/sci/ - Science & Math

[Anonymous No.7207546]

Conjecture: There is not a single person on /sci/ who can solve this.

if X, Y and Z are randomly placed in [0,1], what is the probability that they are at least a apart form each other
|X-Y|>a and |Y-Z|>a and |X-Z|>a

The answer is (1-2a)^3, I solved this with monte carlo simulation, and I am not bad at probability. So, any of you fuckers can solve this question properly?

[Anonymous No.7207558]

Claim Numero 2: If there were 4 randomly placed thingies, the probability would have been (1-3a)^4
or (1- (n-1)a )^n

[Anonymous No.7207721]

It was easy with 2 but I don't know where to go from there.

[Anonymous No.7207729]

>>7207721
(1- (n-1)a )^n
this applies for 2, so (1-a)^2, i can solve that one properly
the 3 point version, seems quite a bit more difficult

[Anonymous No.7208582]

I mean, this is a very difficult question but not one person on /sci/ can solve it?
cmoon

[Fever No.7208598]

>>7207546
Surely you mean |X-Y|>=a? At least that's the wording "at least" implies.

[Anonymous No.7208604]

>>7208598
it doesnt matter, the result would be the same

[Anonymous No.7208753]

>>7208582
If |X-Y|=a then the probability is (1-2a)^2, that seems like it should be relevant.

[Anonymous No.7208833]

>>7207546
All you have to do is find the volume of
{(x,y,z) \in [0,1]^3 : |x-y|>a, |y-z|>a, |x-z|>a }

[Anonymous No.7208838]

>>7208833
Yes, that is the problem but how do you solve it?

[Anonymous No.7208848]

>>7208838

triple integral. child's play.

[Anonymous No.7208864]

>>7207546
>(1-2a)^3

I'd be quite surprised if this is right. Isn't it going to be a compound function... a polynomial if a is in some range, another polynomial if a is in another range... etc.

[Anonymous No.7208893]

<span class="math">P(|X-Y|>a, |Y-Z|>a, |X-Z|>a)[/spoiler]
<span class="math">=6 \cdot P(|X-Y|>a, |Y-Z|>a, |X-Z|>a, X > Y > Z)[/spoiler]
<span class="math">=6 \cdot P(X > Y + a > Z + 2a)[/spoiler]
<span class="math">=6 \int_{0}^{1-2a} \int_{z+a}^{1-a} \int_{y+a}^1 dxdydz[/spoiler]
<span class="math">=(1-2a)^3[/spoiler]

[Anonymous No.7208914]

>>7208893
what is this 6 ?

[Anonymous No.7208917]

>>7207546

not 100% certain been a while since I did this
choose x>y>z, originally 6 ways to order this so multiply result by 6. this choice means
x-a>y-a>z and x-a>y which results in

<span class="math">6 \cdot \int_{0}^{1}dx\int_{0}^{x-a}dy\int_{0}^{y-a}dz=(1-2a)^{3}[/spoiler]

[Anonymous No.7208922]

>>7208917
I meant <span class="math">(1-3a)^{2}[/spoiler] so it's not what you got and probably wrong
>>7208914
that's where that anon got his 6 though

[Anonymous No.7208923]

>>7208914
factorial(3)

[Anonymous No.7208929]

>if X, Y and Z are randomly placed in [0,1]

Randomly according to what law ? Do they even have the same law ? If we assume they all have uniform law, are they independant ? If X=Y there is pretty little chance they're going to be apart from at least a>0.

So let's assume (X,Y,Z) is a 3-sample with uniform law. The probability will depend on wether, say, X is at distance at least a from 0 (or 1). So you have a different case for each d<a where d is min(d(X,0),d(X,1)). You know the probability of this happening for each case. Then, in each case, you know the probability of something being at at least a from X, that gives you the probability of |X-Y|>a. Then you have (again) to make a case distinction depending on how Y is close to 0 and 1.

You simulation can only take into account one case, OP. For instance if X is in [0,a/2], the probability of |X-Y|>a is 1-(3a)/2, but if X is neither in [0,a] nor in [1-a,1], then the probability is (1-2a). Only "difficult" thing is minding those bounds problems.

[Anonymous No.7208936]

>>7208893
are the integrals in the right order wrt to dxdydz ?

[Anonymous No.7208939]

>>7208893
Nice job, but you're assuming a is rather small. (a<1/2 at least).

[Anonymous No.7208941]

>>7208939
It is impossible if a > 1/2.

[Anonymous No.7208944]

>>7208939
If a is > 1/2 the probability is 0

[Anonymous No.7208986]

Geometric solution:
|x-y| > a
if y =< x then x-y > a
y < x-a
if y > x then y-x > a
y > x+a

These linear inequalities plotted on each plane form three trirectangular tetrahedrons with leg length 1-a

The intersection of these planes forms 6 smaller trirectangular tetrahedrons with leg length 1-2a

The area of these divided by the total area of the 1x1x1 cube is 6((1-2a)^3/6)/1 = (1-2a)^3

QED

[Anonymous No.7208990]

Geometric solution

|x-y| > a
if y =< x then x-y > a
y < x-a
if y > x then y-x > a
y > x+a

These linear inequalities plotted on each plane form three trirectangular tetrahedrons with leg length 1-a

The intersection of these planes forms 6 smaller trirectangular tetrahedrons with leg length 1-2a

The area of a trirectangular tetrahedron is the product of the legs divided by 6

The area of these divided by the total area of the 1x1x1 cube is 6((1-2a)^3/6)/1 = (1-2a)^3

QED

[Anonymous No.7209080]

>>7208893
thats nice as fuck
I could have found the ordered probability
I didn't think of it

>>7208929
simulation definitely works

[Anonymous No.7209087]

>>7207546
>Conjecture: There is not a single person on /sci/ who can solve this.
Proven wrong

[Anonymous No.7209089]

>>7209087
the funny thing is, if the question a simple conditional probability coin toss question, 90% of the answers would have been wrong.

[Anonymous No.7209093]

>>7209089
The apparent difficulty of the question has selective pressure