Thread 16728100

You should be able to solve this.

>>16728100 (OP)
I hate statistics and probability

assuming the ducks can occupy the same space, (1/2)^4 * 2. next.

>>16728128
Good explanation. Also, you're wrong.

>>16728136
bad explanation. also, fuck you

>>16728100 (OP)
2 ducks 100% 3 ducks 75% infinity close to 0% im too tired to reason for 4 good night

>>16728145
Correct so far as it goes

>>16728136
I think he's right

>>16728128
The probability would be much higher than that.
Assuming the ducks are point-like, the probability at d=2 approaches 1 since the only case where you couldn't draw such a diameter is if the ducks both are lying directly on the same diameter.
In worst case scenario, where both ducks are arbitrarily close to said diameter but on opposite sides of the center, d=3 would be 1/2 (remember this is the worst case). Same probability for the 4th duck and so on...

So the chance of success is higher than 1/4 but by how much I can't be fucked to calculate.

>>16728100 (OP)
50%
It either happens or it doesn't

I'm not sure if my solution is correct but I tried it like that: assign labels 1,2,3,4 to the ducks. Let [math] \alpha [/math] be the small angle between ducks 1 and 2, and let [math] \beta [/math] be the small angle between ducks 3 and 4. Fixing the ducks 1,2, one sees that there is only a sector of angle [math] \alpha + \beta [/math] where duck 3 cannot be placed if the ducks should be on the same semicircle. Using the random variables [math] X = \alpha / \pi [/math], [math] Y = \beta / \pi [/math], the probability for having all ducks on the same semicircle, where [math] X=x; Y = y [/math] are fixed, is therefore given by [math] p(x,y) = 1 - \frac{x}{2} - \frac{y}{2} [/math]. Integrating [math] p(x,y) [/math] for [math] 0 \leq x,y \leq 1 [/math] results in the probability [math] \frac{1}{2} [/math].

duck1 is angle 0
duck2 is angle x
duck3 is angle y
duck4 is angle z
x,y and z must all be either between 0~180 or 180 and 360
2 ducks: x must be between 0~180 or 180~360, sums to 1
3 ducks: x & y must either be between 0~180 or 180~360, (0.5^2)*2=0.5
4 ducks: x, y & z must all be between 0~180 or 180~360, (0.5^3)*2 = 0.25

>>16728223
This is correct.
>>16728252
>x,y and z must all be either between 0~180 or 180 and 360
This is wrong.

Call the event in question [math]A[/math].
Pick a random duck, bisect the circle through that duck, and see whether all other ducks are in the half-circle that is counter-clockwise from the one you picked. Call this event [math]B[/math].
The probability of [math]B[/math] happening is [math]P(B) = 1/2^{N-1}[/math], since the individual events are independent. But it is also
[math]P(B) = \frac{P(A)}{N}[/math],
because [math]B[/math] happens exactly if [math]A[/math] happens and we then choose the single correct duck that bisects the circle.
Solving, the answer is
[math]P(A) = \frac{N}{2^{N-1}[/math].

>>16728223
This is correct.
>>16728252
>x,y and z must all be either between 0~180 or 180 and 360
This is wrong.

Call the event in question [math]A[/math].
Pick a random duck, bisect the circle through that duck, and see whether all other ducks are in the half-circle that is counter-clockwise from the one you picked. Call this event [math]B[/math].
The probability of [math]B[/math] happening is [math]P(B) = 1/2^{N-1}[/math], since the individual events are independent. But it is also
[math]P(B) = \frac{P(A)}{N}[/math],
because [math]B[/math] happens exactly if [math]A[/math] happens and we then choose the single correct duck that bisects the circle.
Solving, the answer is
[math]P(A) = \frac{N}{2^{N-1}}[/math].

>>16728278
Pretty slick solution, I like it.

>>16728278
Thinking about it again, there's not really a need to randomly pick a duck. Just let [math]C_i[/math] be the event "the [math]i[/math]-th duck can be used to bisect the circle as required."
Then again [math]P(C_i) = 1/2^{N-1}[/math], we have
[eqn]A = \bigcup_i C_i,[/eqn] and we can just add up the individual probabilities as
[eqn]A = \sum_{i} P(C_i) = \frac{N}{2^{N-1}}.[/eqn]
This is allowed because the [math]C_i[/math] are disjoint (you can use inclusion-exclusion if you really want to).

