Who can solve this?

[Anonymous No.9390779]

Hey /sci/, I have a math problem for you.

Let's say we have a circle with a radius of 1.
We start on the edge of the circle.
Now, every step we move with a distance of 1 in a random direction.

How many steps on average does it take to leave the circle?

[Anonymous No.9390783]

1, because I'm always lucky and a wizard.

[OP !!pKAZJ5QmYaG No.9390788]

>>9390783
wrong

[Anonymous No.9390789]

What is the pdf nigger.

[Anonymous No.9390792]

>>9390779
On average it only takes one step to leave the circle. You have a zero percent chance to land on the circle on your first step since you can only land at the two points 1 rad away and you have an infinite number of points (the interval) of all points 1 away that aren't on the circle

[Anonymous No.9390793]

185

[Anonymous No.9390796]

>>9390788

[OP !!pKAZJ5QmYaG No.9390803]

>>9390792
No, you don't understand.
We're not just talking about being exactly on the edge of the circle here.
You're inside the circle when you're INSIDE the circle.

>>9390793
wrong...

[OP !!pKAZJ5QmYaG No.9390819]

>>9390789
What are you talking about?

[Anonymous No.9390820]

>>9390803
Okay, how much distance from the edge exactly? You know you can arbitrarily get close to the boundry of circle righr? Also you failed to proviee a probability measure.

[Anonymous No.9390823]

>>9390819
Probability density function, aca, you cannot say random like it means something brainlet.

[OP !!pKAZJ5QmYaG No.9390829]

>>9390823
Sorry, I'm not that well versed in math.
Uniform I guess?

[Anonymous No.9390833]

>>9390779
Nah, if he's standing on the edge he has 1/3 chances of moving into the circle. Assuming uniform probability distribution with the angle of his movement being the random variable

[Anonymous No.9390835]

>>9390833
meant to quote
>>9390792

[OP !!pKAZJ5QmYaG No.9390836]

>>9390820
>how much distance from the edge exactly?

I don't understand what you mean.
You're inside the circle when you're inside the circle...

[OP !!pKAZJ5QmYaG No.9390842]

>>9390833
>uniform probability distribution with the angle of his movement being the random variable
Yes, this.
Thank you for expressing it for me.

[Anonymous No.9390852]

>>9390836
Yea, but a starting point is well, a point... You need to define it, exactly in te edge works fine.

[OP !!pKAZJ5QmYaG No.9390858]

>>9390852
Yeah the starting point is exactly on the edge.

[Anonymous No.9390861]

In case anyone is wondering, this is what the distribution looks like.

[Anonymous No.9390864]

>>9390779
1.5

The proof is trivial and left as an exercise to the reader.

[OP !!pKAZJ5QmYaG No.9390866]

>>9390861
Ok if we're gonna do simulations, I might as well tell you the approximate decimal answer I got by simulating it:

~1.55981

[Anonymous No.9390867]

>>9390823
not understanding the basic shit he said makes you the brainlet.

[Anonymous No.9390870]

>>9390867
Hey, for all I know he coukd assigned weights to some steps:^)

[Anonymous No.9390871]

>>9390789
>>9390823
Uniform, obviously.

[OP !!pKAZJ5QmYaG No.9390873]

>>9390864
wrong

[H No.9390886]

>>9390823
He clearly means the uniform distribution.

[Anonymous No.9390889]

>>9390861
>>9390866
Same.

[Anonymous No.9390902]

You can't take 1.5 steps.

You can take 1 step or 2 steps.
1 step is still on the circle's edge.
2 is the correct answer you brainlets.

[OP !!pKAZJ5QmYaG No.9390904]

>>9390889
Nice, but can you increase the accuracy?
Cause your answer almost makes it seem like the answer is pi/2, but I don't think that's the answer. I think it's slightly less than that.

[Anonymous No.9390905]

>>9390902
You can't have 1.5 kids but that's still the average for American households.

[OP !!pKAZJ5QmYaG No.9390930]

>>9390902
I don't think you're understanding the problem.
Just to clarify, here's a random example where it took 6 steps to leave the circle.

(The line segments are supposed to be the same size as the circle's radius, but I didn't really measure them.)

