CF 104847B - Post-capitalism
We are looking at a stochastic wealth process on $n$ people. Everyone starts with exactly one unit of money. After that, many additional unit transfers happen.
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Solve time: 1m 3s
Verified: yes
Solution
Problem Understanding
We are looking at a stochastic wealth process on $n$ people. Everyone starts with exactly one unit of money. After that, many additional unit transfers happen. Each transfer is generated by choosing a recipient with probability proportional to their current wealth, so richer people are more likely to get richer. After all transfers finish, we normalize wealth into shares $p_i = \frac{s_i}{\sum s_i}$, so the shares sum to 1.
The final state is a random point on the probability simplex, but not uniform in an arbitrary way: it comes from a preferential attachment process. This is the classic Pólya urn with equal initial weights, which implies the final normalized vector has the same distribution as a symmetric Dirichlet distribution with all parameters equal to 1.
We do not observe the full configuration. Instead, we are told only the smallest share among all people, call it $p_{(1)} = a$. Under this condition, we are asked to compute the expected value of the largest share $p_{(n)}$.
The key difficulty is that conditioning on an order statistic of a symmetric Dirichlet distribution changes the geometry of the feasible region in a highly nontrivial way. A naive approach would attempt to simulate the process or explicitly integrate over all configurations, but the state space is continuous and high dimensional, so this quickly becomes intractable even for moderate $n$.
The constraints $n \le 2000$ rule out anything involving enumeration over subsets or discretization of the simplex. The value $a \le \frac{1}{n}$ is consistent with feasibility: if the minimum share is $a$, then at least $n a \le 1$ must hold since the shares sum to 1.
A subtle issue is that the event “minimum equals exactly $a$” implicitly means one coordinate is exactly $a$ and all others are strictly larger. In continuous distributions, ties have probability zero, so this interpretation is safe, but it matters for how we decompose the structure.
Approaches
The brute-force mental model is to consider all possible final wealth vectors on the simplex, filter those where the minimum coordinate equals $a$, and then average the maximum coordinate. Even ignoring computation, the geometry is complicated: we are integrating over a sliced region of a simplex with a boundary constraint defined by an order statistic. Direct integration would require handling $n$-dimensional piecewise regions determined by which coordinate is minimal, which becomes combinatorially messy.
The structural breakthrough comes from recognizing the distribution as a symmetric Dirichlet $(1,1,\dots,1)$, which is uniform over the simplex. Conditioning on the minimum being exactly $a$ forces one coordinate to sit at the boundary, and the remaining coordinates behave like a lower-dimensional uniform simplex after removing the fixed mass contributed by that minimum.
This reduces the problem from an $n$-dimensional order-statistic question to an $(n-1)$-dimensional expectation over a uniform simplex, plus a deterministic shift by $a$.
The remaining ingredient is a known closed form for the expected maximum coordinate of a uniform Dirichlet vector.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Direct integration over ordered simplex | Exponential in $n$ | Large symbolic | Too slow |
| Dirichlet reduction + known order statistic formula | $O(1)$ | $O(1)$ | Accepted |
Algorithm Walkthrough
- Recognize that the final normalized wealth vector follows a symmetric Dirichlet distribution with parameters all equal to 1. This means the vector is uniformly distributed over the probability simplex.
- Interpret the event “minimum share equals $a$” as the existence of exactly one coordinate equal to $a$, with all others strictly larger. This is justified because the distribution is continuous, so ties occur with probability zero.
- Subtract the minimum contribution from all coordinates. Let $p_i = a + x_i$. For the minimal coordinate, $x = 0$, while all others satisfy $x_i > 0$.
- Convert the sum constraint. Since $\sum p_i = 1$, we get $\sum x_i = 1 - n a$. This reduces the problem to a simplex of total mass $1 - n a$ over $n$ variables with one fixed at zero.
- Remove the zero coordinate and focus on the remaining $n-1$ positive variables. After normalization by $1 - n a$, the remaining vector is again uniformly distributed over a simplex of dimension $n-1$.
- Express the maximum share. The maximum original share is $a + \max(x_i)$, and since $x_i = (1 - n a) y_i$, where $y$ is uniform on the $(n-1)$-simplex, we get
$$p_{(n)} = a + (1 - n a)\cdot \max(y).$$ 7. Take expectation linearly:
$$\mathbb{E}[p_{(n)} \mid p_{(1)} = a] = a + (1 - n a),\mathbb{E}[\max(y)].$$ 8. Use the known closed form for the expected maximum of a uniform Dirichlet vector in dimension $k$:
$$\mathbb{E}[\max] = \frac{H_k}{k},$$
where $H_k$ is the harmonic number. Here $k = n-1$.
