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I’m not very good at maths, so I didn’t understand the whole post. However, does the size of the whole population affect the “3/n” thing? For example, if I’ve r

I’m not very good at maths, so I didn’t understand the whole post. However, does the size of the whole population affect the “3/n” thing? For example, if I’ve read 200 pages of a 201 page book and not discovered a typo the chances are 3/200 if I’ve understand the post correctly. If the book has 20000 pages, is the probability still 3/200?

geetfun 8y ago
N is the sample size that you’ve sampled already in your observation. If you’ve sampled all the pages, then we are talking about certainty which this wouldn’t apply.
- blowski 8y ago
  In my example, there’s either 1 page or 19800 pages left. But the post suggests the probability of a typo is the same in both cases.
  - babygoat 8y ago
    It's the probability of finding a typo on a page, not the remainder of the book.
    - blowski 8y ago
      Ah now I see the difference, thank you.
- Koshkin 8y ago
  The assertion was that the probability is less than 3/N.
mirimir 8y ago
> It says that if you’ve tested N cases and haven’t found what you’re looking for, a reasonable estimate is that the probability is less than 3/N. So in your example, according to the rule, the probability of errors is less than 3/200. Which it is, either 0/201 or 1/201.
paulsutter 8y ago
After 200 pages you could estimate the probability that the next PAGE has a typo. You could use the number of pages left in the book to estimate how many pages may have typos.
deleted 8y ago
[deleted]
yellowstuff 8y ago
The article is confusing, and the rule doesn't actually have anything to do with the odds of never finding an error. This rule lets you calculate a likely ceiling on the probability of there being an error on the next page, assuming you make some assumptions that nothing tricky is going on. So after reading 20 pages with no errors and having no other information the odds of finding an error on the next page are no more than 1/60. If you have 19,800 more pages left you'd expect to find no more than about 330 errors.
- jrumbut 8y ago
  I believe the reason for bringing in the idea of there being no typos in the entire book is to show how you aren't able to reason well without some change in your approach. Of course, the comments here suggest that it just makes the whole concept confusing to an audience who should either already know about confidence intervals or be well prepared to learn about them.
olooney 8y ago
The rule of three gives an upper bound estimate that each page has a 3/200 or 1.5% chance of having a typo. For the 20,000 page book, the probability of typo per page is still 1.5%, but you have to roll the dice 19,800 more times, each time you turn a new page. The formula for the upper bound on probability of getting through all the remaining pages without a typo is 1 - (1-3/200)^19800 Which of course is basically 1. The lower bound is 0, the upper bound is almost 1, so the confidence interval is [0, 1). Basically, no information at all. So its much to early to conclude that you're not going to find any typos in the entire book. The best way to state your conclusions is to say "after sampling 200 pages at random, we are 95% confident that the true typo-per-page rate is in the interval 0% - 1.5% or equivalently the total number of typos in the 20,000 page book is between 0 and 300."
- jdmichal 8y ago
  Minor correction to your otherwise excellent final statement: "... or equivalently the total number of [pages with] typos in the 20,000 page book is between 0 and 300." We're not measuring the number of typos on a page, but only whether the page contains a typo or not. So we can't speak about the number of typos.
  - PeterisP 8y ago
    The math in that article is based on the assumption that typos are scarce, because that's the situation where it makes sense that you've seen 0 events. If that probability is close to zero, the probability of two typos per page is insignificantly small and can be discarded, and the expected number of typos in the book is approximately equal to the expected number of pages with typos. So in the context of these assumptions it's not necessary to consider this difference. Just as the article itself states regarding the approximation "Since log(1-p) is approximately –p for small values of p" which relies on discarding p-squared terms as insignificantly small.
    - jdmichal 8y ago
      There's no reason to be imprecise and potentially sloppy over a two-word difference. Especially when the comment is being targeted to explain something to someone who did not have a solid grasp of the material, which is obviously necessary to make a leap such as yours. And I won't even bother with the argument that presence of one typo on a page might actually increase the probability of another typo on the same page or nearby pages. The idea that this principle is based on uniform distribution is discussed enough in other threads.

geetfun 8y ago

N is the sample size that you’ve sampled already in your observation. If you’ve sampled all the pages, then we are talking about certainty which this wouldn’t apply.

blowski 8y ago
In my example, there’s either 1 page or 19800 pages left. But the post suggests the probability of a typo is the same in both cases.
- babygoat 8y ago
  It's the probability of finding a typo on a page, not the remainder of the book.
  - blowski 8y ago
    Ah now I see the difference, thank you.
Koshkin 8y ago
The assertion was that the probability is less than 3/N.

mirimir 8y ago

> It says that if you’ve tested N cases and haven’t found what you’re looking for, a reasonable estimate is that the probability is less than 3/N. So in your example, according to the rule, the probability of errors is less than 3/200. Which it is, either 0/201 or 1/201.

paulsutter 8y ago

After 200 pages you could estimate the probability that the next PAGE has a typo. You could use the number of pages left in the book to estimate how many pages may have typos.

deleted 8y ago

[deleted]

yellowstuff 8y ago

The article is confusing, and the rule doesn't actually have anything to do with the odds of never finding an error. This rule lets you calculate a likely ceiling on the probability of there being an error on the next page, assuming you make some assumptions that nothing tricky is going on. So after reading 20 pages with no errors and having no other information the odds of finding an error on the next page are no more than 1/60. If you have 19,800 more pages left you'd expect to find no more than about 330 errors.

jrumbut 8y ago
I believe the reason for bringing in the idea of there being no typos in the entire book is to show how you aren't able to reason well without some change in your approach. Of course, the comments here suggest that it just makes the whole concept confusing to an audience who should either already know about confidence intervals or be well prepared to learn about them.

olooney 8y ago

The rule of three gives an upper bound estimate that each page has a 3/200 or 1.5% chance of having a typo. For the 20,000 page book, the probability of typo per page is still 1.5%, but you have to roll the dice 19,800 more times, each time you turn a new page. The formula for the upper bound on probability of getting through all the remaining pages without a typo is 1 - (1-3/200)^19800 Which of course is basically 1. The lower bound is 0, the upper bound is almost 1, so the confidence interval is [0, 1). Basically, no information at all. So its much to early to conclude that you're not going to find any typos in the entire book. The best way to state your conclusions is to say "after sampling 200 pages at random, we are 95% confident that the true typo-per-page rate is in the interval 0% - 1.5% or equivalently the total number of typos in the 20,000 page book is between 0 and 300."

jdmichal 8y ago
Minor correction to your otherwise excellent final statement: "... or equivalently the total number of [pages with] typos in the 20,000 page book is between 0 and 300." We're not measuring the number of typos on a page, but only whether the page contains a typo or not. So we can't speak about the number of typos.
- PeterisP 8y ago
  The math in that article is based on the assumption that typos are scarce, because that's the situation where it makes sense that you've seen 0 events. If that probability is close to zero, the probability of two typos per page is insignificantly small and can be discarded, and the expected number of typos in the book is approximately equal to the expected number of pages with typos. So in the context of these assumptions it's not necessary to consider this difference. Just as the article itself states regarding the approximation "Since log(1-p) is approximately –p for small values of p" which relies on discarding p-squared terms as insignificantly small.
  - jdmichal 8y ago
    There's no reason to be imprecise and potentially sloppy over a two-word difference. Especially when the comment is being targeted to explain something to someone who did not have a solid grasp of the material, which is obviously necessary to make a leap such as yours. And I won't even bother with the argument that presence of one typo on a page might actually increase the probability of another typo on the same page or nearby pages. The idea that this principle is based on uniform distribution is discussed enough in other threads.