On average, how often does any given hydrogen nucleus run into another hydrogen nucleus in the Sun? I think there is a misprint in an article. I will include the link if you don't mind and cut paste the sentence that I think is a misprint. Here is the link
https://www.abc.net.au/science/articles/2012/04/17/3478276.htm#:~:text=The%20power%20output%20of%20the,can%20it%20be%20so%20low%3F
...and here is the sentence in the article"

" On average, any given hydrogen atom will run into another hydrogen atom only once every five billion years."

Overall I think it is an interesting article but maybe I am not grasping what the author has intended with this sentence.
By the way I wasn't specifically looking for this but ran across it by accident while trying to find the rate that fusion occurs in the sun's core and compare that to man made fission or fusion events I haven't finished researching that and then I ran into this and became puzzled...
 A: This statement is indeed misleading.
For a given hydrogen atom, the frequency of collisions in the center of the Sun is enormously high, some $10^{17} s^{-1}$ or thereabout. So it will "run into and collide with other protons" all the time.
In contrast, the Sun's livetime is of order $10^{10}$ years, so it will burn through half its hydrogen inventory over a time of order $5\times 10^9$ years, by means of nuclear fusion. Thus, it will take a given hydrogen nucleus billions of years to "run into another proton and fuse with it".
You can cross-check that: On average that's $10^{17}*10^7*10^9=10^{33}$ collisions before the Coloumb barrier (a few MeV) is overcome. The $10^7$ Kelvin at the center of the Sun correspond to 1keV. Particles follow a Boltzmann distribution, which at high energies goes like $\sim exp(-1/T)$. You just need to integrate this distribution (still just an exponential...) to find that tail where the integral is $10^{-33}$, which for a characteristic temperature of 1keV, should be about 1MeV.
A: If “run into” means takes part in a fusion reaction then this is correct as an order of magnitude approximation. Wikipedia says:

... each proton (on average) takes around 9 billion years to fuse with one another using the PP chain

As noted in comments, if the average lifetime of a lone proton were very much shorter than this then the Sun would have already used up most of its hydrogen. However, we know that the proportion of hydrogen in the core of the Sun is still around $40\%$ and the Sun is only about half way through its main sequence life span.
