How To Solve The Alien Extinction Riddle - HARD Job Interview Question

A single alien lands on Earth. Every day after that, each alien on Earth undergoes a transformation, which could be any of the four equally likely events: (a) the alien dies, (b) the alien does nothing, (c) the alien replicates itself (2 aliens total), (d) the alien replicates itself twice (3 aliens total). With bad luck the alien race might die out quickly. But with good luck the alien race might survive indefinitely. What is the probability the alien race eventually dies out and goes extinct? This video presents a solution to this problem, which has been asked as a technical job interview question.
Blog post: wp.me/p6aMk-4Sm
This puzzle was emailed to me by Anand Gautam. It has been asked as a job interview question in technical jobs.
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[@titanspirit7238]

Those McDonald's interviews are intense...

[@amoghopprasad8286]

💀

[@bbsamazon]

😂😂😂😂😂😂

[@MindYourDecisions]

Some people think the aliens will always go extinct. I did not anticipate this reaction, so let me clarify and expand a bit about it. When there are N aliens, there is always a (0.25)^N chance that all of them could die the next day. However, this probability goes to 0 as N increases. Every new day we expect 1.5 aliens, so over X days, we expect (1.5)^X aliens. So the alien population tends to increase over time-if it grows large enough, it is likely enough will replicate to outweigh the number that die out. Think about it in terms of passing along a family last name, like Smith, which is customarily passed on by male children. At first Smith has to have at least 1 male child to pass along the name. In the next generation, the male descendants have to marry and also have male children. At any given generation, there is a chance of the descendants not getting married or having male children. However, after some generations, there might be enough Smiths to keep the name alive and “avoid extinction.” The problem also has an application in physics for modeling chain reactions and in biology for modeling mutations of genes.

[@judgeomega]

In finite time i agree that we approach the 41% probability. But if we look at infinity proper, the non zero chance that all aliens go extinct seems certain.

[@judgeomega]

There is only one possible end state. There is nothing that excludes the possibility of all aliens going extinct. No matter how many aliens exist it will always be a finite number. Regardless of low a probability it (aliens going extinct) will be in finite time, it is certain in infinity when it is the only end state (and not excluded by some other condition). When you assert a possibility that they will never die, you are injecting error. Because there is only one possible end state (extinction). Not 2. Regardless, we rarely deal with infinities in reality. So all of this is more of a mental exercise than anything.

[@judgeomega]

***** Having more aliens isnt an end state. The only thing that would stop the iteration is extinction.

[@RQstarbar3]

judgeomega While it is true that a finite number of aliens will go extinct in an infinite time, it is incorrect to say that "there will always be a finite number of aliens" unless you are also talking about a finite amount of time. The certainty of extinction becomes cloudy when you consider that even with infinite iterations, an infinite number of aliens may never trigger the extinction because "there will always be one more alien". This isn't the terminating "end state" you're looking for, but one could argue that an infinitely increasing explosion of the alien population is also an "end state". It is a state where one infinity is too strong for the other infinity. In the same way that 2 + 0.2 + 0.02 + 0.002 + ... = 2.222... will never be greater than 3, even though the number is infinitely increasing. Its "end state" converges to a finite number because another force of infinity is keeping it in check (namely that the number increases by progressively smaller magnitudes each time).

[@alexq2411]

the probability could never reach 0 as N increases as there will always be a chance that they all die at once

[@eskimoprime09]

If they continue to populate for "infinity", will we have -1/12 aliens?

[@diamonddave2622]

LOL. I'm going with Riemann too.

[@tomhazell983]

the -1/12 thing was a slight of hand. A intentional 'misunderstanding' of how series and Reinmann works, which is useful for quantum physics because, well, quantum physics is weird, but mathematically it's incorrect. The sum of integers is infinity.

[@akolyadin]

Sadly, infinity is not a number.

[@katzen3314]

The sum of integers is -1/12, but just because it's infinite, it doesn't mean that it's the sum of all the integers.

[@coolnobodycares]

+Андрей Колядин No, it's a quantity, and all quantities can be measured in numbers. Therefore you can put an infinite amount of numbers in the equation of infinity. :P

[@Justin.JM.McNeil]

this isn't a riddle, it's a math problem.

[@orekihoutarou6107]

Justin McNeil More specifically, it is a Statistics problem.

[@Justin.JM.McNeil]

right. a riddle is something anyone can get with a little bit of outside the box thinking. this is just a tad more advanced. anyone who disagrees would have to be a bit pretentious.

[@orekihoutarou6107]

Justin McNeil I imagine that I could have solved this if I the statistics class I am just about to take, but at this point, I couldn't solve it, I just decided that since I wouldn't be able to actually calculate something like this that I would decide that the chances were 100% after an infinite amount of days.

[@phteve1900]

"Can you solve this MATH PROBLEM?" Gets less hits than "Can you solve this RIDDLE?" Both get more than "Watch me Solve a Math Problem I Probably Saw A Professor Solve in Class Today"

[@Joepvn]

he's written at least five books, any chance he himself is the professor and he just likes to share knowledge? 😉

[@theginginator1488]

100% because eventually the universe will proceed along with the second law of thermodynamics to equilibrium making life impossible.

[@Sethalapod]

but this problem did not state any acceptance of physics and thus exists in a state where there is only the replications of aliens, thus there is no reason to conclude extinction due to outside sources.

[@jamesmcdonald1889]

this isn't proven so who knows what will happen, also it's aliens they might be able to survive in different conditions to us

[@skya6863]

+Seth Zebrack yes, why think of outside sources when it's a mathematical world, we don't have any info of the outside so the outside doesn't exist (in this problem)

[@dariobarisic3502]

To apply second law of thermodynamics you have to make sure certain conditions are satisfied. Universe as a whole might not be what we call thermally isolated system. When we consider thermally isolated system, we think of something that is bounded by a wall but still interacts slightly with it's surroundings in a manner that cannot be predicted. That is a basis for using statistics in calculations of such systems from which we derived the whole of thermodynamics. Otherwise, if there's no any interaction with surroundings system that we observe is in principle deterministic and using statistical methods will get you the wrong results. I know that comment you posted was a joke but still many people consider it true and that is a problem.

[@Joaking91]

Dario Barišić the observable universe is a closed system

[@BallJuiceOfZeus]

*Nods head as though he understands all of this..

[@boomcomedyent]

I always thought that *(inert action here) was always in 1st person.

[@katzen3314]

KZbin gramemr lesson everyone: Putting a single asterisk before a word is a correction to someone's spelling or grammar. Putting words between asterisks *like this* make the text go bold, signifying an action. Now practise this by correcting my comment appropriately.

[@BallJuiceOfZeus]

Katzen4u *Doesn't know how to use bold letters so does it this way

[@katzen3314]

fantana Just do the same thing but with another asterisk at the end...

[@BallJuiceOfZeus]

Katzen4u *uses bold working*

[@mihailghinea]

These aliens are expected to reach a population of 637,621,500 by day 50. Don't underestimate them!

[@matthewgiallourakis7645]

I have an interesting follow up question: In the case that the alien race does eventually die out, what is the expected number of total decendants from the original alien?

