As many Barys as baryons.

Recent thoughts on the applicability of models made me think about if it was valid that we use a model of people’s height which gives a non-zero probability to a person with a negative height.

Trying to calculate the probability of a man having a negative height has its challenges because most normal distribution tables don’t include such extreme Z numbers. I used the numbers for the average hight and standard deviation of a man from this site.

We can convert our normal distribution to a Z score using the following formula:

“Adult men in the United States are approximately normally distributed with a mean of 70 inches and a standard deviation of 3 inches.”

As we are looking at a raw score of not negative x will be 0. Mu is 70 and sigma is 3.

That gives us a Z score of about -23.

Since most Z score tables don’t get even to 5 standard deviations from the mean, we will need to calculate being 23 out ourselves.

With a little bit of help in finding out how to use scipy to do numeric integration on the normal PDF from this page, we get a truly incredibly small probability.

The variable z representing probability here, a poor choice of name.

This gave me the idea of putting this number in context, the 117th power reminded me of the number of barons in the universe generally considered to be around the 88th power mark.

I then set out to answer the question, if there were as many Barrys as baryons in the universe what is the probability, using the unbounded normal distribution model, that none of them will have negative height?

NB: I have confused myself and instead of 10^88 Barrys only considered 88.

The first thing I tried was to simply run the calculation through in python, however, the large numbers involved mean that a floating-point representation wouldn’t handle the calculation well.

This confirms that for all intents and purposes using an unbounded model of height is fine.

I wanted to go further though. I wanted to bypass the limitations of the floating-point representation and get an understanding. To do this I had to break the calculation up into its constituents and calculate them in a different order.

Given this calculation is of the form (1-x)^n we can use the formula for a binomial expansion:

$(a+b)^{n} = \sum_{k=0}^{n}\binom{n}{k}(a)^{n-k}(b)^k$

Since we have x=1 we can simplify this to:

$(1-x)^{88} = \sum_{k=0}^{88}\binom{88}{k}(-x)^k$

Now because powers of x can become vanishingly small I separated out the exponent from the mantissa.

I could then extract for each value of k, the binomial coefficient, the parity (Which alternates as we are taking powers of negative x), and the exponent of x.

$(1-x)^{88} = \sum_{k=0}^{88}\binom{88}{k}(-1)^k(x_{mantissa})^k(10^{x_{exponent}k})$

What this results in is the following table:

This shows the first element in the summation to be 1 followed by the negative of our original probability. The terms continue to become vanishingly small, the third term being about 10^120 times smaller than the second. This vanishing size of the terms comes from the increasing exponent of -117*k which dominates the size of the number.

The fact that this probability has remained at basically 1 even while taking such a large sample means that using an unbounded normal distribution is warranted.

For comparison, if the probability of being a negative height was 0.1 then the chance that 88 people not having negative height would be:

0.00001

Realising my 88 vs 10^88 mistake.

I thought it was worth publishing despite my mistake. I did follow up by taking a look at how to do the calculation for 10^88 Barrys.

$(1-x)^{10^{88}} = \sum_{k=0}^{10^{88}}\binom{10^{88}}{k}(-x)^k$

Now that summation is intractable. So is running the combinatorics calculation through a standard library function.

The problem of the size of the combinatorics can be solved using the Stirling approximation a derivation of which is explained well here.

$ln(n!) = \sum_{j=1}^{n}ln(j)\approx n\ ln(n) -n \#\ by\ stirling\\ ln \left (\binom{10^{88}}{k}\right ) = 10^{88}ln(10^{88})- 10^{88} - \left ( k\ ln(k)k \right ) \left ( (10^{88}-k)ln(10^{88}-k) -(10^{88}-k)\right )\\\\ 10^{88}ln(10^{88})- 10^{88} - k\ ln(k) + k- (10^{88}-k)ln(10^{88}-k) + 10^{88}-k\\ 10^{88}ln(10^{88})- 10^{88} - k\ ln(k) + \cancel{k}- (10^{88}-k)ln(10^{88}-k) + 10^{88}-\cancel{k}\\ 10^{88}ln(10^{88})- \cancel{10^{88}} - k\ ln(k)- (10^{88}-k)ln(10^{88}-k) + \cancel{10^{88}}\\ \therefore\\\\ ln \left (\binom{10^{88}}{k}\right )=10^{88}ln(10^{88}) - k\ ln(k)- (10^{88}-k)ln(10^{88}-k)\\$

That doesn’t help us much though as the sum over 10^88 elements is still intractable. The derivation of Stirling’s estimation for the log of the factorial gave me an idea. In the derivation, the summation is approximated with an integral. We could do the same thing here:

$(1-x)^{10^{88}} = \sum_{k=0}^{10^{88}}\binom{10^{88}}{k}(-x)^k= \sum_{k=0}^{10^{88}}\frac{\Gamma(10^{88}+1)}{\Gamma(k+1)\Gamma(10^{88}-k+1)}(-x)^k\\\\ \#\ Using\ the\ integral\ approximation \\ \sum_{k=0}^{10^{88}}\frac{\Gamma(10^{88}+1)}{\Gamma(k+1)\Gamma(10^{88}-k+1)}(-x)^k\approx \int_{0}^{10^{88}}\frac{\Gamma(10^{88}+1)}{\Gamma(x+1)\Gamma(10^{88}-x+1)}(-x)^x dx$

This at least gives us a formula, albeit a complicated one to solve. In fact, given wolfram alpha says the following I don't have much hope for it being solvable:

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As many Barys as baryons.

Realising my 88 vs 10^88 mistake.

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