[QUOTE=shanecruise;308402]if 6.67 is the probability of one prime in a file then 1:6.67^3 is probability of finding 1 prime in each of 3 consecutive files.[/QUOTE]

You may be right on this... :unsure:

Meanwhile I digged out the formula for computing the odds for a single test to yield a prime, depending on k, n, and the sieve depth:

p = 1.781*log(sievedepth) / (log(k) + n*log(2))

Taking n = 1080000, an average k = 6000 and sievedepth = 100,000,000,000,000,000 (100P) we get p = 0.000093127.

Inserting this as "x" into your Bernoulli formula and taking N=1479 (the number of candidates in your file) we get the following:

P(0) = 0.871324, or a chance of 1:1.15
P(1) = 0.120023, or a chance of 1:8.33
P(2) = 0.008261, or a chance of 1:121
P(3) = 0.000379, or a chance of 1:2640
P(4) = 0.000013, or a chance of 1:76818

P(3+) = 1 - (P(0) + P(1) + P(2)) = 0.000392, or a chance of 1:2550

On average there should be one prime in 10743 tests ( = 1/p), which is a bit more than my rough estimate of one in 10000.

:beer: perfect

Thomas provided the per-test formula I had wished to look up, and the probability calc was exactly what I intended, since each prime test is independent.

2550-to-1 are steeper odds than I would have guessed for that event!
-Curtis

1215*2^1295400-1 (389958 digits)

177*2^1775674-1 (534534 digits)

From the 10th drive:
24217*2^1304085-1 (392574 digits)

263*2^1587306-1

Hi Oscar,
The server at Top-5000 says that above number is composite.
Can you check your submission, maybe you mistyped, or submitted a wrong number.
Or, maybe, but hopefully not, you have a hardware issue.

Thanks

You are right Kosmaj, I misstyped it.
I resubmitted as 263*2^1587302-1 (477828 digits).
This one should be the good one.

225*2^1177945-1 (354600 digits)

4037*2^1000136-1 (301075 digits)