[OP !!pKAZJ5QmYaG No.9390935]

>>9390930
And here's a random example of where it took just 1 step to leave the circle.

[Anonymous No.9390941]

Trying to solv this mess, but I'm having difficulty considering the probability of staying in the circle when a point is inside the circle. I'm sure it must depend solely on it norm, but I'm not sure, for a point outaide the circle I used the tangents, but here i don't know.

[Anonymous No.9390957]

>random direction

Meaning everything to the right of this red line is also a likely direction, all of which still take 1 full distance/step, if the starting point is the edge where the arrow is..

[OP !!pKAZJ5QmYaG No.9390961]

>>9390957
yes, correct

[Anonymous No.9390970]

>>9390779
So I'm guessing first find the distribution function in terms of distance r from the center of the circle. Then use that to find the probability of remaining in the circle given that you currently are some point within the circle. After that solve for the expected value by Negative Binomial Distribution. The distribution even fits the plotted graph.

[OP !!pKAZJ5QmYaG No.9390976]

>>9390970
idk sounds good

[Anonymous No.9391002]

I would think that the chance of the second step staying inside the circle is the average of 1/3 and 1/2. But the simulation seems to say not. It seems to say this chance is 2/5 instead of 5/12. So Either my analysis is wrong or the simulation is wrong.

[Anonymous No.9391012]

>>9391002
Ah OK, I think I have it. The chance of the nth step staying inside the circle is the average of the (n-1)th and (n-2)th steps.

2/3+2(1/3)(1-(1/3+1/2)/2)+3(1/3)(5/12)(1-(5/12+1/3)/2)+4(1/3)(5/12)(15/24)(1-(15/24+5/12)/2)...

[Anonymous No.9391017]

Does anyone know any methods to solve for g(x)?

I get ~1.56 with simulations too

[OP !!pKAZJ5QmYaG No.9391019]

>>9391012
Are you saying the chance of still being in the circle after 2 steps is 2/3+2(1/3)(1-(1/3+1/2)/2)+3(1/3)(5/12)(1-(5/12+1/3)/2)+4(1/3)(5/12)(15/24)(1-(15/24+5/12)/2)... ?

[Anonymous No.9391025]

[math]1+\gamma[/math]

[Anonymous No.9391034]

>>9391025
fuaaarrrrrkkkk explain

[OP !!pKAZJ5QmYaG No.9391035]

>>9391025
uhh what's γ?

[Anonymous No.9391037]

>>9391019
No, that is the answer to the question you posted. The chance of still being in the circle after 2 steps is 2/3+(1/3)(1/3+1/2)/2

[Anonymous No.9391042]

Made a nice picture :)

[OP !!pKAZJ5QmYaG No.9391043]

>>9391025
>>9391035
https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant

Oh shit
Congrats dude, you found the answer

[Anonymous No.9391044]

>>9391035
euler mascheroni constant

[Anonymous No.9391052]

Hey guys looks at all the answers I found durr

[OP !!pKAZJ5QmYaG No.9391053]

>>9391037
I don't know if it's correct or not
Does it come out to about 1.57721?

[Anonymous No.9391060]

>>9390861
>>9390889
look at these code monkeys being so smart and solving problems

[OP !!pKAZJ5QmYaG No.9391061]

>>9391042
nice

[Anonymous No.9391063]

>>9391053
On the 5th step it is at 1.57889...

[Anonymous No.9391064]

>>9391025
how the fuck did you get it

[OP !!pKAZJ5QmYaG No.9391069]

>>9391063
Wow good job dude

[OP !!pKAZJ5QmYaG No.9391087]

Wait now I'm starting to doubt if this >>9391025 is even the right answer.

Because my simulation says it's something like 1.5598 instead of 1.57721

[Anonymous No.9391088]

>>9391064
Probably entered the numeric mean into Wolfram Alpha.

Prove me wrong by posting a proof.

[Anonymous No.9391101]

>>9390779
every fucking step is random!? fuck you op this isn't determinable

[Anonymous No.9391102]

>>9390930
this is the most braindead shit I've seen in a long time

[Anonymous No.9391109]

>>9390779

[Anonymous No.9391110]

>>9391102
?