[@BubuSnow93]

Shouldn't they eventually die all after an infinite amount of days?

[@AntoshaPushkin]

on average if you run a lot of infinite situations like this, you'll find out that 41% of them die after infinite amount of days but others don't

[@patrickmagee7111]

How do you run infinite situations? No matter how large their numbers got theres a chance they all die on a given day. Assuming infinite days the probability they all die eventually is 1.

[@AntoshaPushkin]

Patrick Magee the idea is that the expected number for 1 alien to evolve is (0 + 1 + 2 + 3) / 4 = 1.5 so every step on average 1 alien becomes 1.5 aliens, which means that on average after N steps there will be 1.5^N aliens which is really a lot. And probability for all of them to suddenly die is (1/4)^(1.5^N) which becomes smaller and smaller at every step really rapidly (for N = 10 it is 1.9 * 10^(-35), really small number, and for N = 20 it is about 10^(-2002), it's very tiny probability, boi, there are 2002 zeros before first non-zero digit) so the probability that they all die after infinite number of steps is not 1. The fact that you don't have physical ability to run infinite number of iterations doesn't mean that you can't use this mathematical concept, after all, you use calculus to solve engineering tasks which involves infinite tasks to calculate integral or derivative.

[@BubuSnow93]

Ye but the point of infinity is that no matter how small a probability is it will happen, sooner or later they will all die on the same step since it's the only "finishing" state of the system.

[@AntoshaPushkin]

BubuSnow93 nope, the speed of decreasing of probability for all of the aliens to die is so high it "beats" the probability to die at some point. It feels counter-intuitive, but it's maths, in maths "well kinda sooner or later dat aliens will die" doesn't work, probability that all aliens will be dead after infinite number of steps is sqrt(2)-1, all the maths of the process are shown in the video. Actually better way to understand is the second way with P_k - probability to die after k steps

[@Therm6000]

We know three things that we can use to solve this problem, and though it's not the same approach as the one in the video, it's still perfectly valid and logical. A. On any specific day, the chance that EVERY alien dies at the same time will always be >0, no matter how many aliens there are. B. There are an infinite number of days to be considered. C. Day number infinity can't be reached, and therefore there can never be infinite aliens in order to make the probability of instant extinction 0. Therefore, all remaining aliens will eventually die at the same time, given enough time. The chance always exists, so it _must_ happen at some point within infinity. That's simply what always happens with nonzero probabilities.

[@trulyUnAssuming]

The chance decreases over time since you have more and more aliens every day as the expected value is greater than one. The mathematical theory behind this is, that an infinite series can converge. Let's take 1/2 then 1/4 then 1/8 then 1/16 then 1/32 and so on - you can visualize that by drawing a square and half it in two equal pieces and fill one out. If you now take half of the unfilled space and fill it you added a quarter, now you take half of the unfilled space - you just added 1/8, and you can continue until infinity and will never leave that square. Now that series converges towards 1. If you remove the first bit it converges towards 1/2 if I add a 1 at the front it converges towards 2. And so on. This concept can be applied to this example aswell. The probability might be greater than 0 at any day - but that doesn't mean the probabilites will add up to 1 - even over infinity. They might. But they don't have to. It depends how fast the probability goes down to zero. The harmonic series for example does not converge. I hope that helps. Infinity is a weird thing to try and wrap your head around. It gets even weirder if you realize there are different sorts of infinity - like countable and uncountable... but now I am just rambling on :-p

[@Atrasoir]

KEine Ahnung Just due to how probability works, mathematically they will at some point go extinct. However in reality they may not, seeing as their is a limited amount of time in the universe in which life will be able to sustain itself.

[@Therm6000]

KEine Ahnung I think this whole thing ends up being a question of mathematical philosophy... The view I have, using the three facts I stated, can be defended. It can be a valid solution, and it can overturn the video's approach. But the video's approach can overturn my view, too. It's just a matter of how we look at it. After all, it's perfectly valid that all things with a _non-zero_ probability will eventually happen given infinite time, no matter how fast said probability decreases. And nobody can deny that 'iteration number infinity' will never be reached. And only on that hypothetical 'iteration number infinity', IF and ONLY IF the aliens have survived for an infinite amount of time first, the probability of extinction would be 0. That's just how iffy things get when we try to work with something that breaks math in a lot of ways, so we're forced to make tons of new rules around it, mostly because it doesn't really exist... You know, that thing called infinity. We can argue that all non-zero probabilities eventually happen given infinite time, no matter the circumstances, and we can argue that there are special cases in which some non-zero probabilities can be avoided forever, even within infinite iterations. In the end, even though the math may check out to overturn my view, I still think that the facts I stated are infallible, because I just see 'infinity' combined with 'non-zero probability', and there's a general rule for that; it'll happen eventually.

[@charlesfowler4308]

If you draw it out as a graph the chance of the Aliens dying out gets closer to one every day but never quite reaches as the graph is an asymptote.

[@Sarastro404]

It is true that given an experiment, if we repeat an infinite time the experiment then every non-zero probability outcomes will be realized. That's the very definition of a probability actually. But that principle can't be applied the way you're using it... In our case what is the experiment? If you think carefully, the experiment is ""seeing what happened to the aliens after an infinite amout of days"". So if you wish to apply the above principle, it can only be done on THAT experiment repeated an infinite number of time NOT on the infinite number of days. There are two infinite to consider and you arrived to your (wrong) conclusion because you mixed them up. A principle can only be used if the conditions are met, if you misuse them you'll get wrong results. And even if it was a philosophical question you'll still be wrong for misuing a principle (also there is the fact your reasoning contradicts itself at some point)

[@dramawind]

Isn't the probability of all of them dying 100%. When an alien dies nothing can follow, while when an alien is alive, there is a 25% chance of dying, so at any given moment there is a probability no matter how small that all aliens die at the same time. So if we consider time as infinite, there will be a point when they will all die at the same time.

[@TeachAManToPhish]

This was my answer as well. I wonder, if it is not the correct answer, why it is not correct.

[@seppet0ni]

It would seem logical, and I thought so too at a glance, but keep in mind how infinity works. When you've entered the scope of infinity, the time is infinite and the odds of each outcome becomes infinite too. Infinitely many aliens will die every day, infinitely many aliens will also be replicated and infinitely many will do nothing.

[@dramawind]

Ignoring the concept of infinity, at every given moment, if there are aliens, then there will be a next moment. If in the next moment the amount of aliens increases or decreases (not becoming zero), then nothing has changed in the sense that aliens exist, and there is still a chance of disappearing (smaller or bigger than before, it doesn't matter), and the number of aliens was finite before, and is still finite now. If the alien population reaches zero, however, then it remains zero forever, while time keeps going.

[@seppet0ni]

Dramawind yup, and that is the chance explained in the video

[@dramawind]

seppet0ni What I'm saying is that there's 100 percent chance, not √2-1.

