In a mean field Schelling model, agents of type A are assigned neighborhoods of size
𝑘
=
8.
The fraction of type A agents among all occupied sites is
𝑝
𝐴
=
0.6
. An agent of type A is happy if at least 5 of its neighbors are also type A agents (i.e., the happiness threshold is
𝜆
𝐴
=
5
8
=
0.625

What is the probability
𝑃
happy
(
𝐴
)
that a rand omly chosen type A agent is happy?

Question

In a mean field Schelling model, agents of type A are assigned neighborhoods of size 
𝑘
=
8.
 The fraction of type A agents among all occupied sites is 
𝑝
𝐴
=
0.6
. An agent of type A is happy if at least 5 of its neighbors are also type A agents (i.e., the happiness threshold is 
𝜆
𝐴
=
5
8
=
0.625

What is the probability 
𝑃
happy
(
𝐴
)
that a rand omly chosen type A agent is happy?

BlackTom AI · Accepted Answer

First, let's restate the problem setup in my own words to frame what we are calculating. We have a mean-field Schelling model with neighborhood size k = 8 for type A agents, and a global fraction p_A = 0.6 of A-type agents among all occupied sites. An A-type agent is happy if at least 5 of its 8 neighbors are A-type, so the happiness threshold is 5 out of 8, i.e., λ_A = 5/8 = 0.625. The question asks for the probability that a randomly chosen A agent is happy, given these numbers.

Option analysis, option by option:
- Option: ∑_{i=5}^{8} (8 choose i) (0.6)^{i} (0.4)^{8−i}
  - This expression is a classic tail probability for a Binomial(n, p) distribution. Here, we consider the number of A-type neighbors among the 8 neighboring sites. If we let X ~ Binomial(n=8, p=0.6) represent the number of A neighbors, then P(X ≥ 5) is exactly the probability that at least 5 neighbors are A-type. This aligns perfectly with the happiness condition (requiring ≥5 A neighbors to be happy). The summation from i = 5 to 8 of the binomial probability mass function binom(8, i) (0.6)^i (0.4)^{8−i} is the correct expression for P(X ≥ 5).
  - In other words, this option directly computes the fraction of A-type agents whose neighbor count of A-type neighbors meets or exceeds the threshold, using the binomial model with p = 0.6 and n = 8. There is no extra factor or alteration needed because we are conditioning on an agent being type A; the internal randomness is entirely in the composition of its 8 neighbors, each neighbor independently being A with probability 0.6 (given the mean-field assumption).
  - A potential conceptual pitfall would be to try to use a normal approximation or a different distribution, but since n = 8 is small, the exact binomial tail sum is the correct and straightforward method. Another common mistake would be to use P(X ≥ 5) with p = 0.4 (the probability that a neighbor is not A), which would be inappropriate here since we are counting A neighbors, not non-A neighbors.
  - The structure of the expression mirrors the combinatorial counting: for each possible number i of A neighbors from 5 to 8, there are binom(8, i) ways to choose which neighbors are A, each such configuration has probability (0.6)^i (0.4)^{8−i}, and summing over i from 5 to 8 aggregates all configurations that make the agent happy.

Why other hypothetical alternatives would be wrong (if they appeared):
- If someone wrote P(X ≥ 5) with p = 0.4, that would treat the probability of a neighbor being non-A as the success probability, which would be conceptually misaligned with the event of interest (number of A neighbors, not non-A neighbors).
- If someone used P(X ≥ 5) with n ≠ 8, or integrated over a continuous distribution instead of the discrete binomial, that would fail to capture the exact combinatorial structure of the 8 fixed neighbor slots.
- If an option attempted to compute P(X = 5) or P(X ≤ 5) instead of P(X ≥ 5), it would describe a different event than the happiness condition requires (the event “at least 5 A neighbors” vs. “exactly 5” or “at most 5”).

In summary, the listed expression exactly represents the probability that a randomly chosen A agent has at least 5 A neighbors among 8, under a binomial model with p = 0.6 for each neighbor being A. It integrates the combinatorial counts with the correct per-neighbor success probability to yield P(happy | A).

5284AGBM6Y Agent-Based Modeling Practice Quiz

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