into ranges of n=5K and add together the indvidual pieces.

No promises...I'm pretty busy right now and probably won't have enough time to do it. Nonetheless, it would be somewhat interesting idea. :smile:

As for your new estimate for the n-range this conjecture might need to go to: :shock: Man, we'd better get some BIG advances in computer technology between now and then! I'm hopeful that in 10 years we'll be able to do n=100M (base 2) tests in hours rather than in years, which would go a long way towards making this conjecture (and others) provable, but of course there's no way to predict the future.