[@jamesmersh1770]

Assuming an infinite number of times that this randomization happens, though the percentage for any given day gets progressively smaller, as long as the possibility is above 0 (which it always will be) they can always all die the next day. With no limit on days, eventually they would all die. Simple answer is : 100% (unless they have to die out before a certain time)

[@exylophone1]

James Mersh what i thought

[@IFrancyISantosI]

yeah.. was thinking the same.. but some sense in my says no.. i dont know why

[@wotchadave]

I initially thought the same, but I think I've found a discrepancy between your answer and his after thinking about it for a while: For the recursive function P(k+1), notice he took the limit to infinity, which is not the same as P(infinite). If you plug in P(infinite) into the recursive equation, you actually get the solution as P(infinite) = 1 simply because 1 is the only number than will work for P(infinite). So to sum up, your answer is literally for an *infinite* number of days, where as his describes what the probability heads towards over an unbounded finite number of days. So really, it all depends about how you interpret "eventually"... xD

[@DrGerbils]

wotchadave The recursive function P(k) is defined only for integers. Infinity is not an integer. What is P(infinite) supposed to mean and how do you calculate it?

[@kokainum]

Not really. How do you know it's not bounded? Maybe part of them dies and part reproduces but number of aliens doesn't go to infinity? Or maybe limsup is inf but liminf is not inf?

[@DobesVandermeer]

why not 100 percent, if you keep going eventually they will die out by amazing fluke. this can only be truly calculated if you limit the number of days.

[@AntoshaPushkin]

lim pk = sqrt(2)-1, boi

[@turun_ambartanen]

but one day, by chance all aliens die out. it becomes less and less likely the more aliens there are, but the chance is still there. and if you have an infinite number of days it will happen.

[@cheongziyong8871]

+Turun Ambartanen That would happen after an infinite amount of days, as that probability is (1/4)^n, where n is the number of aliens

[@turun_ambartanen]

yes, so the chance of instantly dieing out never disappears.

[@Huntracony]

+Turun Amdartanen So shouldn't the chance be 100% of the species eventually dying out, as that's the only possible stable outcome after infinite days?

[@pk13910]

Well suffice to say I didn't get the job.

[@jessstuart7495]

Yeah, I didn't get this one either, so I decided to read-up on branching stochastic processes. You have to know about Probability Generating Functions to be able to solve this. You've either seen it before and know how to solve it, or you don't. Getting the right answer by working from first principles isn't really an option (for most people).

[@JohnRandomness105]

It took me a moment to realize that I could express P in terms of P for the following day. The resulting equation is cubic, and it was very lucky for me that I was able to find the solution P=1 by inspection. Although I knew it couldn't be the solution, I had to use that solution. Divide the polynomial by P-1 (using long division) to get a quadratic polynomial, which I solved to get sqrt(2) - 1.

[@bergennj7270]

More simple solution: Let P is the probability we are looking for. Then P = 0.25 + 0.25*(P) + 0.25*(P^2) + 0.25 *( P ^ 3) = 0.25(1 + P + P^2 + P^3) Therefore P^3 + P^2 - 3*P + 1 = 0 which is (P - 1)(P^2 + 2P -1) = 0 Possible solutions are P = 1, -1 + sqrt(2), -1 - sqrt(2) Answer : P = -1 + sqrt(2) = 0.414

[@phteve1900]

To answer the "Riddle", 100%. All life will eventually cease to exist on Earth. Period. Stop referring to Math problems as Riddles.

[@FluffyPanda586]

lol thats what i thought My thought exactly "Well eventually it has to die, so 100%"

[@SergeofBIBEK]

The aliens replicate at an average of 1.5 per day... so... it's not 100%.

[@americatorres3978]

+SergeofBIBEK what would they feed on?

[@SergeofBIBEK]

America Torres Who said they need to feed on anything?

[@americatorres3978]

+SergeofBIBEK they are LIVING organisms

[@marouaniAymen]

This problem looks like the Uranium fission process, a neutron can fly without hit, can be absorded by a nucleus or can trigger a split with 3 more neutrons.

[@bibop224]

6:57 An increasing sequence bounded by a real M does not necessarily converge to M

[@laughingdaffodils5450]

I definitely didn't calculate it out to the same degree you did, and would have taken more time to do so were it necessary. But I did come up with two off-the-cuff answers in less than 30 seconds. The first was an at-a-glance approximation, (1/3)

[@urvinius4153]

so 42 is the answer, just say 42

[@oldbutgold3389]

😂😂😂 i died

[@boohoo334]

Urvinius 3 years later and this comment is right on 42 likes, good job people.

[@sanchesseli]

I didn't make any calculations and said myself at the very start "it should be about 40%". I was close :)

[@TyphoidBryan]

"Good news, Mr. Johnson! You got that alien question right, so you got the job! Welcome to the Wal-Mart family, you will be working in the shoe department."

[@douglasbyrd2944]

Given that the question is "eventually" without a solid timescale, the probability is 100%. Since every individual has a 25% chance of dying on any given day, there will come a time when every member of the species on the planet will end up getting the death result. It might take a long time to happen, depending on how large the population is on any given day, but it will happen eventually.

[@Sethalapod]

Not if the series converges, ie the sum of 1 + 1/2 + 1/4 + 1/9 + .... 1/n^2 + .... converges to a number less than 1.7 Such, although the sum of the probability that the species will die out by a given day increase with each day, it converges to the sqrt(2) - 1; that's the point of the second half. Although every alien will die eventually, there exists a chance of ~ 58% that there exist aliens after a finite amount of time.

[@douglasbyrd2944]

Seth Zebrack The original question though does not set a time limit. It just says "eventually". He added the time limit in the later half arbitrarily.

[@lloydgush]

No, that exists a chance bigger than ~58% for any given time, even infinite time.

[@douglasbyrd2944]

lloydgush as long as the odds of all aliens dying at once is a number higher than zero, even if it is vanishing small, it is inevitable that givin infinite time, that result will occur eventually.

[@lloydgush]

Douglas Byrd That's the intuition, but that's not what the function converges to as k goes to infinity.

[@Jimpozcan]

The tricky bit, which was skipped in the video, was getting from P = 0.25(P^3 + P^2 + P + 1) to P = -1-2^0.5, 1 or -1+2^0.5. There is, however, a simple observation which makes this easy. Take the equation P = (P^3 + P^2 + P + 1)/4 and rewrite it as P^3 + P^2 + 1 = 3P If P = 1, this becomes 3 = 3, so P = 1 is a solution. Then divide P^3 + P^2 - 3P + 1 = 0 by P - 1 This can be done by a method like long division just instead of 10s you have Ps. (P^2)(P - 1) is P^3 - P^2 Subtract this from P^3 + P^2 - 3P + 1 and you get 2P^2 - 3P + 1 2P(P - 1) is 2P^2 - 2P Subtract this from 2P^2 - 3P + 1 and you get - P +1 -1(-P +1) is - P + 1 Subtract this from - P + 1 and you get 0 So you can factorise P^3 + P^2 - 3P + 1 as (P - 1)(P^2 + 2P - 1) Thus (P - 1)(P^2 + 2P - 1) = 0 So either P = 1 or P^2 + 2P - 1 = 0 This is a simple quadratic equation but here's an easy solution. Add 2 to each side. P^2 + 2P + 1 = 2 Now it's easy to factorise. (P + 1)^2 = 2 Thus P + 1 = ± 2^0.5 Thus P = - 1 ± 2^0.5 Including the previous solution gives P = - 1 ± 2^0.5 or 1

[@tuxino]

A funny thing happens with randomness and infinities. For any unbounded but finite number of days, the probability that the aliens would go extinct approaches ~41%, but for an infinite number of days, the probability is 1. Any event that is possible will happen given an infinite number of attempts.