[OP !!pKAZJ5QmYaG No.9391121]

>>9391109
You found the probability of leaving the circle in 1 step, good job.
But the question was how many steps it takes on average to leave the circle.

[Anonymous No.9391131]

The probability of being within the circle at step n is given by 2/3^n. This means that the expected value of n is given by [math]\sum^{\infty}_{n=0} \frac{2n}{3^n}[/math]. This sum converges to 1.5.

[Anonymous No.9391132]

>>9391131
>The probability of being within the circle at step n is given by 2/3^n
wrong

[Anonymous No.9391134]

>>9391132
Shit, I only considered the cases when your step puts you on the edge of the circle. My bad.

[Anonymous No.9391198]

I don't fucking know.

[Anonymous No.9391201]

Alright, I just did some extremely shitty math and made a bit of progress. I think I found the probability of leaving the circle in one step from point (x,y). First, find the distance between (x,y) and (0,0), which is given by [math]\sqrt{x^2+y^2}[/math]. Then, plug this value in to the equation to find the y-coordinates of the touching points of two intersecting circles. This gives us touching points with a positive y-coordinate of [math]\frac{1}{2}\sqrt{-(x^2+y^2)+4}[/math]. Now, take the arcsin of the positive root and divide by pi to get the probability of leaving the circle in one step: [math]P(x,y)=\frac{\arcsin(\frac{1}{2}\sqrt{-(x^2+y^2)+4})}{\pi}[/math]. Can anyone verify my math?

[Anonymous No.9391208]

I think this means it's twice as likely to land in the region on the left as the parts on the top right and bottom right.

[Anonymous No.9391218]

The chance of the nth step staying inside the circle is the average of the (n-1)th and (n-2)th steps.

So the average number of steps is

[math]\sum_{j=1}^{s} ((1-\frac{2 a_{j-1}-(-1)^{j-1}}{3*2^{j-1}}) \prod_{k=1}^{j-1} \frac{2 a_{k-1}-(-1)^{k-1}}{3*2^{k-1}})[/math]

[math]a_0 = 2[/math]

[Anonymous No.9391222]

>>9391218
[math]\sum_{j=1}^{s} (j(1-\frac{2 a_{j-1}-(-1)^{j-1}}{3*2^{j-1}}) \prod_{k=1}^{j-1} \frac{2 a_{k-1}-(-1)^{k-1}}{3*2^{k-1}})[/math]

a0 = 2

[OP !!pKAZJ5QmYaG No.9391236]

>>9391201
>Can anyone verify my math?
If we choose a point on the edge, for example x = 1, y = 0, then the probability of leaving the circle in one step should be 2/3.
But your formula gives arcsin(sqrt(3) / 2) / pi, which is not 2/3.

[Anonymous No.9391237]

[eqn]\sum_{j=1}^{s} (j(1-\frac{a_j}{3*2^{j-1}}) \prod_{k=1}^{j-1}\frac{a_k}{3*2^{k-1}})[/eqn]
[eqn]a_n = 2 a_{n-1}-(-1)^{n-1}[/eqn]
[eqn]a_0 = 2[/eqn]

[Anonymous No.9391240]

[eqn]\sum_{j=1}^{s} (j(1-\frac{a_j}{3*2^{j-1}}) \prod_{k=1}^{j-1}\frac{a_k}{3*2^{k-1}})[/eqn]
[eqn]a_n = 2 a_{n-1}-(-1)^{n-1}[/eqn]
[eqn]a_1 = 2[/eqn]

[OP !!pKAZJ5QmYaG No.9391246]

>>9391240
hold on let me verify this

[Anonymous No.9391249]

>>9391246
Thanks for having an actually interesting thread, OP

[Anonymous No.9391254]

>>9391246
Hmm some corrections:

[eqn]2/3+\sum_{j=2}^{s} (j(1-\frac{a_j}{3*2^j}) \prod_{k=1}^{j-1}\frac{a_k}{3*2^k})[/eqn]
[eqn]a_n = 2 a_{n-1}-(-1)^{n-1}[/eqn]
[eqn]a_0 = 2[/eqn]

[Anonymous No.9391256]

>>9391254
Fucking hell, last line should be

[eqn]a_1 = 2[/eqn]

[Anonymous No.9391261]

I don't think you know what you're doing.

