Daily math

100% given enough time.

>>8793476
No, that would mean that they _always_ go extinct. But it is possible for the population of object to grow indefinitely.

>>8793375
holy shit

hm on the first day that probability would be 0.25 and from there it would exponentially falloff...I don't know how to solve it but it seems like some logarithmic function..right??

I remember that was solved involving some stupid polynomial...

I remember thinking about the written out solution, more or less understanding it, but not leaving with a real good intuition.

>>8793492
My line of reasoning was that, given enough time, there will exist a step where every object in the population will die.

>>8793555
>I remember that was solved involving some stupid polynomial
That's the solution I know.

>>8793604
Yes, I don't consider your answer stupid.

You mean some day as an arbitrary day?

>>8793672
I still can't see what's wrong with my logic. Do you have an argument, or did i miss the point of your reply? Genuinely intrested. Keep up the daily problems, my dude.

>>8793375
Let F(N) be the probability of a collection of N objects eventually disappearing entirely. Then we have
F(1) = 1/4 + 1/4 F(1) + 1/4 F(2) + 1/4 F(3)

But if there are multiple objects, the probability of them all eventually dying off, is the same as the probability of each of them dying off independently:
F(N) = F(1)^N for N>1

Substituting that into F(1), we get:
F(1) = 1/4 + 1/4 F(1) + 1/4 F(1)^2 + 1/4 F(1)^3

Defining X = F(1) for ease of notation yields:
X = 1/4 + 1/4 X + 1/4 X^2 + 1/4 X^3
4X = 1 + X + X^2 + X^3
X^3 + X^2 - 3X + 1 = 0

Solving this polynomial yields X = 1 or X = sqrt(2) - 1 or X = -sqrt(2) - 1. Thus, the probability of one object dying off eventually is either 1, sqrt(2) - 1, or -sqrt(2) - 1.
Of these, -sqrt(2) - 1 is not a probability at all for it is negative; leaving 1 and sqrt(2) - 1 as the two possibilities for the probability we're after.

And here I am stuck. It seems to me that the probability cannot be 1, and that therefore it must be sqrt(2) - 1, but I don't know how to rigorously argue this.

>>8793744
The expectation value of the grow rate is 1.5, so you would expect the population to grow (and therefor the probability, that they will survive > 0.5, which in fact turns out to be true).

>>8793762
Correct, gold medal goes to you my friend!

Originally, the "objects" are aliens. Follow the link for a detailed solution (there is also the answer to you p=1 problem, has to do with the expectation grow rate of 1.5):

https://mindyourdecisions.com/blog/2016/10/02/can-you-solve-the-alien-extinction-riddle-tough-job-interview-question-sunday-puzzle/

or youtube (don't know if it's good, didn't watch the whole video):

https://www.youtube.com/watch?v=A5-Q2GdD5xw

>>8793794
>https://mindyourdecisions.com/blog/2016/10/02/can-you-solve-the-alien-extinction-riddle-tough-job-interview-question-sunday-puzzle/
Ah, neat!

The growth rate argument seems fishy to me. Yes, it is strongly indicative, but it is not a rigorous proof on its own and I see now obvious way to turn it onto one.

The induction argument is excellent, though. The limit of the probability of extinction after N days as N goes to infinity is the real answer we are after, and this algebraic solution of mine is just a clever trick in that direction; when imported back into the limit setting, distinguishing 1 from sqrt(2) - 1 becomes quite easy.

Suppose for [math]r_k\in\mathbb{R}>0[/math] and [math]-\pi<\theta_k<\pi[/math], the following are true:
[eqn]\sum_{k=1}^3 r_k\sin{\theta_k}=0,\;\;\;\sum_{k=1}^3 r^2_k\sin{2\theta_k}=0,\;\;\; \sum_{k=1}^3 r^3_k\sin{3\theta_k}=0[/eqn]
Show that [math]\theta_k=0 [/math] for at least one value of [math]k[/math], and if, say, [math]\theta_1=0[/math], then [math]\theta_2=-\theta_3[/math]

>>8793811
>The growth rate argument seems fishy to me. Yes, it is strongly indicative, but it is not a rigorous proof on its own.
You are right. Should not have mentioned it. It's math, *bitchslap myself*.

>>8793375
on the first day, there is a 1/4 chance that there are no objects left by the second day.
on the second day, there is a 1/4 chance there's 1 object, 1/4 chance theres 2 objects, and 1/4 chance theres 3 objects.
So on the second day, there's a 1/16+1/64+1/256 chance everything goes extinct by the 3rd day
if 1 object:
1/4 0
1/4 1
1/4 2
1/4 3
if 2 objects:
1/16 0
2/16 1
3/16 2
4/16 3
3/16 4
2/16 5
1/16 6
if 3 objects:
0 1/64
1 3/64
2 6/64
3 10/64
4 12/64
5 12/64
6 10/64
7 6/64
8 3/64
9 1/64

on day 3:
0 85/256= 1/4+1/16+1/64+1/256
1 27/256
2 34/256
3 42/256
4 24/256
5 20/256
6 14/256
7 6/256
8 3/256
9 1/256

after infinite days, the chance of 0 is 1/4+1/16+1/64+1/256+....=n
1/4(1+1/4+1/6+1/64+...)=n
4n=1+n
3n=1
n=1/3
there is a 1/3 chance there will be no objects left someday. I'm guessing by symmetry theres a 1/3 chance of there being infinite objects left and a 1/3 chance there being 1 or finite objects left.