Not homework. Winning (final) question in math competition.
If they’re sitting in a circle?
In a line?
As a big old radially symmetric bunch of birds in a space defined by the stacking of of chicken-standing-floor-area-normalized-to-neck-length-to-peck dimensionality?
That last bit, someone please help with how this is stated correctly, or ask for clarification.
Not a math person, but I’m assuming that the reasoning for “in a circle” being 25 is that “each chick has a 50% chance of being pecked from their left, and 50% chance of being pecked from their right, so only a 25% chance of being unpecked.” Similar application for the “in a line” except the two chicks on the ends only have a 50% chance of being pecked, rather than 75%, so its 25.5 expected non-peckees instead.
No idea what you mean by scenario #3.
That appears to be correct, but I wasn’t sure that reasoning was valid because the chicks’ pecked/unpecked statuses aren’t all independent of one another. So I ran a random computer simulation which convinced me that (your spoilered answer) is indeed correct.
I agree with magnusblitz reasoning.
I’m going to guess that the translation from Leo Bloomian into English for #3 is the chicks are arranged in a hexagonal grid. In that case, each interior chick has only a 1/6 chance of being unpecked. The ones on the edge have a better chance, depending on how many neighbors they have. The exact result will depend on exactly how the 100 are arranged, since 100 chicks don’t fit into a neat symmetrical hex grid.
They are not independent. For example, if chick n is unpecked, then it is certain that chicks n-2 and n+2 are pecked.
Presumably the answer still comes out to 25, but we lack a rigorous proof of that. I suspect that there’s a simple line of reasoning, but I haven’t come up with it yet.
Yes that’s along the lines of the case, but I hoped to generalize it, even though I started out with radial symmetry (of what order unspecified).
That is, take an overhead view of the chickens and calculate optimum space filling in a circle, or, for mathematical interest, other 2-D planes.
I don’t know the modulus for chickens for which a ring-spacing (for example) might go to keep chickens at unit 1, but, looking down on the ring: 1 chicken at the bull’s eye, ringed by 3, which are ringed by 5, etc.
The neck-length-peck criterion I added was to add a change from a 3-D packing problem–ie, with sphere packing, the “pecking opportunities/neighbors” would be the points at which the solids touch; chickens are not spheres, but I’m imagining the problem like so:
This bunch of chickens arranged arbitrarily on a barn floor can peck their neighbors in their lateral direction as well as between the shoulders (chickens have shoulders) to peck at a “neighbor” in a different ring (to stick with the radial overhead view).
And since chickens aren’t spheres, their “contact points” must be calculated as a discrete point on each of them.
Am I getting closer to explaining what I think might be interesting?
ETA: Generalizations to chickens-in-a-barn-in-microgravity packing would be a natural expansion of the problem.
The question only asked what is the expected (average) value of the number of unpecked chicks. So it makes no difference whether or not they are independent. The argument is completely rigorous.
In the line scenario, wouldn’t the second to last have a 100% probability of being pecked, since the last one only has one neighbor to peck?
Yes, my brain is creaking with rust. You can add expected values, you are right. Nothing else is needed.
Yes.
So 25 in that case as well.
Apropos of nothing, there are chicken blinders that can be used to reduce pecking.
And here’s a guy who sold them. https://www.youtube.com/watch?v=hEpSVCYok8s#t=4m
I read that as “chicken blenders” and thought “Yup, after you run them through the Chicken-o-Matic the answer will reduce to zero”.
Incidentally, here’s another cute fact: the probability that precisely x chickens go unpecked is the same as the probability that precisely 50 - x chickens go unpecked. And thus, by symmetry, the mean number of unpecked chickens is 25.
This symmetry works with the total number of chickens being 100. It doesn’t work analogously with all other possible total numbers of chickens. Can you figure out under what conditions it works and why?
To be clear, the mean number of unpecked chickens is n/4 for any n total chickens (n > 2), by the same “linearity of expectation” argument. It’s just that, for some n, we also get this interesting additional symmetry, which provides an alternate (less generalizable) route to the same answer, and at any rate, is interesting on its own, though it is not available for all n. See if you can find why this symmetry arises and when.
What if they’re “all adjacent to all?” Every chick has a chance of pecking every other chick. (Maybe they all promenade past each other, and, at some random point, peck.)
If I’m doing it right, the prob of not getting pecked is 49/50 ^ 50 – and that as the number of chicks rises, the limit of this prob turns to 1/e.
If I’m not doing it right, never mind!
Indistinguishable, if I’m reading you right, the interesting property of 100 is just that 100 is even. Certainly, the property you mention can’t apply to any odd number.
Chronos, you’re right that the condition you give is necessary, but it is not sufficient.
Really, the interesting question I should have just asked is, why do we have this symmetry for the 100 chicken problem? What explains it? And then, in discovering that explanation behind the symmetry, you will also discover what generality it does and doesn’t hold in.
Aw, no takers…
Does your symmetry
have to do with the fact that if I take a cyclic string of length 2k, say eg LLRRLLLL, and I count the number of times it alternates (t=2 in my example), then I flip every other direction (obtaining LRRLLRLR), the resulting string will obviously have 2k-t (=6) alternations? If so, your “interesting additional symmetry” is that n is divisible by 4, by reversing the pecking directions of every other pair of adjacent chicks. However, I do not think Chronos is mistaken that the probability of x unpecked chicks is equal to the probability of n/2-x chicks whenever n is even.