>>16728145
>2 ducks 100%
Even 1 duck isn't 100%, it can be right in the center. And with 2 ducks it would be impossible if they're aligned on a diameter.

if they can be at any point with equal probability, wouldn't it be infinite because you can form an infinite set of varying diameters along each semicircle that don't overlap?

>>16728100 (OP)
They're ducks.
100%.
>they think this is statistics

>>16728100 (OP)
I'm gonna say half pi

just sounds spiffy

>>16728318
https://en.m.wikipedia.org/wiki/Almost_surely

for a given circumference angle, the probability is 1/8. integrate that over all possible angles from 0 to pi.

>>16728100 (OP)
Duck my balls professor
>*blap*
too slow

>>16728100 (OP)
50%
1 duck = 100%
2 ducks you draw around duck one, and move the diameter to include duck 2 (always possible), so 100%
3 ducks? 50% chance it's in the initial 2-duck line, PLUS: you can move the first diameter through (180 - difference between first two ducks angle) which is a uniform distribution
So for first 2 ducks at 0 degrees ... 100% chance of being able to fit a third duck
90 degrees: 75%
179.999 degrees: 50%
So, P(3 ducks) = 75%
4 ducks:
Assume duck 1 is the furthest. Distance 12 and 13 then are both unform random vars in (1,180)
p(max_angle) = (max_angle/180)^2
average over 180 degrees to get average angular movement left, we get 60 degrees.
Hence, 75% x (180 + 60) / 360 = 50%


50% chance

>>16728100 (OP)
50%
Either it happens (A) or it doesn't (B).
/sci/, you're slipping...

>>16728456
2/3
Either it happens or it doesn't.

>>16728100 (OP)
1-(the probability that they can’t)

ez next question

>>16728456
>>16728180

if you can't answer it with bayes' theorem you're wasting hours of your life on this board

>>16728100 (OP)
[math]2\int_0^1 x\int_x^1(1-z)\,dz\dx + 4\int_0^1\int_x^1\int_y^1z \,dzdydx=7/12[/math]

>>16728100 (OP)
Draw a random diameter across a circle.
There is a 100% chance that Duck 1, if randomly placed, will be in the half with Duck 1 in it.
There is a 50% chance that Duck 2, if randomly placed, will be in the half of the circle with Duck 1 in it.
There is a 50% chance that Duck 3, if randomly placed, will be in the half of the circle with Duck 1 and 2 in it.
There is a 50% chance that Duck 4, if randomly placed, will be in the half of the circle with Ducks 1 thru 3 in it.
There is a 12.5% chance of finding all four ducks in the same half of the circle.

>>16728665
>>16728294

>>16728665
>There is a 50% chance that Duck 2, if randomly placed, will be in the half of the circle with Duck 1 in it.

Ran 50 different sets of trials with 30k groups of 4 ducks:
p = 0.3752 +/- 0.0029

Seems to be converging to about 3/8ths.

>>16728776
>using R
You might be a redneck. I mean retarded

>>16728776
10 million samples gives something close to 1/2. Other simulations also give results that follow n/2^(n-1):
https://godbolt.org/z/b7cWWrhcd

>>16728780
nta but, R is great for many statistical things. Many non-standard statistical procedure only have packages in R. The fact that you call somebody retarded for it suggests you never went deep enough into statistics for R to be the better option. Wasn't there also a basic error for years in a common statistical function in sklearn or numpy or something?

>>16728992
>the fact you call out retardation suggest you not retarded
Thank you?
>R has many good pqckages for women
Thank you for proving my point?

>>16728780
>using R
Excel

>>16728127
fpbp
>>16728128
pretty sure this is the correct answer. when you draw a line, there is a 1/2 chance that a duck ends up on a specific side of that line. Assuming each duck ends up on the "left" side, that is [math]1/2 \times 1/2 \times 1/2 \times 1/2[\math]. We can also account for the possibility all duck end up on the right side aswell. This leaves: [math](1/2 \times 1/2 \times 1/2 \times 1/2) \times 2 = 1/8 = 12.5%[\math] as this anon says.

>>16728433
>100% chance of being able to fit a third duck
It isn't. Imagine the ducks form an equilateral triangle around the center of the circle.

