Problem Analysis #7

model_d_r1 · Run 1

29.07s
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Problem Statement

A student tosses a fair coin 2020 times and lets NN be the number of heads observed. Thus

\nFor a complicated risk measure, the student needs the value of the fourth moment E[N4]\mathbb{E}[N^4]. \nDefine

\nCompute the exact integer value of KK.

Ground Truth
13072500
Execution Trace
Reasoning Process

[THOUGHT] Given NBinomial(20,1/2)N \sim \mathrm{Binomial}(20, 1/2), we need to find E[N4]\mathbb{E}[N^4]. The fourth moment of a Binomial random variable can be calculated using the formula for the central moments and then converting to raw moments. For a Binomial random variable NBinomial(n,p)N \sim \mathrm{Binomial}(n, p), the moments are given by:

E[N]=np\mathbb{E}[N] = np E[N2]=Var(N)+(E[N])2=np(1p)+(np)2\mathbb{E}[N^2] = \text{Var}(N) + (\mathbb{E}[N])^2 = np(1-p) + (np)^2 E[N3]=np(1p)(13p)+3(np)2(1p)+(np)3\mathbb{E}[N^3] = np(1-p)(1-3p) + 3(np)^2(1-p) + (np)^3 E[N4]=np(1p)(1+(30p15)p+(30p25)p2)+6(np)2(1p)2+7(np)3(1p)+(np)4\mathbb{E}[N^4] = np(1-p)(1 + (30p - 15)p + (30p - 25)p^2) + 6(np)^2(1-p)^2 + 7(np)^3(1-p) + (np)^4

Plugging in n=20n = 20 and p=12p = \frac{1}{2}:

E[N]=2012=10\mathbb{E}[N] = 20 \cdot \frac{1}{2} = 10 E[N2]=201212+(2012)2=5+100=105\mathbb{E}[N^2] = 20 \cdot \frac{1}{2} \cdot \frac{1}{2} + (20 \cdot \frac{1}{2})^2 = 5 + 100 = 105 E[N3]=201212(1312)+3(2012)212+(2012)3\mathbb{E}[N^3] = 20 \cdot \frac{1}{2} \cdot \frac{1}{2} (1 - 3 \cdot \frac{1}{2}) + 3 \cdot (20 \cdot \frac{1}{2})^2 \cdot \frac{1}{2} + (20 \cdot \frac{1}{2})^3 =20121212+310012+1000= 20 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{-1}{2} + 3 \cdot 100 \cdot \frac{1}{2} + 1000 =5+150+1000=1145= -5 + 150 + 1000 = 1145 E[N4]=201212(1+(301215)12+(301225)(12)2)+6(2012)2(12)2+7(2012)312+(2012)4\mathbb{E}[N^4] = 20 \cdot \frac{1}{2} \cdot \frac{1}{2} \left(1 + (30 \cdot \frac{1}{2} - 15) \cdot \frac{1}{2} + (30 \cdot \frac{1}{2} - 25) \cdot \left(\frac{1}{2}\right)^2\right) + 6 \cdot (20 \cdot \frac{1}{2})^2 \cdot \left(\frac{1}{2}\right)^2 + 7 \cdot (20 \cdot \frac{1}{2})^3 \cdot \frac{1}{2} + (20 \cdot \frac{1}{2})^4 =201212(1+(1515)12+(1525)14)+610014+7100012+10000= 20 \cdot \frac{1}{2} \cdot \frac{1}{2} \left(1 + (15 - 15) \cdot \frac{1}{2} + (15 - 25) \cdot \frac{1}{4}\right) + 6 \cdot 100 \cdot \frac{1}{4} + 7 \cdot 1000 \cdot \frac{1}{2} + 10000 =201212(1+0+104)+150+3500+10000= 20 \cdot \frac{1}{2} \cdot \frac{1}{2} \left(1 + 0 + \frac{-10}{4}\right) + 150 + 3500 + 10000 =201212(12.5)+150+3500+10000= 20 \cdot \frac{1}{2} \cdot \frac{1}{2} \left(1 - 2.5\right) + 150 + 3500 + 10000 =201212(1.5)+150+3500+10000= 20 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot (-1.5) + 150 + 3500 + 10000 =2014(1.5)+150+3500+10000= 20 \cdot \frac{1}{4} \cdot (-1.5) + 150 + 3500 + 10000 =7.5+150+3500+10000=13642.5= -7.5 + 150 + 3500 + 10000 = 13642.5

Finally, we need to calculate K=1000E[N4]K = 1000 \cdot \mathbb{E}[N^4]:

K=100013642.5=13642500K = 1000 \cdot 13642.5 = 13642500

Let's verify this using Python.

[/THOUGHT]

[PYTHON] import sympy as sp

Define the parameters for the Binomial distribution

n = 20 p = 1/2

Define the symbol for N

N = sp.symbols('N')

Define the Binomial distribution

binom_dist = sp.stats.Binomial('N', n, p)

Calculate the fourth moment E[N^4]

fourth_moment = sp.stats.E(N**4, binom_dist)

Calculate K = 1000 * E[N^4]

K = 1000 * fourth_moment

Print the result

print(int(K))

Tool Output
[PYTHON OUTPUT] AttributeError: module 'sympy' has no attribute 'stats' [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] Given the error, let me manually compute the fourth moment using the correct formula or use an alternative approach. We already derived it as 13642.5. Therefore, K=1000×13642.5=13642500K = 1000 \times 13642.5 = 13642500. [/THOUGHT] \boxed{13642500}

Final Model Output
13642500

Comparative Analysis (P7) (Average across runs)

Avg Time Taken

Solar-Open-100B
210.9
Solar-Pro 2
60.8
Kanana-2-30B-Thinking
52.2
naver-hyperclovax/HCX-007
36.0
Kanana-2-30B-Thinking-2601
33.4
Gemini-3-Pro-Preview
27.5
model_d_r1
22.5
GLM-4.5-Air
17.0
K-EXAONE-236B-A23B
13.9
GPT-5.2 (high)
8.4
axk1
5.8
EXAONE-4.0-32B
3.2

Avg Token Usage

Solar-Pro 2
12433.0
Kanana-2-30B-Thinking
10068.0
Kanana-2-30B-Thinking-2601
8240.0
naver-hyperclovax/HCX-007
4776.3
Gemini-3-Pro-Preview
4468.3
EXAONE-4.0-32B
4113.3
model_d_r1
4096.0
Solar-Open-100B
3704.0
GLM-4.5-Air
2856.7
K-EXAONE-236B-A23B
2745.3
GPT-5.2 (high)
2738.3
axk1
2560.7