[Anonymous No.9391265]

Here's what I have so far:

Let [math]d_t[/math] denote the distance from the origin after [math]t[/math] steps. Then [math]d_0 = 1[/math] and [math]d_{t+1} = \lvert d_t + \exp i \theta \rvert = \sqrt{d_t^2 + 2 d_t \cos \theta + 1}[/math] where [math]\theta \sim \mathcal{U}(0, 2\pi)[/math] and thus [math]\cos \theta[/math] has an arcsine distribution.

[Anonymous No.9391271]

>>9391265
I meant [math]\theta_t[/math] since the angle depends on [math]t[/math].

[OP !!pKAZJ5QmYaG No.9391280]

>>9391254
ehh if I run this with s = 10, I get 46.28

[Anonymous No.9391292]

Circle has no edges.

[Anonymous No.9391307]

>>9391280
I must be writing it wrong, because on s = 8 I have 1.5534

[Anonymous No.9391308]

>>9391280
2/3+2(1/3)(1-(1/3+1/2)/2)+3(1/3)(5/12)(1-(5/12+1/3)/2)+4(1/3)(5/12)(3/8)(1-(3/8+5/12)/2)+5(1/3)(5/12)(3/8)(19/48)(1-(3/8+19/48)/2)+6(1/3)(5/12)(3/8)(19/48)(37/96)(1-(37/96+19/48)/2)+7(1/3)(5/12)(3/8)(19/48)(37/96)(25/64)(1-(37/96+25/64)/2)+8(1/3)(5/12)(3/8)(19/48)(37/96)(25/64)(149/384)(1-(149/384+25/64)/2)

[OP !!pKAZJ5QmYaG No.9391390]

>>9391307
>>9391254
No, it was my mistake.
Your solution is correct! :)

I have no idea how you did it though,
or how you came to the insight that the chance of the nth step staying inside the circle is the average of the chances of the (n-1)th and (n-2)th steps staying inside the circle.

[Anonymous No.9391418]

>>9391390
Go back to the picture in >>9391002 which shows the possible 2nd steps are simply ranging uniformly over the internal angles 2pi/3 to pi, from starting at the edge of the circle to the middle of the circle. Dividing the internal angle by 2pi gives the probability of staying in the circle. So the probability of staying in the circle at the 2nd step is 5/12.
Then on the 3rd step it ranges from starting in the same arc as the 2nd step (5/12) to starting at the edge of the circle (1/3). And the 4th step ranges from starting at the 3rd step to starting at the 2nd step, etc.

[Anonymous No.9391435]

>>9390779
0.72

[OP !!pKAZJ5QmYaG No.9391445]

>>9391435
no, sorry
this guy already got it >>9391254, it's 1.557943...

I'm leaving now bye

[Anonymous No.9391914]

>>9390779
what I thought it was
((pi+e)*0.1)+1

kek

[Anonymous No.9391924]

>>9390861
What? How is there 10% probability that you will need 3 steps! The maximum is always 2 steps. Garbage in, garbage out.

[Anonymous No.9391928]

>>9390861
I fucking hate Python I get why it's so useful but I just don't like it at all. The syntax, the white space, ugh

[Anonymous No.9391961]

>>9390779
well the edge of the circle is round anon so its probably more than 50% chance to leave on the first step

[Anonymous No.9391973]

>>9391924
There is no upper limit; it could take an infinite amount of steps. Here are just two ways you could theoretically be trapped forever, no matter how small that chance.

>bounce between the edge and the origin forever
>bounce across six equidistant points along the edge
>Combination of the two, also

So the answer has been posted
>>9391025

Can it be derived using pen&paper?
I'm a brainlet trying to learn more, and maths make the most sense to me in this way.

[Anonymous No.9391978]

>are we assuming we can pass through the circle?
>if so
shouldnt be more than 2.

moving without leaving the circle would mean you make a straight line directly to the otherside on your first try. barring this 1/360 chance doesnt occur a second time consecutively, you would leave the circle. i guess if we're going on rational likelihood instead of pure probability, the answer would be one.