>>16729107
>for first 2 ducks at 0 degrees
i.e. assuming the first 2 ducks are in the same place

The formula is:

N * (N - 1) * integral_{0}^{1/2} x^(N-2) dx
= N / 2^(N - 1)

So for N=4 we have 4 / 8 = 1 / 2

>>16728100 (OP)
Well it depends, you are not countung the retardation factor of ducks.

>>16729007
You insult somebody for using R, I counter that by saying it has legit use cases and you just expose yourself as unknowing of those use cases.
Then you have nothing to say to that so you fall back to a weird comment where non-standard statistical procedures are for women somehow and women are retarded?

Class post mate.

>>16728318
Point-like ducks.

>>16728100 (OP)
the trick is to lure them to one side with bread pieces
the ducks like it too so everybody happi

This is probably a retarded answer but 1/16?

>>16728100 (OP)
Try this instead

ok so if we assume that the axis can pivot at will then the first two have a 100% chance of being on the same side, the second 75% taking into account the fact that the axis can always pivot so a 25% chance that it does not fall on the same side and the last 25% + 12.5% so 32.5%, adding the two and subtracting it from 100 there would be a 42.5% chance that they all fall on the same side

>>16729896

*the third

>>16729897

no fuck it's 25%+37.5% so 37.5% chance that it falls on the right side, it's 5am at my place

Why aren't these banned? It's thinly veiled homework threads. Is this a janny post? I didn't bump this.

>>16729985
Are you mad you don't know how to integrate?

>>16729985
Are you saying you can't solve it?
You should be able to.

Please explain to me why this isn't a simple 0.5^4 x 100% probability and how this meaningfully differs from flipping 4 coins

>>16730112
There are many diameter lines that cross through the center.

>>16730122
oh okay I get it

>>16730129
Do you?
Consider this, if all of the diameters had the odds of 1 in 2^n, then why would the ratio vary?

[eqn]P(N) = \frac{N}{2^{N-1}}[/eqn]
For [math]N=4[/math] ducks, you get:
[eqn]P(4) = \frac{4}{2^{4-1}} = \frac{4}{8} = \frac{1}{2}[/eqn]

>>16729811
50%

>>16730190
>P(2) = 1
And if one of the ducks is at the center of th circle?

>>16730194
If one duck were exactly at the center, it would lie on every diameter. Therefore, it could never be in a single semicircle, and the probability of success in that specific case would be 0.
But the problem states the ducks are at any point in the circle with equal probability, which implies a uniform distribution over a 2D area. The probability of a duck landing on any single point (which has zero area), including the exact center, is zero.
[eqn]P(\text{duck at center}) = \frac{\text{Area of center point}}{\text{Area of circle}} = \frac{0}{\pi r^2} = 0[/eqn]
It's a measure zero event. We can ignore it for the exact same reason we ignore the possibility of two ducks landing in the exact same spot. The radial position is irrelevant; only the angle matters.

>>16730200
>It's mathematically impossible a duck can be at the center of a circular pond
Lol and this is why scientists mock the field

>>16730203
>Misunderstands post he's making fun of
>Exposes lack of elementary probability knowledge
>"and this is why scientists mock the field"
Probability 0 does not equal impossible, in case you need it spelled out. I suggest you stick to the IQ and schizo poster threads.

>>16730236
>Probability 0 does not equal impossible
So it can happen, but with probability zero. Do you not understand how this contradicts itself?

>>16730243
>So it can happen, but with probability zero.
There is no contradiction here. You've decided to equate the layperson and imprecise meaning of probability with the mathematical and precise definition of probability. In this case, they disagree. If you want to have a discussion about language or if you have a suggestion of a different axiomization of probability, that is fine. But don't pretend to be doing anything else.

>>16728100 (OP)
the angle from the center is all that matters
change problem to picking points on the boundary uniformly
first two at some shortest arc t1 in (0, pi)
p1= dt1/pi
second two at some shortest arc t2 in (0,pi)
p2= dt2/pi

probability they are admissible = (2pi - t1 - t2)/2pi
you get that from looking at the offset angle you may place the second arc relative to the first. the second arc must not overlap with the antipole arc of the first arc.
Integrate
Final probability = 1/2

>>16730254
The the formal and precise meaning of probability as prescribed by mathematicians does not map to reality and is therefore not predictive. Not exactly the argument you want to make, kiddo

>>16730283
2n/2^n is obvious generalization
2n ways to choose n-1 consecutive wedges
2^n ways to make n duck/negaduck choices