[@elpachanga]

yeah but you are not trying every problem an infinite number of times, you are trying an infinite number of problems 1 time each, and each of those less likely than one before it, yes there are infinite attempts but the chance is also infinitely small, so small that if you multiply both you actually get sqrt(2)-1

[@helloitsme7553]

Now I'm nervous for my job interview that's like in 8 years probably 😂

[@justyourfriendlyneighborho2061]

halfway there

[@alexandermcclure6185]

three quarters of the way there

[@ninjahappysquid]

Okay so one thing I'm curious about - why was the "do nothing" option included? It made there be a nice neat "p^0, p^1, p^2, p^3" in your formula, but it doesn't actually effect anything, mathematically or conceptually: Mathematically - solving the equation p = (1/3)*(1 + p^2 + p^3) still provides the answer of 0.414 Conceptually - "doing nothing" is just delaying the choice of reproducing or dying, it won't impact which direction it ends up going. So why was it included in the problem? Just to throw people off or something?

[@alexandermcclure6185]

Only for aesthetics.

[@memelorddabble632]

Not a riddle, just a statistics problem. Such click bait, much wow.

[@mesplin3]

That was quite enlightening. I always assumed that in the long run, assuming probabilities remain constant, implied extinction for these types of problems. It always seemed strange to assume that something wouldn't end given a probability of death.

[@blockhead134]

The question was a riddle but you answered it with math. the answer is 100% because there was no end to the time period and they will die out eventually.

[@thedystopyansociety]

isn't there an insane chance that they always replicated 2 or 3 times?

[@blockhead134]

The Dystøpyan Society no. Since the sequence ends when there is no more aliens but does not end if there are still some left, evenually they WILL all die out.

[@maxenis1577]

"my answer is correct because I don't care about the math that disproves my answer, I didn't come here for math after all"

[@blockhead134]

Jeasker S I didnt come here for math. The title says RIDDLE not PROBLEM

[@maxenis1577]

you're at the wrong channel. go watch cat videos or something

[@PurpleViking221]

To those who think they will always go extinct eventually: Every day, there will always be at least one option that is not Extinction. Therefore, the odds of Extinction can never be 100%. Consider this scenario: Day one- Alien does nothing. Day two- Alien does nothing. Day three- Alien does nothing. Every day afterwards- Alien does nothing. In this endless scenario, every single day, extinction does not occur. The fact that this scenario alone is technically possible, according to the rules, proves that extinction is not a 100% guarantee.

[@valentinziegler1649]

The probability of that specific branch goes to zero, therefore the probability of your scenario NOT happening is 100%

[@jediflamaster]

Considering an infinite time limit shouldn't they all eventually die out?

[@hommelwijf]

i was thinking the same thing

[@larrytroxler7017]

@jediflamaster yeah this one of his most challenging videos.

[@vijuarez3859]

Not for math. Given an infinite amount of time, the aliens are expected to grow in number. As the number of aliens grow, the possibility of all of them dying in the same generation becomes smaller and smaller. It's true that you have infinite tries, but after each try the chances of all dying out is smaller. The possibility is (1/4)^n, where n is the number of living aliens. When n grows to infinity, the possibility becomes smaller and smaller, getting closer to zero. This is called a limit and it's a basic tool for calculus.

[@jediflamaster]

Vicente Juárez I know. It's just so counterintuitive, cause, you know, they only need to die out once.

[@flipsflips1653]

Vicente Juárez You are saying that there is a chance. You also said that there are infinite tries which means you WILL have it at some point. Which then means that the aliens WILL die. It's not a tool at all. It's logic

[@alanharper23]

This may be wrong, but I feel as though both p=1 and p=0.414 are valid. Over a finite period of time, there's an extinction probability of 41.4%, but over an infinite period of time, its certain that they will die out, hence p=1. Given that the question asks for the probability that they "eventually" die out, there's no finite timeframe given, so I'm leaning towards p=1. I feel like this relates to a Markov chain, where the only absorption state is extinction.

[@Stonefallow]

Can someone please explain how he went from P = 0.25(1+ P + P^2 + P^3) to P = √2 - 1

[@marche45]

Its 1/4(p+p^2+p^3+1)=p. So basically you wanna find the roots of p^3+p^2-3p+1=0 That gives you the three roots

[@prim16]

Same I'm so lost

[@Stonefallow]

marche45 That doesn't answer my question at all.

[@kejtos5]

+marche45 yes, it exactly did.

[@Stonefallow]

Jan Rejthar No, it exactly didn't.

[@Crazycolorz5]

I'd've done it with generating functions, where a_n = the probability of n aliens eventually going extinct (a_0 = 1). The presented solution is simpler, though.

[@Josh-cd3zf]

I did that too. After doing it, though, I wound up with the cubic equation that mind your decisions gets. The way that I realized this was by writing it out for a_2. Once I did this, I realized that the probabilities behaved like powers (a_n is a^n) and did the cubic, with the answer from the cubic being sqrt2 - 1. Did you wind up using the cubic too?

[@dustinbachstein]

I was actually asked this question in my interview for a technical job. I looked at my interviewer and replied "well your race has a 58.6% chance to survive, yet you're expected to be dead in 4 days." The interviewer threw off his human mask, I got the job and meanwhile I invented a potion that turned him and all his descendants into humans so they could live longer. So yes - the overpopulation is actually my fault, and I'm proud of it! ;)

[@yoavshati]

Am I the only one that hears "piece of cake" every time he says "P sub k"?

[@andrewmirror4611]

Ha! P(iece) of K(ake)

[@MitchBurns]

I didn't do all that work, but just looking at it quickly I figured out it was around 40%. Since there was a 1/4 chance it dies each day that means it must be at least 25%. Since it does nothing one day that means there is an additional 25% * 25% chance it dies. This becomes something of an infinite sequence. I can see that it ends up being close to a 25% * 50% chance, which is a 12.5% chance. If you then add that to 25 you get 37.5%. Then, there is also a small chance death after replication. This adds just a little bit more, and since 40% is the next nice pretty number, you can see 40% is a good guess. Maybe not the correct way to solve it, but the correct way to ESTIMATE it for sure.

[@reubensamms2866]

Okay, I'm still having trouble grasping why the probability isn't 1. With an infinite amount of time, all possible outcomes should arise eventually, right. And since the alien's population is a defined number, regardless of how big it is, shouldn't it eventually come to zero, even if it took an amount of time that is not comprehensible to us, as it is infinite. My issue is with the amount of time being infinite and the number of aliens not being infinite, I suppose. Thanks in advanced for all your insight.

[@Lexington365]

My first thought was probability 100%. If there is some probability of something happening, even if it gets smaller by the day, then over a long enough time line it will happen.