[OP !!pKAZJ5QmYaG No.9392046]

>>9391973
>>9391978
sorry, maybe I should have explained it clearer
see >>9390930 and >>9390935

but I think this anon already solved it btw >>9391254

[Anonymous No.9392156]

>>9392046
Personally I don't think that anon got it. His answer doesn't make sense to me and his premise that the probability of it taking n steps is the average of the probabilities of it taking (n-1) and (n-2) steps is just plain wrong.

I think he just stumbled on a series that gave a somewhat correct looking answer after much messing around. Closed form expressions that approximate arbitrary truncated decimals are not difficult to come by as shown here >>9391052

[Anonymous No.9392179]

infinity

[Anonymous No.9392254]

Shouldnt it just be 1.5 exactly?
The probably of leaving on the nth step is (2/3)*(1/3)^(n-1) and the number of steps itll take is n times that.
So according to wolfram, the sum from 1 to infty of n*(2/3)*(1/3)^(n-1) = 1.5

[Anonymous No.9392258]

>>9392156
> the probability of it taking n steps is the average of the probabilities of it taking (n-1) and (n-2) steps is just plain wrong.
That was not my premise. My premise is that the probability of staying in the circle at the nth step is the average of the probabilities of staying in the circle at the (n-1)th and (n-2)th step. This is apparent if you visualize the possible ranges for each step. The formula I posted directly follows from that premise:

>>9391254
>>9391256

[Anonymous No.9392264]

>>9392254
>The probably of leaving on the nth step is (2/3)*(1/3)^(n-1)
The probability of leaving on the 2nd step is the probability of not leaving on the first step multiplied by the probability of leaving on the second step. The range of possibilities for the second step can be seen here >>9391002 and ranges uniformly between internal angles 2pi/3 and pi. This means that the chance of staying in on the second step is (1/3+1/2)/2 = 5/12. So the chance of taking two steps and being outside the circle is (1/3)(1-5/12) = 7/36 which means your formula is wrong.

[Anonymous No.9392416]

>>9391928
Why? It looks like pseudocode which is great.

[Anonymous No.9392425]

>>9392258
>the probability of staying in the circle at the nth step is the average of the probabilities of staying in the circle at the (n-1)th and (n-2)th step

Why?

[Anonymous No.9392525]

>>9392425
See >>9391418

[Anonymous No.9392528]

>>9390779
infinite steps?

[Anonymous No.9392626]

>>9392528
Are you retarded?

[Anonymous No.9392659]

>>9391025
C L E O

[Anonymous No.9393191]

[Anonymous No.9393198]

>>9392659

Fake news it's not an integral

[Anonymous No.9393216]

>>9390779
Place the starting (1st) position on the edge. Because we can only assume the direction he moves is uniformly random, then the curve defining all possible 2nd positions is another circle of radius 1. The probability of stepping outside the circle on the first step (1st to 2nd) is equal to the proportion of the 2nd circle that is outside of the first circle. Calculate that probability.

In the event you do not step outside the circle on the first go, you perform essentially the same move again as a probability conditioned on the first. If a closed form for this sequence exists (even maybe if it doesn't), you can derive the Probability Mass Function as this sequence. Then, you can just use that PMF to derive the Mean number of steps.

Not gonna bother working it out tho.

[Anonymous No.9393679]

>>9393216
>Not gonna bother working it out tho.

[Anonymous No.9394571]

>>9393216
It's easy, it's 1 step.

There is exactly 1 angle/direction that won't take you out of the circle on the first step. Since there's an infinite number of directions to choose, the probability of hitting that 1 direction is 0. Therefore, the answer is 1 step.

[Anonymous No.9394622]

>>9394571
radius, not diameter

[Anonymous No.9394817]

>>9394571
>There is exactly 1 angle/direction that won't take you out of the circle on the first step.

[Anonymous No.9394836]

>>9390961
Probably worships Egyptian faggotry.
Isis, muh all seeing eye, derp, all that jazz

[Anonymous No.9394871]

>>9390779
Faulty assumption: given the construction of this problem in a rational framework, the application of the concept of "random direction" is disjunctive and causes contradiction - we cannot rationally determine what "random" is.

We assume that something for which there is no known pattern is "random" - however, if we had enough information, we would be able to determine a strange attractor, and thus better predict the emergent shape along a vector of spacetime.

