Are you smarter than a 13y/o?

>At most there can be 2 unpecked chicks
Wrong.

Let [math]\rightarrow[/math] denote a chick that pecks right and let [math]\leftarrow[/math] denote a chick that pecks left.

Now, we can build a structure like this: [math]X = \rightarrow \rightarrow \leftarrow \leftarrow \leftarrow[/math] where we have 5 chicks and 2 are unpecked.

Now, [math]\underbrace{X \ldots X}_{20}[/math] has 100 chicks and 40 of them are unpecked. Note that chicks in different X's never peck each other.

>There you go. This took me half a second, thus making me smarter than

At MOST a third could be unpecked, as every third chick happens to be between two that are pecking the other way.

Wrong, using this but instead with 4 chicks we can have 2 out of 4 being unpecked as long as the amount of chicks is a multiple of 4.

does being pecked on the beak count as a peck

Here's what I came up with.

What if there are 3 chicks in a ring?

On average 0.75 chicks get pecked.

the answer may well be 1/4 but the logic here seems to be faulty:

Yes, the chance for 1 chick to get pecked can be computed, but if you have circle, the chances are correlated. If you're a chick and the chick on the left pecks away from you, then this says something about the chick two left from you.

>If you're a chick and the chick on the left pecks away from you, then this says something about the chick two left from you.
no it doesn't

Yeah it does. But it doesn't matter so far as calculating expectations goes.

What the hell no it doesn't. They're independent identically distributed random variables.

they're not independent you dense brainlet. if the chicks are in line ABC and B pecks A then we know B doesn't peck C

It does for n=3

1

Isnt this just a hairy ball?

>if the chicks are in line ABC and B pecks A then we know B doesn't peck C
>if the first die doesn't roll a 6 then the first die rolls a 1, 2, 3,4 or 5
thanks for this wonderful insight

>chicks are sitting peacefully
>one chick violates the NAP, pecking the chick next to it
>this causes a the precariously fragile web of alliances and treaties to unravel in a violent, thrashing display of both defense of honor and political upheavals
>backs are stabbed, promises forgotten, feathers plucked
>in the wake of this devastation, how many innocent souls can be expected to crawl from the ashes, weary to rebuild the house of cards that is society?
0. The system works.

25

Let [math]X_i = 1 [/math] if the [math]i[/math]-th chick gets pecked and [math]0[/math] otherwise.
Then [math]P(X_i = 1) = \frac{1}{4} [/math] for all [math] i [/math]. So
[eqn] E\left[ \sum_{i=1}^{100} X_i \right] = \sum_{i=1}^{100} E[X_i] = \sum_{i=1}^{100} P(X_i = 1) = \sum_{i=1}^{100} \frac{1}{4} = 25 [/eqn]

>p(xi = 1) = 1/4
Show your work. This isn't obvious for example for n=2 or n=3.

>n=2 or n=3
There are 100 chicks.
All you need is that the i-1-th chick picks left and the i+1-th chick picks right. Both are independent events with an implied 50% chance.

it's actually obvious for n=3 I don't know what I was thinking

You messed up pecked and unpecked chickens

25

13 yo eternally BTFOed

Let me extend this answer to make it formally correct.
Define [math]\left ( \Omega = \left \{ L,R \right \}^{100}, \mathbb{P}\left ( \Omega \right ), \text{Unif}(\Omega )\right )[/math] as the probability space.
Then, [math]X_i = \left\{\begin{matrix}
0 & \omega _{i-1} = R \: \vee \omega _{i+1} = L\\
1 & else
\end{matrix}\right.[/math].
As there are , for a given set of adjacent chickens, 4 total possible combinations of L and R and only one results in no peck, [math]\forall i \:\text{P}[X_i = 1] = 0.25[/math], and thus your result follows.

25
Unless the question is retarded and we have to assume that two chicks cant peck each other at the same time as their mouths would meet or something making the answer 0.

to be unpecked in a circle where all chicks are facing inwards, the chick to the left must peck to its left while the chick to the right must peck to its right, the chances of this are 1/4

100/4=25

0.5 chance to get pecked by each side. 0.25 chance for each to be unpecked. Overall expected 25 unpecked chickens

The 25posters are correct.

Nowhere in the problem does it say that p = 0.5

0, because there is no time limit. Eventually, every chick will be pecked.

Let's make it more interesting:

1. Add a third "do-nothing" state - the chicken pecks neither side. (All three states occur with equal probability.)

2. Any unguarded peck is fatal, but mutual pecking cancels out. (Pecks occur simultaneously, so a fatally pecked chicken can still also deliver a fatal peck.)

What is the expected number of living chickens at the end?

Less than 1 second. Who was he, the fucking flash? It takes longer than a second to read it.

It was a quiz show and the time only started counting after the question was fully read.

>25%

You are not adding the fact that if a chick gets unpecked it necessarily means that another chick gets pecked, its not indepedent experiments so the probability of getting pecked is more

Do I start filling out a probability transition matrix ? Do I Markov Chain here ?

The events aren't independent. If [math]X_i = 0[/math], then [math]X_{i-2}[/math] and [math]X_{i+2}[/math] are 1

That doesn't matter for the expected value.
It would if you were to specify the distribution of unpecked chickens, but that's not what the question is asking.

I wonder what distribution this follows.

but we don't know anything about which bird C pecks or whether C was pecked by D

If all chicks pecks to the right, then 100% of chick get pecked.

If all chicks pecked alternately, they'll still all be pecked.

So there is zero unpecked chicks.

Hmm... I don't think that's quite right.

The surviving chickens from Out[5] should be 4, 5, 6, 7, 9, 10 based on the rules. That doesn't match up with Out[7].

Whoops, forgot some cases. I think this is right. Also forgot to change it to 100 chicks for the histogram part.

it's necessarily bounded, so that limits it a bit. the final value is a result of summing repeated bernoulli trials, so maybe a compound poisson? or some type of beta distribution?

>summing repeated bernoulli trials
So a binomial.

bleh you're right, it's been a while since my stats class

So what was the actual answer and reasoning in the end?