[@Minecraftster148790]

I have never heard of the theory of statistical branching processes and have no clue where u got your answer from. Couldn't u just solve the equation?

[@AntoshaPushkin]

p = 0.25 (1 + p + p^2 + p^3) 4p = 1 + p + p^2 + p^3 p^3 + p^2 - 3p + 1 = 0 p^3 + p^2 - 2p - p + 1 = 0 p(p^2 + p - 2) - (p - 1) = 0 p(p^2 - p + 2p - 2) - (p - 1) = 0 p(p(p - 1) + 2 (p - 1)) - (p - 1) = 0 p(p + 2)(p - 1) - (p - 1) = 0 (p - 1)(p(p + 2) - 1) = 0 p - 1 gives solution 1 and p(p + 2) - 1 is a quadratic equation which gives sqrt(2)-1 and -sqrt(2)-1 or you could just wolframalpha it and p here is a probability for a newly-born alien to die somewhen in the far future with all his descendants dead. try to think for a while and watch video again, it makes sense

[@fixpontt]

i dont think too many people can solve this in an interview....

[@captainobvious9581]

what, can't you guys just do this in your head? /s

[@luckydannumber2]

This is Discrete maths which has combinatorics and induction, statistical branching has nothing more than to solve the initial part of the question to get the equation you need to solve. Bar the initial equation it's got nothing to do with it.

[@Minecraftster148790]

Thanks you :) studying discrete maths right now actually, we move on to game theory today

[@dlbattle100]

I took a graduate course in Markov Processes, and you know what, the professor had it wrong! He said that any branching process with a non-zero probability of zero offspring means it terminates with probability 1. But now I see looking into it further, the probability of extinction can be less than 1 if the expected number of offspring is greater than 1 on average.

[@nychold]

Interestingly, I set the problem up incorrectly, but got a very close answer. I followed the same logic as in the video, but instead of saying p^2 and p^3, I said it was "half" and a "third" as likely to happen, so: p = 1/4 (1 + p + p/2 + p/3) p = 1/4 (1 + 11p/6) 4p = 1 + 11p/6 24p/6 = 1 + 11p/6 13p/6 = 1 p = 6 / 13 ~ 0.46 or 46% Strange that you can be so wrong, and yet so close to the right answer.

[@ektadebnath8297]

2050 Aliens invade the earth and the first thing they do is delete the video Me be like I always knew I didn't have to answer it for a job

[@Pining_for_the_fjords]

Can you explain how 1+p+p^2+p^3 = sqrt2-1? This equality wasn't explained at all.

[@johnathantran9228]

P = 0.25(1+P+P^2+P^3) 4P = 1+P+P^2+P^3 P^3+P^2-3P+1 = 0 This is a cubic equation with roots P = 1, -sqrt(2)-1, sqrt(2)-1 Easy solution: use a calculator (One of the) more difficult solution(s): 1 is an obvious root to the equation --> divide the cubic by (P-1) to get the others You get P^2+2P-1 = 0 Quadratic formula gets you P = -sqrt(2)-1, sqrt(2)-1

[@leif1075]

@@johnathantran9228 so it's not possible to solve this problem if you didn't know this little stats rule about square root of 2 crap and didn't know that limit..? Not a true test of intelligence or reasoning then just kniwledge

[@pinkegg3179]

@@leif1075 this is not a test for intelligence or logic. This is a pure math problem.

[@npip99]

7:15 Just because it's bounded by sqrt(2)-1 and it's increasing, doesn't mean it approaches sqrt(2)-1. It might approach sqrt(2)-1.0001 instead. The characteristic polynomial of the recursive function does let you use process of elimination though on the roots. it's not -sqrt(2)-1 because it's increasing and it's not 1 because it's bounded below that, and thus it's sqrt(2)-1.

[@aaronboothproduction]

Aaaaaaaaaand you lost me.

[@KS-en8hn]

How did you write p=0.25(1+p+p^2+p^3)?

[@androidkenobi]

I couldn't follow a thing :(

[@ebojfmdboojoh4023]

Me neither my brain has exploded

[@peterbonnema8913]

I was solving p=0.25(1+p+p^2+p^3) for p and had to do complex number algebra and the likes (did not yet succeed) but the actual answer also stumps me.

[@randomgeekery1535]

Solving for the cubic can be a pain. The key is seeing that P=1 is a solution for the equation, so [P-1] can be factored out. P = (1/4)(1 + P + P^2 + P^3) 1=(1/4)(1+1+1+1). First subtract P from each side, giving: (1/4)P^3 + (1/4)P^2 - (3/4)P + (1/4) = 0 Factor out (P-1) from the equation: [P-1][something] = (1/4)P^3 + (1/4)P^2 - (3/4)P + (1/4) For the unit term, [-1][something] = (1/4), so [something] contains -(1/4) For the P term, P[-(1/4)]+[-1][something P] = -(3/4)P -(1/4)P - [something] = -(3/4)P, so [something] contains (1/2)P. For the P^2 term, P[(1/2)P]+[-1][something] = (1/4)P^2 (1/2)P^2 - [something] = (1/4)P^2, so [something] contains (1/4)P^2 and finally for the P^3 term, P[(1/4)P^2]=(1/4)P^3, which works. Putting all the terms together: [P-1] [(1/4)P^2 + (1/2)P - (1/4)] = 0 From there, it's a quadratic equation and solved with the quadratic formula: (1/4)P^2 + (1/2)P - (1/4) = 0 A = 1/4, B = 1/2, C = -(1/4) The solutions turn into -1 + sqrt(2) and -1 - sqrt(2). The only positive solution is sqrt(2) - 1, which is your answer.

[@androidkenobi]

thanks! that's one "mystery" solved. The other for me is how to even get started with probabilities. Here, it's the "conditional probability" column, specifically: why is the first one a 1?

[@JBergmansson]

A little late, but the conditional probability is how likely the condition (in this case alien extinction) is in the forthcoming of each possible outcome each day. In the first outcome, we know the alien does die, which is represented by a 1. In the second outcome, we arrive back at one alien, and since we already represent the probability of one alien dying out as P, the conditional probability for outcome number two is P.

[@jablecherman8433]

I did this differently, and I think your answer might be flawed but I'm probably wrong so correct me. I found that there was a 25% chance that the alien would go extinct the next day, or 4^-1. I then found the next soonest and most likely extinction situation, in which the alien stays the same and then dies the second day, so 1/4x1/4 or 1/16 or 4^-2. Then, I found the next soonest and most likely extinction of a duplication and then both die, or 1/4x1/4x1/4, or 4^-3. I found this led to an infinite sequence of 4^-1x4^-2x4^-3...=P. So the answer is 100% because although the probability of extinction deviates towards zero, something that goes on for an infinite amount of time must eventually reach extinction.

[@darealpoopster]

His solution is not flawe

[@zakariasbindesbllandersen6458]

0th

[@joonanieminen5721]

dnf

[@thephysicistcuber175]

joona, are you a cuber?

[@Neo-lc9wu]

I am :D

[@thephysicistcuber175]

Cool

[@ManBearPiglet]

When there's only one alien the chance of extinction is one third (33%) but interestingly the probability that they all die out becomes practically negligible once there are 5 aliens (0.4%) and it just keeps decreasing as the number of aliens grows. So if they haven't died out after a few days, we can assume they're never going to die out.

