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In my previous post I set up the problem of gun statistics and planet statistics (where I mean math-problem, rather than trouble-problem). There's a question of the number of guns per capita, versus the fraction of the population with a gun (fraction of citizenry that are gun-owners). Also in there is the number of guns per gun-owner.

Similarly, there's the question of the number of planets per star throughout the galaxy, the fraction of stars with a planetary system, and the number of planets per system.

To illustrate this mathematically (and this involves nothing but multiplication and division, so stay with me!), let's set up two scenarios. Both scenarios have five stars and five planets:

Now let's introduce some mathematical terms. The first is the total number of stars in our sample, $N_\star$. Next is the number of planets, $n_p$. In both of the cases in the figure above, $n_p = 5$ and $N_\star = 5$. These two terms allow us to assess the number of planets per star:

$f_p = n_p/N_\star = 5/5 = 1$

So both cases give 1 planet per star. In terms of bulk numbers this is interesting. But it doesn't really give you a lay of the land. If you are searching for planets, the two cases pose very different situations with very different problems to solve. In Case 1, it's a hunter's bonanza: every target star has a planet to detect! In Case 2, it's 4 misses and one hit, but the one hit has a very complicated signal to decipher.

For planet-hunting, we want the fraction of stars with a planetary system $f_S$, where a system is one or more planets. Clearly the two example cases in the figure have very different values of $f_S$. This number is given by the number of systems $N_S$ divided by the number of stars $N_\star$. By inspection, Case 1 has $N_S = 5$ and $f_S=5/5=1$. Case 2 has $N_S = 1$ and $f_S = 1/5 = 0.2$.

A little more thought should reveal that $f_p$ and $f_S$ are related to one another. They're related by the average number of planets per system, given by $R$ (I'm running out of symbols!). To find the relation, let's figure out how one would use the various symbols above to come up with the number of planets in the sample. The number of planets is given by the number of stars, times the number of systems per star, times the number of planets per system:

$n_p = [{\rm Stars}] \times \left[\frac{{\rm Systems}}{\rm{Star}}\right]\times \left[\frac{{\rm Planets}}{\rm{System}}\right] = N_\star f_S R$

$\frac{n_p}{N_\star} = f_S R$

$f_p = R f_S$

Notice that if $R=1$, then $f_p = f_S$ because there is only one type of planetary system: those with a single planet. Of course, this isn't a likely scenario for planetary systems (just look at ours!), and it certainly isn't the case with most gun-owners. I'd hazard a guess that once you get to the point that you like guns enough to own one, you'll probably own several.

Also, note that $f_S$, or equivalently the fraction of gun-owners, must be less than or equal to one ($f_S \leq 1$). However, $f_p$ can be more than one, but for that to be the case, $R$ has to be bigger than one.

Anyway, the upshot is that the math is fairly simple, but the statistics of gun ownership and planet occurrence are somewhat subtle and you can get tied into knots unless your question is well-phrased. For the answer to the number of planets per star, stay tuned for our press release and final publication of our paper. For gun ownership, let's check out the numbers.

In the U.S. there are 89 guns per 100 civilians, or 0.89 per capita. We're number 1 in the world based on that statistic. (U-S-A! U-S-A! Suck it Serbia!). Thus $f_p = 0.89$.

But only 40-45% of adults own a gun (only!), or $f_S = 0.43$ to pick the average. This means that the number of guns per gun-owner is:

$R = f_p / f_s = 0.89/0.43 \approx 2$

This means that there are about 2 guns per gun-owner, on average.

So this is probably why Ta-Nehisi Coates had such a hard time squaring the various numbers for gun ownership. On the one hand, there are 0.89 guns per citizen. But only 45% of people own a gun, and when they do, they tend to have 2 guns.

In my previous post I set up the problem of gun statistics and planet statistics (where I mean math-problem, rather than trouble-problem). There's a question of the number of guns per capita, versus the fraction of the population with a gun (fraction of citizenry that are gun-owners). Also in there is the number of guns per gun-owner.

Similarly, there's the question of the number of planets per star throughout the galaxy, the fraction of stars with a planetary system, and the number of planets per system.

To illustrate this mathematically (and this involves nothing but multiplication and division, so stay with me!), let's set up two scenarios. Both scenarios have five stars and five planets:

Now let's introduce some mathematical terms. The first is the total number of stars in our sample, $N_\star$. Next is the number of planets, $n_p$. In both of the cases in the figure above, $n_p = 5$ and $N_\star = 5$. These two terms allow us to assess the number of planets per star:

$f_p = n_p/N_\star = 5/5 = 1$

So both cases give 1 planet per star. In terms of bulk numbers this is interesting. But it doesn't really give you a lay of the land. If you are searching for planets, the two cases pose very different situations with very different problems to solve. In Case 1, it's a hunter's bonanza: every target star has a planet to detect! In Case 2, it's 4 misses and one hit, but the one hit has a very complicated signal to decipher.

For planet-hunting, we want the fraction of stars with a planetary system $f_S$, where a system is one or more planets. Clearly the two example cases in the figure have very different values of $f_S$. This number is given by the number of systems $N_S$ divided by the number of stars $N_\star$. By inspection, Case 1 has $N_S = 5$ and $f_S=5/5=1$. Case 2 has $N_S = 1$ and $f_S = 1/5 = 0.2$.

A little more thought should reveal that $f_p$ and $f_S$ are related to one another. They're related by the average number of planets per system, given by $R$ (I'm running out of symbols!). To find the relation, let's figure out how one would use the various symbols above to come up with the number of planets in the sample. The number of planets is given by the number of stars, times the number of systems per star, times the number of planets per system:

$n_p = [{\rm Stars}] \times \left[\frac{{\rm Systems}}{\rm{Star}}\right]\times \left[\frac{{\rm Planets}}{\rm{System}}\right] = N_\star f_S R$

$\frac{n_p}{N_\star} = f_S R$

$f_p = R f_S$

Notice that if $R=1$, then $f_p = f_S$ because there is only one type of planetary system: those with a single planet. Of course, this isn't a likely scenario for planetary systems (just look at ours!), and it certainly isn't the case with most gun-owners. I'd hazard a guess that once you get to the point that you like guns enough to own one, you'll probably own several.

Also, note that $f_S$, or equivalently the fraction of gun-owners, must be less than or equal to one ($f_S \leq 1$). However, $f_p$ can be more than one, but for that to be the case, $R$ has to be bigger than one.

Anyway, the upshot is that the math is fairly simple, but the statistics of gun ownership and planet occurrence are somewhat subtle and you can get tied into knots unless your question is well-phrased. For the answer to the number of planets per star, stay tuned for our press release and final publication of our paper. For gun ownership, let's check out the numbers.

In the U.S. there are 89 guns per 100 civilians, or 0.89 per capita. We're number 1 in the world based on that statistic. (U-S-A! U-S-A! Suck it Serbia!). Thus $f_p = 0.89$.

But only 40-45% of adults own a gun (only!), or $f_S = 0.43$ to pick the average. This means that the number of guns per gun-owner is:

$R = f_p / f_s = 0.89/0.43 \approx 2$

This means that there are about 2 guns per gun-owner, on average.

So this is probably why Ta-Nehisi Coates had such a hard time squaring the various numbers for gun ownership. On the one hand, there are 0.89 guns per citizen. But only 45% of people own a gun, and when they do, they tend to have 2 guns.

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