It takes only one step to leave the circle, but the probability is unknowable and comes down to a single question: are you moving toward or away from it?

The solution is that you are both moving toward and away from it at the same time, but until you can measure this distance from an observable point outside the circle, you don't know which it is.

You have to be outside the circle to solve the problem (which you are if you're thinking about it as an abstraction, looking down from above as the diagram suggests, but you also aren't, because you're inside another circle, the boundaries of which you cannot see - this is the circle of your knowable universe).

[Anonymous No.9394886]

>>9394871
We already resolved that buddy boy :)

[Anonymous No.9394905]

>>9390779
let [eqn]
r_0 = \left\{(0,0)\right\} \\
r_n = \bigcup\left\{\partial\mathcal{B}_1(p) : p \in r_{n-1}\right\} \\
r_n' = r_n \cap \mathcal{B}_1(0,0) \\
\mathbb{P}_n = \frac{\int_{r_n'} f(r_n', p)\ \text{d}\mu(p)}{\int_{r_n} f(r_n, p)\ \text{d}\mu(p)} \\
\begin{align}
\text{where } &f: \mathscr{P}(\mathbb{R}^2)\times\mathbb{R}_2 \to \mathbb{N} \\
&f(U, p) := \#\left\{ p' \in U : p \in \partial\mathcal{B}_1(p') \right\}
\end{align} \\
\text{and } \mathcal{B}_r(p) := \left\{p' : \left\|p-p'\right\| \le r\right\}
[/eqn] then [math]\mathbb{P}_i[/math] is the probability of making a transition that stays inside the circle on the [math]i[/math]th step. i think. maybe.

don't know what to do next. if there's some version of the negative binomial with a success probability that depends on the number of successes theretofore, i'd take the mean of that.

[Anonymous No.9394925]

>>9394905
Could it be a product of the probability of the previous successes?

[Anonymous No.9394933]

>>9394925
if you did [math]\prod\limits_{i=0}^{n}\mathbb{P}_i[/math] i think that would be the probability of being inside the circle by the [math]n[/math]th transition.

[Anonymous No.9394986]

>>9391254
>>9391256
>>9391390
Hmm I now realize my answer is only a close approximation based on an assumption of uniform distribution of probabilities over each arc when the distribution is not exactly uniform.

The probability of staying in the circle at the second step is (3/pi)(the integral from 0 to pi/3 of arccos(sqrt(2-2cos(x))/2)/pi dx ~ 0.3789 which closely matches the simulation.

[Anonymous No.9395084]

>>9394871
Shut the fuck up

[Anonymous No.9395112]

>>9394986
OK so

Let d_n be the distance from the center at step n.

Let x_n be the angle of your next step at step n relative to the line between the center of the circle and your position.

d_0 = 1

Then d_(n+1) = sqrt(1+(d_n)^2-2(d_n)cos(x_n))

In order to stay inside the circle x_n must be less than arccos(d_n/2)

The probability of staying inside the circle after the nth step is therefore (1/arccos(d_n/2))(integral from 0 to arccos(d_n/2) of (arccos((d_(n+1))/2)/pi)dx)

So the answer is way too messy to write out.

[Anonymous No.9395132]

>>9395112
Jesus Christ dude, use LaTeX.

[Anonymous No.9395210]

>>9391254
don't wanna bother anyone but does anyone know of a way of putting this into GeoGebra?

[Anonymous No.9395224]

>>9395132
I would of I wasn't on my phone.

[Anonymous No.9395229]

>>9395210
Lol no. But I wrote that and it's wrong.

[Anonymous No.9395251]

1.0000000001 steps. Ha

[Anonymous No.9395316]

Is being on the edge of the circle defined as in?

If so then you have a 50/50 chance of stepping out of the circle

if not then
you are sol

[Anonymous No.9395418]

>>9395112
Iv'e been trying to find the expected value for any given d from 0 to 1.
Let d be your starting point.
The probability of the first step going out is 1-arccos(d/2)/pi

Let E(d) be the expected number of steps to leave from d.

Let D(d,t) = sqrt(1 + d^2 -(2d)cos(t)) = |d-exp(it)| where t t goes from -arccos(d/2) to arccos(d/2).
D tells you what radial distances are reachable from d.