[@hunterstalisman2615]

I evaluated and guessed about 40% in about 20 seconds. By ignoring the option B, since it accomplishes/changes nothing it gives a 1/3 chance of all events. Since, there are more opportunities for a species to go extinct past the first chance, it must be slightly less than 10% more than the original 33%, leaving about 36%-43% as the only option.

[@malignor9035]

I (initially) think 100% The key word is "eventually", which means the entire sample space of possible options is exhausted, including 0.25^x where x is the population. I think a better question would be "what is the average expected population after 100 days, and what is the standard deviation?" But... If you sum the infinite sequence (0.25^x, x>=1), you get 1/3 EDIT: The reason I summed the sequence is because for any given population (X), the probability that they all die at once is 0.25^X. U just summed the probability.

[@edsznyter1437]

On the second day, there is a 25% chance that the aliens have gone extinct and a 75% chance that they have at least maintained their population. Thus, trivially, 0.25

[@Tehom1]

4:30 I just used the fact that the final expected population of aliens is infinite, so the probability of extinction must be less than 100%. This observation rules out the answer p = 1. We know that the asymptotic expected population is infinite because the expected population on day N+1 equals the expected population on day N times 6/4.

[@mikefox20]

i know this may seem like a weird question to ask, but during the video, you kept saying "an alien and all it's descendants" As you say this, are you meaning to say extinction? or do you mean that if the original alien dies, any and all aliens that it has created die on that day? or just that one, with each alien having a 25% chance of dieing each?

[@sinchaos5493]

Is it correct to simplify the problem by assuming option (b) and (d) cancel each other, and are the same as option (c)? So you have 25% chance for option (a) and 75% chance for option (c).

[@martijnbouman8874]

No, not at all. It would be correct to delete option (b) altogether, though, because (b) changes nothing to the situation.

[@shikhanshu]

Please correct my if I am wrong, just trying to translate the equation into English. The reason why P is the sum of all possible branches is that either one of those branches can lead to extinction, and hence we take the sum. If ALL the branches were required for extinction, we would do product. Correct?

[@vtvtify]

Pretty easy, p=1/4(1+p+p^2+p^3) 1-3p+p^2+p^3=0 p=1 -> trivial solution (p-1)(p^2+2p-1)=0 p=1 --> obviously not true (might prove that observation later) p=sqrt (2)-1 -->V p=-sqrt (2)-1. --> negative -->X p=sqrt (2)-1

[@st-gelaisrene3287]

Sorry to be bad in math and in english. I did figure out the 1,5 alien possible progression rate but I cannot see that at any moment the chance of extinction being above 25%. Imagine every alien throwing a 4 sided dice everyday ( the dice would be a 4 sided pyramid including the bottom.). The chance of one alien dying is 1 out of 4 possibilities, or 1 out of 4 exponent 1, 25%. If they become 2 aliens, the probability of both dying is 1 out of 16, or 1 out of 4 exponent 2, 6,25%. If they become 3, that day the chance of extinction is 1 out of 4 exponent 3 (64), or 1,5625%. If they become 4 at any day, the possibility of extinction is 0,39%. If they become a million, the chance of extinction is 1 out of 4 exponent 1 000 000. If 999 999 die that day, the loner goes back to a 25% chance of death and extinction the next day. The curve of chance of extinction relative to the number of aliens goes from 25% maximum at one alien to drop rapidly very close to zero as the number of aliens augments ( due to the 1.5 average augmentation rate). Still the chance is always there for extinction at any day. As their number increase, the possibility diminish but remains. In an infinity of throws, they are extinct. I see the problem as the maximum extinction possibility at any day ( 25% ) but with a certainty to reach extinction in infinity. I am wrong?

[@hughmacdonald5585]

Right.... Fantastic question. Completely silly premises btw. Your response to this question to any place that asks this during a job interview is to smile politely , stand up and say, "My interview is over. I don't work with silly people asking pointlessly dumb hypothetical questions. Good day. And good luck!"

[@reubenfrench6288]

I got sqrt(2)-1, with extraneous solutions of 1 and -sqrt(2)-1.

[@wohlgangerc]

The question they ask is different than the one they solve for. They solve for "What is the probability that the race will be dead at some arbitrary point in the future?" Given an infinite amount of time, the alien race will go extinct. Their claim is equivalent to saying that the Martingale betting strategy for Blackjack will not result in a loss after an infinite amount of time. The only endpoint to both systems is 0.

[@cgalon6781]

this is interesting philosophically a population in which every individual has only a 25% chance of death and a 50% chance of doubling or tripling itself per iteration, and is able to survive indefinitely with no external threats, still has a 41% chance of going extinct then again a 59% chance of living for eternity is also quite impressive

[@julienbongars4287]

M8, can you please explain how you derived p = sqrt(2)-1? Thank you.

[@Sapphire04]

I guess the reason many people thought the answer was 1, was that they took the limit, and as the number of days gets reaaaally large, don't the chances of all aliens dying out equal 1? And so you can argue, that if the aliens get reaaaaaally lucky and survives for an infinitely long time, then they will definitely go extinct. That arguement is false. Because although infinity is the limit for the number of days in this case, you can never reach an infinite number of days because infinity is a concept defined by "never-ending". This implies that the probability of all aliens dying out can never equal 0.

[@AlexanderVRadev]

The math is good, but if you look at the question in its context, you are missing the most important point. It is a job interview question for technical jobs. As someone that is an engineer, I can tell you that questions like that are asked, to see if people have good logic and out of the box thinking. And the logical answer is we don't know. 1 alien landed on Earth, but you can't ignore all other aliens that didn't. They are probably living happily on their own planet, so the probability is 0. Unless this was the last alien from their race, so then we solve the problem, but we don't know if it is the last one. Lastly the universe will eventually end, so the ultimate big picture answer is 100%. So we can't know, because we have insufficient information, that explains only small part of a bigger system.

[@weckar]

Here's the part that confuses me. You say P is the chance that an alien and all its descendants die out, and that this probability is 1 if the current alien dies. However, this does not include any descendants it may already have produced. Explain, please?

[@patsk8872]

I worked in math for a while. And if anyone got this during an interview I'd know they just looked up the answer in advance. A quick analysis of just the D and N branches indicate just those seem to have a limit of 1/3 chance of extinction. I'm surprised it's all the way up to over 41% looking at all branches.

[@philipcross1586]

apart from not understanding this at all (please don`t bother trying to explain it), i do have a question about part d in the first question, the alien replicates itself twice, assuming that the alien replicates asexually alien 1 would juring the first relication cycle have 1 offspring, but then the alien goes through a second cycle if the offspring also goes through that cycle, you have 4 aliens each of which can change the answer, shouldn`t that be factored in with the question.

[@mattm6430]

This is better. Too many easy ones lately. Would have liked you to do a numerical solution too, as I couldn't get one of your steps.