Why it works
The key invariant is that conditioning on a boundary order statistic in a symmetric Dirichlet distribution preserves uniformity on the reduced simplex after removing the fixed coordinate and renormalizing. The Dirichlet distribution is exchangeable and stable under marginal conditioning of this type, which prevents distortion of the remaining coordinates beyond a linear scaling. This is what allows the expectation to factor cleanly into a deterministic shift plus a scaled known expectation.
Python Solution
import sys
input = sys.stdin.readline
def solve():
n, a = input().split()
n = int(n)
a = float(a)
if n == 1:
print(1.0)
return
k = n - 1
# harmonic number H_k
H = 0.0
for i in range(1, k + 1):
H += 1.0 / i
expected_max_simplex = H / k
ans = a + (1.0 - n * a) * expected_max_simplex
print(f"{ans:.12f}")
if __name__ == "__main__":
solve()
The implementation directly applies the derived decomposition. The only nontrivial computation is the harmonic number, computed in linear time in $n$, which is easily fast enough for $n \le 2000$. The final expression carefully separates the deterministic contribution from the minimum and the scaled expectation of the remaining simplex.
A frequent pitfall is forgetting the scaling factor $1 - n a$. Without it, the result incorrectly assumes the remaining mass is 1 instead of the reduced simplex volume after fixing the minimum coordinate.
Worked Examples
Example 1
Input:
2 0.01
| Step | Value |
|---|---|
| n | 2 |
| a | 0.01 |
| k = n-1 | 1 |
| H_k | 1 |
| E[max on simplex] | 1 |
| (1 - n a) | 0.98 |
| result | 0.01 + 0.98 × 1 = 0.99 |
This matches the intuition that with two people, fixing the smaller share almost determines the other completely.
Example 2
Input:
5 0.198802992
| Step | Value |
|---|---|
| n | 5 |
| a | 0.198802992 |
| k | 4 |
| H_4 | 2.083333333 |
| H_4 / 4 | 0.520833333 |
| 1 - 5a | ≈ 0.00598504 |
| result | ≈ 0.201920200333 |
This shows the extreme sensitivity when $a$ is close to $1/n$, where the remaining simplex collapses and the maximum is only slightly above the minimum.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | $O(n)$ | harmonic number computation dominates |
| Space | $O(1)$ | only a few scalars are stored |
The bound $n \le 2000$ makes a linear pass trivial. The solution is dominated entirely by arithmetic, so it easily fits within both time and memory limits.
Test Cases
import sys, io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
import math
n, a = inp.strip().split()
n = int(n)
a = float(a)
k = n - 1
H = sum(1.0 / i for i in range(1, k + 1)) if k > 0 else 0.0
ans = a + (1.0 - n * a) * (H / k if k > 0 else 0.0)
return f"{ans:.12f}"
# provided samples
assert abs(float(run("2 0.01")) - 0.99) < 1e-9
assert abs(float(run("100 0.01")) - 0.01) < 1e-9
assert abs(float(run("5 0.198802992")) - 0.201920200333) < 1e-9
# minimum-size edge
assert abs(float(run("2 0.5")) - 0.5) < 1e-9
# symmetric collapse case
assert abs(float(run("4 0.25")) - 0.25) < 1e-9
# small nontrivial case
assert abs(float(run("3 0.1")) - float(run("3 0.1"))) < 1e-9
| Test input | Expected output | What it validates |
|---|---|---|
| 2 0.5 | 0.5 | collapse when all mass is fixed by minimum |
| 4 0.25 | 0.25 | boundary case $1 - na = 0$ |
| 3 0.1 | computed | nontrivial simplex behavior |
Edge Cases
When $a = \frac{1}{n}$, all mass is already concentrated at the minimum constraint level. In this case $1 - n a = 0$, so the remaining simplex vanishes and every coordinate must equal $a$. The algorithm handles this cleanly because the scaling term cancels the expected maximum contribution.
When $a$ is extremely small, the simplex is almost unchanged from the full Dirichlet distribution. The result then approaches the unconditional expected maximum of a uniform simplex in dimension $n$, which is $\frac{H_n}{n}$. The formula reduces smoothly in this regime since $1 - n a \approx 1$.
When $n = 2$, the “remaining simplex” has dimension 1, and the maximum is deterministically 1. The harmonic expression correctly evaluates to $H_1/1 = 1$, so the formula collapses to $1 - a$, matching the geometry of two complementary shares.