Calculating chance of life on another planet

Paul Paul

New member
Joined
May 8, 2018
Messages
3
Hi All,

I'm hoping to get some help with a little project I've been pondering for a while. I've been thinking about calculating the chance of life on another planet given that we know it has happened at least once. Essentially what I'm trying to boil this down to is calculating the odds of life occurring on 1 and only 1 planet, versus occurring on at least one planet. Then I can put lower and upper restrictions on what I think the reasonable probability could be then subtract one from the other.

As it has been many years since I was in school a lot of statistics math has left me. So instead of doing the math correctly I 'brute forced' the math by running simulations in excel (crashing excel a few times as I had too many numbers).

So on a galaxy size of 1000 planets I made the probability for each planet 1/1,000,000. Then using the random number generator, generated 1000 planets. Then did this for 702 simulations ('galaxies' if you like). Then I had a probability distribution for the likelihood of having a galaxy with 0, 1, 2, 3 etc planets with life.

I then ran the above multiple times, only changing the probability of life occurring on a planet; 1/100,000, 1/50,000, 1/20,000... to 1/100.

After all that I plotted the likelihood of '1 and only 1 planet' and 'at least one planet' versus the probability of life occurring on each planet. See below.
life probability graph.jpg

So my questions are:

What math do I need to plot these without having to set up hundreds of simulations? (I think the word binomial rings a bell) I would like to scale this up to billions of planets and odds of life to 1/billions.

How do I calculate the area under the curve for these graphs. The log scale is throwing me right off, I have no idea where to start.

Also, if anyone has seen someone do something similar that I could read up on, and what glaringly obvious problems have I missed with this idea?

Thanks any and all for your help and input.
 
Hi Paul,

You're looking for a binomial probability distribution, which tells you the probability of having \(\displaystyle k \) successes given \(\displaystyle N \) independent trials, each of which has probability of success \(\displaystyle p \). So in this case, \(\displaystyle P(\mathrm{life}) = p = 10^{-6} \) (for example). The equation for the binomial distribution is:

\(\displaystyle \displaystyle \mathrm{Probability~of~getting~k~successes} = P(k) = \left(\begin{array}{c} N \\ k \end{array} \right) p^k (1-p)^{N-k} \)

We can break this expression down as follows. The number of trials \(\displaystyle N \) is the number of planets in your galaxy. The reason we can consider each planet to be an independent trial is because we can think of each planet as being a separate flip of the highly biased/unfair coin that determines whether life develops or not. The parameter \(\displaystyle k \) is the number of planets on which life actually develops. Since probabilities of independent events multiply, the expression \(\displaystyle p^k (1-p)^{N-k}\) is obviously just the probability of getting \(\displaystyle k \) planets with life and \(\displaystyle N-k\) planets without life (i.e. \(\displaystyle k\) successes and \(\displaystyle N-k\) failures). However, that's for a particular selection of \(\displaystyle k \) planets out of the \(\displaystyle N\). We need to multiply this probability by the number of distinct ways of choosing \(\displaystyle k \) objects out of \(\displaystyle N\). We call this factor (N choose k) the "binomial coefficient", and it is also written as \(\displaystyle \left(\begin{array}{c} N \\ k \end{array} \right)\):

\(\displaystyle \displaystyle N~\mathrm{choose}~k = \left(\begin{array}{c} N \\ k \end{array} \right) = \frac{N!}{k!(N-k)!}\)

The probability of getting exactly one planet with life is the above expression for the binomial probability \(\displaystyle P(k) \) with \(\displaystyle k=1\). The probability of getting \(\displaystyle \geq 1 \) planets with life would be the sum of the probabilites for all \(\displaystyle k\) values that fit that condition. So we have:

\(\displaystyle \displaystyle P(k=1) = \left(\begin{array}{c} N \\ 1 \end{array} \right) p (1-p)^{N-1} \)

\(\displaystyle \displaystyle P(k \geq 1) = \sum_{k=1}^N \left(\begin{array}{c} N \\ k \end{array} \right) p^k (1-p)^{N-k} \)

However, a much easier way to compute this second thing would be to note that the only two possible outcomes are 1) getting 0 planets with life, or 2) getting (1 or more) planets with life. So the probabilities of these must add up to 1. Therefore:

\(\displaystyle \displaystyle P(k \geq 1) = 1-P(0) = 1 - \left(\begin{array}{c} N \\ 0 \end{array} \right) (1-p)^N \)


Obviously you then want to consider these quantities as a function of \(\displaystyle p\). My question is: why do you want to compute the area under your curves for P(k=1) and P(k >=1)? What does that tell you?
 
Thanks j-astron.

Great answer. Easy to follow. Well set out.

I'm working from the assumptions of a galaxy with 1 billion planet, at least one planet has life, and each planet has the same unknown probability of having life, lets call it P. None of these assumptions seem very far fetched. So I'm trying to see what what affect changing P has on the odds of having only 1 planet with life, and at least one planet with life.

below is the graph I made:
Life Probability 1 Billion planets.jpg
So we can see that, as intuition would suggest, the highest chance of having only one planet with life is when the odds of life are 1/billion at 37%. At the same time though there is a 63% chance of having at least one, leave a total chance of 37%/63% = 59% chance of just one planet with life given we already assumed there was at least one planet with life. Seems legit. Odds of P below 1/billion means there is less overall chance of life, odds of P above 1/billion means higher chances of more than 1 planet. So this is the sweet spot.

This isn't very stable though. As soon as you go to 1/10Billion, the chance of any life falls below 10%, leaving a 1/10 chance we should exist at all. Going to other way, as soon as you go above 1/500million, the odds of getting 2 or more planets increases and again the odds of only one planet falls below the 10% mark. Overall if P falls within this range, there is about a 50% chance of only 1 planet containing life.

If you want lower the limit on what you consider unlikey, lets see what happens when we go to 99%. We would need the odds of life to fall between approx 1/100Billion and 1/10million - 3 orders of magnitude. If it falls in this range, there is a 10% chance that life would occur on 1 planet only (this is where the area under the curve comes in. 10% of the area is for 1 planet alone, 90% of the area is for >1 planet).

In conclusion, In a galaxy with N planets, we would need to have the probability of life on each planet (P) fall between 2 orders of magnitude smaller than 1/N and 1 order of magnitude larger than 1/N, to have at least 1% probability of only 1 planet with life. If we assume P falls in this range, then there is a 10% chance that there is only 1 planet with life. If P falls outside this range, it is inevitable that there will be life elsewhere. What the likelihood of P is is up to the scientists. However they seem to be only able to put a lower limit on it, as in it is at least 1/million. Which sheds little light on the lower end.

Is my conclusion correct? I may have skipped explaining detail of some of the steps, I'm not usually good at explaining things. Feel free to elaborate further or sdhow where my assumptions may be wrong.
 
Now that I have also had the chance to google "binomial probability distribution life exoplanet" I've discovered I'm not the first to have this idea.

[FONT=&quot]Rossmo, D. (2017). Bernoulli, Darwin, and Sagan: The probability of life on other planets. [/FONT]International Journal of Astrobiology, 16(2), 185-189. doi:10.1017/S1473550416000148


If anyone knows how to access this paper for free let me know. I'm interested to read it, but not "$40 interested" to read it.
 
Top