[@michaelberg9348]

I'm not going to argue the outcome (for obvious reasons) But as for the explanation..., sorry. The 'se assume this is the answer' (for good reason, root of the equation and all), so we're going to fill it in, into the equation and then simplify, doesn't really work. Yes 'this simplification holds' is a necessary condition, but (as shown by the fact the whole 'but 1 is a root of the equation as well, why can't that be the answer?'-thing came up) it is not sufficient. The simplification does also hold for P=1. The 'should the population ever reach 'rule of large numbers'-size, the population would increase exponentially, therefore, sure extinction isn't an option', doesn't do the trick either. (as that's mainly a 'fundamentally wrong, but probably 'good enough' non-mathematical'-model) I have no problem believing that limit converges, but i can't find the answer to ''why not 1? anywhere in the video.

[@tanmansimansi8983]

Hi Presh.. Question was about the probability the alien race EVENTUALY DIES OUT and GOES EXTINCT. in this case in first 2 events fulfills the conditions of DIE OUT and GOING EXTINCT. So probabiliy is 50-50.. in both last events alien will keep on replicating itself so it will DIE but will not GO EXTINCT never ever.... so its 50% probability is the answer... Please explain..

[@iliasmavridis4916]

Can someone tell me if this is correct? One alien on average ends up being (0+1+2+3)/4= 1.5 aliens, so On day 1: 1 alien On day 2: average 1.5 aliens On day 3: av 1.5^2 aliens On day 4: av 1.5^3 aliens So what about extinction chances on each day? Day 1: 1/4 Day 2: 1/(4^1.5) Day 3: 1/(4^1.5^2) And so on.. As for survival chances: Day 1: 1 -1/4 Day 2: 1 -1/(4^1.5) Day 3: 1 -1/(4^1.5^2) And so on.... So the chance for the aliens to survive forever is [1-1/4][1-1/(4^1.5)][1-1/(4^1.5^2)][...]... That means that the chance for the aliens to go extinct is equal to 1 minus the above quantity.

[@tyler89557]

-when you don't take statistic classes yet but try to solve this video anyways -when you fail miserably

[@brendon1689]

Isn't it (represented as a fraction, a percent would take a bit to find) 1/4^a, where a represents the current number of aliens?

[@brendon1689]

(Assuming that the aliens don't a limited life span, that is.)

[@brendon1689]

This is how I constructed this answer (quite simply). -each alien has a 1/4 or 1:4 or .25 or 25% chance to die each day -the chance that the aliens go extinct decreases with every alien (which means that finding a percent would be impossible?) -each day you would have to apply that one fourth (kind of like a dice roll, i suppose) to each alien And: *extinct would mean that all aliens have died and there are no aliens left* So, for example, if there were 24 aliens on the 19th day (which i think that the day is in actuality irrelevant) each alien has a 1/4 chance to die, making the chance that all 24 aliens will die at the turn of the 20th day: 1/281474976710656, or 1/4^24. This makes finding a percentage basically impossible, since the number of aliens is not always going to be constant, though the chance that all aliens will die on any day is the same as the chance of nothing happening to the entire alien population. Thus, it can be concluded that the most definitive and realest answer would be 1/4^a, and that because of this, the closest the aliens will likely ever get to extinction is on the first day.

[@brendon1689]

Or I am missing many things from this that I don't know about, which is likely since I am in my 9th year. I approached this as best I could with the most logical way I could think up while taking into account all relevant factors. I hope you see this (which is way I'm liking all of these comments xd) and correct me if I'm wrong.

[@FKeel1]

Yeah, you're right. Thats the answer for the probability of the aliens to go extinct on a given day. The original question is the overall percentage, so the sum of the probability of each day after landing. The question is worded in a way that also confused me at first.

[@Yonkage]

I did it by just crunching the numbers on the first two days and coming up with a reasonable extrapolation. 25% chance on the second day (only one alien so 25% chance he dies); that's the easy part. The third day it's another 25% chance if there's only one alien because he did not replicate, or 6.25% chance if he replicated once (25% chance squared) or 1.56% chance if he replicated twice (25% chance cubed) averaging those probabilities results in ~11% chance of extinction the third day. We're up to 36% chance, so because there are additional days where the probability of extinction is lesser than each previous day decreasing to an asymptote that approaches zero, I guessed the total would be around 40% I'm terrible at math so I am surprised I got so close!

[@Gef1ManArmy]

Well the people in the comments wanted a hard question. They got one and now many are complaining lol. Very nice puzzle, stumped me.

[@lucasfabisiak9586]

As is so often the case when people vehemently disagree about the answers to questions, the flaw is in the question itself. The question is-"What is the probability the alien race eventually dies out and goes extinct?" The word "eventually" is the source of the confusion. As others have stated, over an infinite amount of time, talking about the probability of an event loses its practical sense because the probability of any actually possible event becomes 1. No matter how many days pass, the number of aliens will always be finite, yet there is a possibility that they will all one day die, and with an infinite number of days for that to happen in, one of those days will necessarily realize that potential.

[@AkshaySakariya]

+MindYourDecisions Hey, aren't you supposed to consider the cases when the world has two aliens and one decides to multiply into 3 and one decides to die? And all such combinations... (Including ones with three aliens in the second day.) In such a case, the answer isn't the same. It actually very quickly goes out of hand. The case that you portray is the case when at each stage, the aliens decide to conform and do the same thing. And as to how the question was framed, I think it should be the assumed that the aliens decide independently decide their fate everyday.

[@arandombard1197]

Basically you can't have infinite days. Its a concept that you have to be very careful with when you apply it in maths. In this context, as the number of aliens increase, their number basically trends towards infinity and the chance of them all dying simultaneously trends towards 0. People using the argument "with infinite time they all die" are wrong because you can't have infinite time but if you could, then you would also have infinite aliens with a 0% chance of dying.

[@orthoplex64]

I'm confused. Here's how I tried it: the average number of aliens alive D days later is 1.5^D and the probability that N aliens will all die at once is 0.25^N, so the probability that all the aliens alive after D days will all die at once is 0.25^(1.5^D). The sum of that from D = 0 to infinity, from WolframAlpha, is around 0.429 which appears like it could be the correct answer but is slightly off and I can't figure out why.

