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For a recent question of mine, an answer was provided suggesting that since there is a limited amount of sapient species in the galaxy, we ourselves would be valuable. This got my gears turning and made me wonder, how many sapient species can I expect to find in the galaxy?

Lets make some fair assumptions;

  • The planet must exist in the habitable zone
  • The solar system must be in the galactic habitable zone
  • The planet must be terrestrial
  • Only 5 percent of the planets above have the possibility of life.
  • Of the 5% with life, only ones older than 4 billion years will have sapient life

Assuming the above is true, how many planets with sapient life can I expect to find in a random Galaxy, where I do not know the exact number of stars?

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    $\begingroup$ Mandatory XKCD comic. $\endgroup$ – Renan Jul 21 '16 at 17:03
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    $\begingroup$ Even with all those constraints, we could still have a fairly high number. There are approx 200-400 billion stars in our Galaxy, so even 1% of 1% of 1% of that is a few hundred thousand. $\endgroup$ – Ryan Jul 21 '16 at 17:17
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    $\begingroup$ So, what's the question. What's a hundred billion stars times all the factors you list? $\endgroup$ – Serban Tanasa Jul 21 '16 at 18:33
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    $\begingroup$ "ones older than 4 billion years" -- you can probably look at this differently, the issue might not be having life for 3.5 billion years, it might be having an analogue to the Cambrian explosion. I don't know whether one per 3-4 billion years is a high or a low estimate for Cambrian explosions in average systems with habitable planets, just that it's the rate we experienced. $\endgroup$ – Steve Jessop Jul 22 '16 at 0:52
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    $\begingroup$ Analogously: "I know the probability per lottery ticket of it being a winner, but how can I come up with the expected number of winning lottery tickets in a randomly selected lorryload of lottery tickets without knowing the exact number of tickets?" Answer, yes, the question makes sense, you need to put a probability distribution on the number of tickets (stars) in a randomly-selected lorry (galaxy) and do the math. What's not clear, and the Drake equation ignores, is whether they're independently distributed, but since they're so sparse the answer is "very nearly" barring FTL or L is huge. $\endgroup$ – Steve Jessop Jul 22 '16 at 1:03
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I need to start by saying, this is the combined work of thousands, But is known by the name "Drake Equation".

from Wikipedia:

"The Drake equation is a probabilistic argument used to arrive at an estimate of the number of active, communicative extraterrestrial civilizations in the Milky Way galaxy."

the most recent version shows as: N = R* • fp • ne • fl • fi • fc • L

where

  • N = The number of civilizations in the Milky Way galaxy whose electromagnetic emissions are detectable.

  • R* =The rate of formation of stars suitable for the development of intelligent life.

  • fp = The fraction of those stars with planetary systems.
  • ne = The number of planets, per solar system, with an environment suitable for life.
  • fl = The fraction of suitable planets on which life actually appears.
  • fi = The fraction of life bearing planets on which intelligent life emerges.
  • fc = The fraction of civilizations that develop a technology that releases detectable signs of their existence into space.
  • L = The length of time such civilizations release detectable signals into space.

Among dozens of papers written about the Drake Equation, some have suggested new considerations for the formula. One such paper stands out for adding well-established probabilistic principles from statistics. In 2010, the Italian astronomer Claudio Maccone published in the journal Acta Astronautica the Statistical Drake Equation (SDE). It is mathematically more complex and robust than the Classical Drake Equation (CDE). The SDE is based on the Central Limit Theorem, which states that given the enough number of independent random variables with finite mean and variance, those variables will be normally distributed as represented by a Gaussian or bell curve in a plot. In this way, each of the seven factors of the Drake Equation become independent positive random variables. In his paper, Maccone tested his SDE using values usually accepted by the SETI community, and the results may be good news for the “alien hunters”. Although the numerical results were not his objective, Maccone estimated with his SDE that our galaxy may harbor 4,590 extraterrestrial civilizations. Assuming the same values for each term the Classical Drake Equation estimates only 3,500. So the SDE adds more than 1,000 civilizations to the previous estimate. (From Astrobio.net)

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  • $\begingroup$ Gracias, Frostfyre...:) $\endgroup$ – Joe Jul 21 '16 at 17:52
  • $\begingroup$ This helped him clarify the question but is not an answer to what he’s asking. $\endgroup$ – JDługosz Jul 21 '16 at 19:10
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    $\begingroup$ @JDługosz - I disagree, this actually answered the question by giving an actual estimate with excellent reasoning and supporting documentation, rather than saying "look up..." $\endgroup$ – user11864 Jul 21 '16 at 19:22
  • $\begingroup$ But he’s not asking for the current best estimate, but the missing real values to go along with his universe’s postulated values so that he can run the calculation himself. You said nothing about how many stars there are to add to his calculation. $\endgroup$ – JDługosz Jul 21 '16 at 19:52
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    $\begingroup$ @JDługosz - There are no "real values". This equation is predicated on the fact that all of its factors are to be filled in with guesses. Until the OP fills in the numbers for every factor other than N (or at least, all of them but one while including N), it will always take three licks to get to the center of a Tootsie Pop. This answer says, here's the formula; do the math. IIRC, for galaxies like ours, reasonable estimates for N range from 0 ~ 1M. $\endgroup$ – Mazura Jul 22 '16 at 2:11
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Assuming the above is true, how many planets with sapient life can I expect to find in a random Galaxy, where I do not know the exact number of stars?

So the other percentages and most of the question is irrelevant. How many suitable stars in a random galaxy?

Note that large clusters will be entirely without galactic habitable zone, I’ve read. So stay away from such dense regions. These contain most of the stars, though, so that pulls the average way down since most galaxies will give you a zero.

Most galaxies are small. Our Local Group has a dew dozen members that have been known since the 20’s or 30’s, 3 large spirals, and a couple hundred tiny satellite dwarf galaxies. In a random choice you will get a dwarf.

The dwarf may not have issues with the center being a non-habitable zone, OTOH it has low occurrence of star formation and low metalicity. So maybe solar systems with planets like ours need a large galaxy. So, a randomly chosen galaxy will have about 0.3% chance of being one of the big ones, and then a 50% chance of not being in a dense central cluster.

So given enough trials the results of galaxies like ours will contribute to the average, which contains one of those and 600 zeros. That is, average together the figure for our galaxy and 600 dead galaxies.

So look up the SETI value for the right kind of star in the right neighbourhood, and then divide by 600.

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  • $\begingroup$ "Divide by 600" ??? $\endgroup$ – user11864 Jul 21 '16 at 19:18
  • $\begingroup$ Yes. 600 galaxies with 0 habitable planets and 1 like ours, averaged together. $\endgroup$ – JDługosz Jul 21 '16 at 19:50
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The Fermi Paradox suggests that it must be a low number indeed. Perhaps < 1. If so, then your assumptions aren't "fair" at all, but rather "impossibly optimistic." :/

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  • $\begingroup$ The Dyson dilemma youtube.com/watch?v=94iDdHRa2X4 suggests that the number is likely less than 1/1,000,000 $\endgroup$ – Donald Hobson Jul 21 '16 at 22:02
  • $\begingroup$ Well, last time I was abducted by aliens (greys) they told me it was 6 including ourselves. $\endgroup$ – JDługosz Jul 21 '16 at 23:00
  • $\begingroup$ @DonaldHobson woow looks promising. At last someone talks about building planets. tnx for link, he made updated version of that video youtube.com/watch?v=QfuK8la0y6s $\endgroup$ – MolbOrg Jul 22 '16 at 5:58

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