I was surprised last week to find this value:
$m_{p_{0000}}=0.16726218229590580987863882056891582636342622102204{}^{1}$
$m_{p_{1959}}=0.167339(31){}^{2}$
$m_{p_{2012}}=0.1672621777(74){}^{3}$
for a formula I had designed for a completely unrelated purpose.
$$\sum _{m=1}^{\infty } \frac{1}{\left(m^2+1\right){}_{2 m}}$$ where $\left(m^2+1\right){}_{2 m}$ is the Pochhammer symbol.
I posted over onPhysics.SE where they beat me up over the definition of mass. I'm not a physicist so it was a bit confusing. But, I don't want to discuss proton mass in this post. I want to discuss the odds of this happening at all!
We have $0.167$ as $3$ digits of accuracy for $1959$ and $0.1672621$ as $7$ digits of accuracy for $2012$. At $10$ numbers per each of the $4$ digits difference, we beat $9999$ to $1$ odds against that happening.
Is there a point after a number of successful steps that we would have some kind of confidence factor that would indicate that the number corresponds to the proton's mass or not?
Footnotes
1: Mathematica code:
NSum[10/Pochhammer[m^2 + 1, 2 m], {m, 1, [Infinity]}, WorkingPrecision -> 50]
2: CRC Handbook for Chemistry and Physics, 12th edition, (1959), Page 16.
3: http://pdg.lbl.gov/2012/reviews/rpp2012-rev-phys-constants.pdf
Clarification
The number $m_{p_{0000}}$ is a percentage. When we multiply the length of the sides of the gram by 10 and then cube it to get kg, we don't cube the number. We change its exponent by 3. Therefore, the number is dimensionless.
I've added a bounty for the best answer that is back on topic.