[@DutchStudent1982]

Ok, two ways I would look at this : Option, common demografics : per generation 1 in 4 aliens dies out, so that gives a mortality rate of 0,25 per generation 4 aliens, produce 7 offspring, so thats gives a birthrate of 1,75 This gives a population growth number of 1,50 per generation. So going by these numbers the number of aliens at any given time will be N*1.50^x (with N being the number of aliens you started with, in this case 1, and x the number of years passed) quite quickly you will have a massive hoard of aliens. so in that case the answer would be : never. however there is another way of looking at this. in generation 1 the chance for extinction ius 25% (1 in 4) IF it survives in generation 2 the chance for extinction is 3,6% (3 in 84) -> this gives that while every generation the chance for total ectinction within that generation gets smaller, there IS a chance for it. as such if number of years goes to infitinty, than anything no matter how small the the chance for ANY outcome to occur at least once is 100%. in that case the answer would be : always. However there is another way to look at it : if we take a any fraction of a whole (x/y)=z and than add the same fraction again (x/y)+((x-(x/y))/y)=z and so on (pardon me I have no knowledge of a formula to formulate that string in one simple equation) than as y->INF than z=>x so the answer than would be they would get infinitely close to 100% probability for extinction but never get there. -however if the fraction is not /y but lets say (x/y*a^y) than we would get an increase or decrease, in this case a steady gradual one. if a is larger than 1, than eventually we would get 100% of whats left, so than the answer would be x, or all die out. if a is exacty 1, than the outcome is the same as (x/y) and this infinitely close to x but never getting there but what if a is lower than 1? (without alowing for a to have negative numbers) than we would get a decreasing curve.. effectively meaning y is schrinking faster than P is increasing.. making it flat of at a certain number. (though currently I am puzzeling at what the formula would be to calculate that number it would flat off at) -> any help here? -> and than there could also be an acceleration or decelleration in the increase or decrease. (and that in itself could be gradual or fixed and so on) (ok and now I hit the limit of where I can follow for now.. I'll just watch the video now) seeing as how the numbers do decrease.. my expectation is that they flat off but where?? -good even your video did not help me to understand, you did not give me the formula that I needed. while I had the right line of thinking...

[@HeyGuysItsMeDinkle]

wait I don't understand, in this "Riddle" you stated that an alien lands on earth. You then stated that "everyday after that, each alien on Earth undergoes a transformation" which is still only one alien. How can you determine the entire races probability based on one alien that landed on earth?

[@mathies3598]

This should be 100%, as this system continues until every alien has died, so eventually it will happen, you just have to keep going

[@clementberthelot33]

I don't understand why the probability P does not depend on the alien you're considering. I mean, if an alien has 1 descendent and another one doesn't have any, P would be equal to 0,25*0,25=0,0625 for the first one, and 0,25, for the other one, wouldn't it ? Sorry for my bad english :x

[@AntoshaPushkin]

P is probability for a newly-born alien to die somewhen in the future with all of his descendants

[@dylanrothmeyer6590]

Or you just take the question and solve it, "A single alien lands on Earth. Every day after that, each alien on Earth undergoes a transformation, which could be any of the four equally likely events: (a) the alien dies, (b) the alien does nothing, (c) the alien replicates itself (2 aliens total), (d) the alien replicates itself twice (3 aliens total). With bad luck the alien race might die out quickly. But with good luck the alien race might survive indefinitely. What is the probability the alien race eventually dies out and goes extinct?" It says every day after that so the option of it immediately dying could go out the window right away, however, it also says that the alien undergoes a transformation. Which means it could not do nothing.

[@NXeta]

I don't understand how you got from the first bolded line to next bolded line at 3:20 (on the left)

[@Oz4Ever]

He just solved for P and skipped the details, but he made it a bit unclear: the 3 solutions of this equations are 1, √2-1 and -1-√2. The two later solutions are being discarded in what follows.

[@NXeta]

oh okay thx

[@MarkFredrickGravesJr]

I paused the video to throw my estimate out there. At first I thought this: There's a 25% chance the aliens will go extinct if the first alien dies. Add that to the probability that the alien will do nothing and then die (6%). Then add the probability that the alien will replicate, then die, then die (1.6%)... This would converge to 33.333...% such as in the equation Ʃ.25^n where n goes from 0 to ∞. But... that doesn't account for recurring probabilities. 25% only occurs once, as does 6%, but 1.6% should occur twice, because the alien could replicate, die, die or he could do nothing, do nothing, die. Then 0.4% occurs 4 times, and 0.1% occurs 16 times. But wait... there's more! That's only accounting for one alien doing all the replicating each cycle, where actually each alien will be diverging into his own set of probabilities. Shit... I give up, I haven't even passed precalc yet. Gonna finish the video, now. lol

[@VoxelProductions]

Couldn’t it be argued that the answer shifts depending on how many aliens there are? To me, it’s closer a logarithmic function than it is to exact percentage the answer would have to represented as an equation Y = 1/ (4^x) where x is the number of aliens. If there’s one alien then 25%, if 2 then 1/16 chance. If you have 3, then 1/64 so on and so forth. It works because the chance of 1/4 is always raised to the power to the number of aliens there are and gives no regard what day it’s on because that information is irrelevant. The days would only come into question if you’re talking about a singular alien on how long they would stay in an assumed period of time and even then the equation be the same but instead of x representing the number of aliens it would represent the number of days.

[@albertlau867]

consider an alternate quiz. each alien has 50% chance to (a) die (b) replicate itself 999 times now start with 1000 aliens. by applying "eventually die" logic, do u think they will 100% extinction? hope that help clearing out some doubt.

[@kradrol]

the probability is 1 in 4. if the first alien lands on earths, and his first transformation is he dies, then the race is extinct.

[@shoulderescape]

#idiotconfirmed

[@kradrol]

***** "it's natural predator, the troll, is approaching. let's see what happens next."

[@imharbinger]

For people who say it must be 1, consider this. if instead of 4 equally likely events. we had 3 equally event, in which the alien dying doesn't exist. so the p should be 0. but we can write p as p=p/3+p^2/3+p^3/3. if you take p value as 1. it fits. but that is obviously wrong. same thing is happening here.

[@thomasr.jackson2940]

Well that was truly interesting. I looked at the first and second days. First I wasted time with binomial theory and combinatorials, but I wasn't making progress. I then realized that the chance of extinction was something higher that an infinite power series of 1/4, i.e. A greater value greater than one, or greater than a hundred percent certainty. That didn't smell right. And I then saw that the probability of survival also represented a power series that was greater than one. Hmmm... So, what I was missing is the theory of statistical branching processes, which has already worked this out and resolved the contradictory results that belie reason. I sort of learned something today. I am not sure why the formula used works, but I know mine didn't, and I have seen the solution. Something for my list of things to check on. Thanks again. Interesting problem. Maybe a little un satisfying in that I don't grasp why the answer works yet, but that would be too long and involved for the video, and I am happy being shown the answer and pointed in the right direction to learn more.

[@marcellomarianetti1770]

What is the conditional probability?

[@SuperDreamliner787]

Can´t the discussion whether it´s 1 or sqrt(2)-1 be simplified by saying that 1 cannot be the solution, since a probability of 1 indicates, that in any case the aliens will die, but of course there are cases, where the aliens don´t die, so it cannot be 1!?

[@HadienReiRick]

I like to think of this in reverse (simply because the math is easier to follow). The probability that a specific alien DOESN'T die today is 100%(all cases) - 25%(death case) = 75% (all survival cases). The probability that the alien lives for two days is 0.75^2 = 56.25%. Continue this ad infinatum and the probability that a specific alien can live for infinite days reaches 0%. Thus the odds that the specific alien will die is (100%(all case) - 0%(never dies)), otherwise stated that there is a 100% chance that the specific alien will "eventually" die. All duplicated aliens follow the same rules as independent cases and independently eventually reach the same conclusion. So each and every alien will "eventually" die and the race will go extinct. the probability is 100% because no time limit was distinctly specified

[@dr.morbius]

If I was asked this question in an interview for a technical job I would get up and walk out. This sounds more like a question that would be asked in a QA or QC position where you do a lot of statistical analysis.