Is it true that E(d) = 1-arccos(d/2)/pi + Integral[1+ E(D(d,t)),t,0,arccos(d/2)]/pi ?

Idk if you can define expected values recursively like that.

The final form is E(d) = 1 + Integral[E(D(d,t)),t,0,arccos(d/2)]/pi.

Then define G(d) = E(sqrt(d))

I get G(d^2) = 1 + Integral[G(1 + d^2 - (2d)cos(t)),t,0,arccos(d/2)]/pi

I might try using a Taylor series for G and work out what the coefficients are. I know it is ill-behaved at 0 and G(0) = 1 + G(1).

[Anonymous No.9395436]

>>9395418
I've also tried change of variables to get a pdf for D(d,t) and eliminate the angle t.

https://en.wikipedia.org/wiki/Probability_density_function#Dependent_variables_and_change_of_variables

It turns out to be more of a mess that way though.

[Anonymous No.9395537]

>>9394886
>>9395084

No, seriously. Obviously, the tacit implication is that the question of whether or not the direction is really "random" is moot to the asker - but it's sort of a big problem.

Probability is based on likelihood. If you're setting a limit as either being inside or outside of the circle, and you're on the edge of it, and you're considering being on the edge as within it (since the question asks how many steps it takes to leave), then essentially after the first step it's either 0% or 100%. The first step is all that matters.

After that, it gets more complicated. If you haven't left after the first step, then your current position is somewhere within a section of the circle that includes the arc of the edge and the remainder of the chord that connects it to the partial circle of whatever radius remains between the edge and the first step.

This means that if it's really "random," then it's fully possible that you never leave the circle because you keep stepping within it.

However, it says "on average." That's the problem. "Average" implies a statistical model based on a rational projection of total steps made and in what direction they'd be made in given a data set.

But there's no data set. We have no idea the number of steps we should consider "average," and therefore it's not a rational question. The instantiation of randomness is inherently irrational, and you can't pair it with a mathematically solid answer without setting a limit by providing a data set from which to draw calculations.

Even if you said "after a million steps, what's the probability that you'd be outside the circle," there'd be an answer we could accept as an objective average number to leave the circle based on that data set...

But you gave no temporal boundary.

Maybe I'm stupid, I dunno - can anyone help a nigga out?

[Anonymous No.9395538]

>>9395537
Just stop.

[Anonymous No.9395558]

>>9395224
>would of
You filthy illiterate.

[Anonymous No.9395563]

>>9395537
>essentially after the first step it's either 0% or 100%
Shut the fuck up and get out of this board.

[Anonymous No.9395566]

>>9390779
the ratio of the length of the transcribed circle segment outside the original to the length within the circle is the chance you're looking for

also you are stuipd

[Anonymous No.9395570]

>>9395566
Are you capable of reading? It's not just the first step dumbass.

[Anonymous No.9395605]

>>9395538
>>9395563
So, no; you can't sufficiently explain this, and you are choosing to call me stupid instead of simply providing the information I lack.

Kthx

[Anonymous No.9395645]

>>9395537
>But you gave no temporal boundary.
There is no "temporal boundary".

[Anonymous No.9395679]

>>9395645
Yeah, there is. As in "what is the probability after x days/months/weeks/years/millennia/fucking any number of ways to literally express a temporal boundary?"

Should I have just said "time limit?" Is that the confusion?

[Anonymous No.9395683]

>>9395679
There is no upper bound to the possible number of steps. OP is just asking for the expected number of steps. Do you know what an expected value is?

[Anonymous No.9395882]

>>9391060
>solving

[Anonymous No.9395953]

>>9395683

Expected value *after how long* though? If you're just taking steps forever, then you could wander forever. Why is this hard?

[Anonymous No.9395961]

>>9395953
Are you dumb.
You can flip a coin forever. The expected number of flips until you get your first heads is 2. Why is this hard?

[Anonymous No.9395973]

>>9395953
Yes, in theory you could wander forever inside the unit disc without ever escaping, just like in theory you could flip a coin forever without producing heads. And yet there is a finite expected number of flips needed to produce heads, just like there is a finite expected number of steps needed to escape the disc.

