8
$\begingroup$

Is it possible to calculate or approximate the probability of something extremely unlikely happening once over a large sample, i.e., in situations where the probability is smaller than machine error?

For example, I was trying to calculate the approximate likelihood of someone sharing my genome. Apparently an individual genome can be losslessly compressed to about 4MB (2^25 bits). So the chance of one of the ~7 billion humans on the planet sharing my genome is about:

$(1-\frac{1}{2^{2^{25}}})^{(7\times10^9)}$.

Or using the Birthday problem method, the likelihood of there being at least two people with identical genomes (let's ignore twins and so forth) is something like:

$\frac{(7\times10^9)!\cdot{{2^{2^{25}}} \choose 7\times10^9}}{({2^{2^{25}})}^{(7\times10^9)}}$

The problem here is that the numbers are so small or so large that it's impossible to guess roughly where they'd lie. So, is there any way of approximating these or similar calculations?

I realise that in some cases the assumptions behind the problems might be out by multiple orders of magnitude, but even being able to approximate to "more likely than not" would be interesting.

$\endgroup$
1
  • 2
    $\begingroup$ Logarithms are appropriate in such circumstances and, if not enough, Stirling approximation gets you close enough to the actual value to evaluate the magnitude of the quantity. $\endgroup$
    – Xi'an
    Commented Jul 25, 2012 at 7:56

2 Answers 2

8
$\begingroup$

In physics, a Fermi problem is an exercise which asks you to estimate an order of magnitude. You can do the same for probabilities. With practice, your intuition should improve.

As Xi'an commented, you can use logarithms. Perhaps you can't see $2^{2^{25}} \gg 10^{10}$ at a glance, but you can see that $2^{25} \gg 10$ (or $10 \log_2 10 \approx 33$), which implies it.

Instead of using complicated formulas to compute exact values you don't need, use estimates which are simple to calculate. For example, the probability there is at least one other person with your genome (ignoring twins) is at most the expected number of people with the same genome, a simple product $\frac {1}{2^{2^{25}}} (7 \times 10^9)$ which you should be able to estimate as very small. Similarly, the probability that some pair of people have the same genome is at most the expected number of pairs of people with the same genome, about

$$ \frac{\frac 12 (7 \times 10^9)^2}{2^{2^{25}}}$$

By the way, I don't accept this model of probability for the genome. I just used your model for examples. This model would predict that the genetic similarity typically found between siblings is astronomically unlikely.

$\endgroup$
4
  • $\begingroup$ Yeah, I know the model is wrong. But non-twin siblings are only likely to share about half of their genes (aside from the ones common to all humans), so I'm not really sure what you mean by genetic similarity between siblings.. $\endgroup$
    – naught101
    Commented Jul 26, 2012 at 2:35
  • $\begingroup$ I'm not sure what you mean by "about half" since the genome is supposed to be compressed. There are places where there are $2$ choices, and places where there are more than $2$. Anyway, the places where they don't have a copy of the same parent's chromosome may be the same anyway by chance, so many more than half of their genes should be the same, and if you pretend that each genome is random you will estimate that the chance of this is astronomically low. In addition, suppose there are no crossovers. Then siblings only need to win $46$ coin flips to have the same genomes, not $2^{25}$. $\endgroup$ Commented Jul 26, 2012 at 3:46
  • $\begingroup$ Interesting stuff, but this is all a bit tangential to the question, for which my overly-simplistic model was only a basic example. If you feel like continuing the genetics discussion, we could do it in the chat room.. $\endgroup$
    – naught101
    Commented Jul 26, 2012 at 4:12
  • 1
    $\begingroup$ This is all basic material. I just didn't want to show the computations within this model without pointing out that the model is bad. $\endgroup$ Commented Jul 26, 2012 at 4:55
3
$\begingroup$

I think this amounts to a problem of estimating the extreme tails of a probability distribution without the extremely large sample size needed to get any or just a small few number of values observed at those extreme values. The only way to do this is by assuming a parametric model which "automstically" assumes a shape for the distributions tails. But if you have justification for the probability model then you can get the estimates you seek by fitting the density from the parametric family and using it to integrate over the tail area to estimate that small probability. If the parametric assumption is wrong the estimate could be way off (by orders of magnitude).

$\endgroup$
1
  • $\begingroup$ Nice, I hadn't thought of it that way. I have no idea what kind of model I would use for this kind of problem though. $\endgroup$
    – naught101
    Commented Jul 26, 2012 at 2:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.