[Anonymous No.9395993]

>>9395961
Because if it were really that expected, they wouldn't use it to start the fucking superbowl because it's literally a symbol of an unpredictable outcome, and trying to find a pattern that could reliably determine it is basically the plot of the movie Pi... am I really stupid? I might be stupid.

[Anonymous No.9396022]

>>9395993
The fuck are you saying man.

[Anonymous No.9396211]

>>9395418
>I know it is ill-behaved at 0
In what way is it ill-behaved?

[Anonymous No.9396227]

>>9395418
Are you missing a factor of two in the final equation? That is, shouldn't it be:
[eqn]G(d^2) = 1 + \frac{2}{\pi} \int_0^{\arccos(d / 2)} G(1 + d^2 - 2d\cos t)\,dt.[/eqn]

[Anonymous No.9396292]

>>9396022
I'm saying that the idea of an "expected value" is only an abstraction, and there is no way to apply it to the physical universe without violating the uncertainty principle. Like, is this not a literal question from OP? Because literally there is an answer, but it's not mathematically satisfying because it results in a recursive paradox that will lead you straight into a non-binary system and out of this universe. And the frustrating part of that is that you can only leave this universe in your imagination, and you still have a physical body. For most people, that's a real buzzkill.

[Anonymous No.9396449]

>>9395418
The equations I posted allow you to express dn as a function of x0 through xn. Then you would simply need to take n integrals with x0 through xn as the variables.

[Anonymous No.9396460]

>>9396022
He's a pseudointellectual with a bad case of Dunning-Krueger; he's not saying a damn thing.

[Anonymous No.9397141]

>>9396227
No, if the integral is from -arccos(d/2) to arccos(d/2), the factor is 1/(2pi), that is, the pdf for t.
When the integral is from 0 to arccos(d/2), the factor is 2/(2pi) = 1/pi. This is just using the fact that cos is even.

>>9396211
It is discontinuous. If you start at zero, no steps go out of the circle on the first step - they all go to the boundary.
Hence, G(0)=1+G(1)
If you start really close to zero, half will go out on the first step and half will go right next to the boundary. As d goes to 0, G(d) goes to (1+G(1))/2 + 1/2.

[Anonymous No.9397198]

>>9396460
Thanks for the explanation. I was both >>9396211 and >>9396227. I mistakenly put the factor of 2 there to try to match G(0) = 1 + G(1).

[Anonymous No.9397243]

>>9396460
No, try reading >>9396292 and proving it wrong. Explain how your statistical model predicts the physical world so well that there's no uncertainty. Or maybe you're using the Dunning-Krueger Effect (which isn't something you can "have a case of," if you actually understood what it means) as a crutch and misapplying it to something you actually haven't taken the time to fully process. It doesn't really matter, of course, but it's worth the consideration.

[Anonymous No.9397250]

>>9397198
Meant to reply to >>9397141.

[Anonymous No.9397255]

>>9397243
shut the fuck up

[Anonymous No.9397270]

>>9397255
Oh, that's a very compelling argument. Thanks for clarifying your position. Brilliant.

[Anonymous No.9398688]

>>9390861
>finite number of trials
As the number of trials tends towards infinity, so does the average number of steps.
To visualise, imagine going forth and back between the same two points forever.

[Anonymous No.9398694]

>>9398688
You really don't understand probability at all, do you.

[Anonymous No.9398695]

>>9398694
I don't get it.

[Anonymous No.9398697]

>>9398688
fucking idiot

[Anonymous No.9399651]

>>9391249
If /sci/ was actually like this, and not some IQ r/iamverysmart meme bullshit, this board would flourish.

[http://i.4cdn.org/sci/15143175(...) No.9399656]

>>9398688

The number of steps can go to infinity but it is increasingly improbable. The average number of steps will not.

[Anonymous No.9399784]

>>9396292
>I'm saying that the idea of an "expected value" is only an abstraction, and there is no way to apply it to the physical universe without violating the uncertainty principle.

You're going to shit the bed when you discover that the uncertainty principle is derived using expected values. Fuck off brainlet.

[Anonymous No.9400364]

>>9398695
Just because a distribution has infinite support (e.g. the random variable can take on arbitrarily large values), doesn't mean the mean of the distribution is infinite.

[Anonymous No.9400367]

1.75
mass/median is 2