Problem Analysis #5

133.86s
23,022 toks
Problem Statement

Consider the following procedure that generates a sequence of random variables that take the value 00 or 11. For an integer n1n \ge 1, we denote the nn-th random variable of a sequence generated by the procedure as XnX_n.

  • X1X_1 becomes X1=0X_1 = 0 with probability 23\dfrac{2}{3} and X1=1X_1 = 1 with probability 13\dfrac{1}{3}.

  • For integers n=1,2,n = 1,2,\ldots in order, the following is repeated until the procedure terminates:

  • The procedure terminates with probability pp (0<p<10 < p < 1) if Xn=0X_n = 0, and with probability qq (0<q<10 < q < 1) if Xn=1X_n = 1. Here pp and qq are fixed constants.

  • If the procedure does not terminate at step nn, then Xn+1X_{n+1} becomes 00 with probability 23\dfrac{2}{3} and 11 with probability 13\dfrac{1}{3}.

When the procedure terminates at n=n = \ell, a sequence of length \ell, composed of random variables (X1,,X)(X_1,\ldots, X_\ell), is generated, and no further random variables are generated.\subsection*{I.}For an integer k1k \ge 1, consider the matrixPk=(Pr(Xn+k=0Xn=0)Pr(Xn+k=1Xn=0)Pr(Xn+k=0Xn=1)Pr(Xn+k=1Xn=1)).P_k =\begin{pmatrix}\Pr(X_{n+k} = 0 \mid X_n = 0) & \Pr(X_{n+k} = 1 \mid X_n = 0) \\\Pr(X_{n+k} = 0 \mid X_n = 1) & \Pr(X_{n+k} = 1 \mid X_n = 1)\end{pmatrix}.

  • [(1)] Express P1P_1 and P2P_2 in terms of pp and qq.

  • [(2)] Express P3P_3 using P1P_1.

  • [(3)] The matrix PkP_k can be expressed in the form Pk=γkP1P_k = \gamma_k P_1 for a real number γk\gamma_k. Find γk\gamma_k. \subsection*{II.}For an integer m2m \ge 2, find the respective probabilities that Xm=0X_m = 0 and Xm=1X_m = 1, given that the procedure does not terminate before n=mn = m.\subsection*{III.}Let \ell be the length of the sequence generated by the procedure. Find the expected value and the variance of \ell. If necessary, you may usem=1mrm1=1(1r)2,m=1m2rm1=1+r(1r)3\sum_{m=1}^{\infty} m r^{m-1} = \frac{1}{(1-r)^2},\qquad\sum_{m=1}^{\infty} m^2 r^{m-1} = \frac{1+r}{(1-r)^3}for a real number rr with r<1|r| < 1.\subsection*{IV.}For an integer k1k \ge 1, find the probability Pr(Xn=0Xn+k=1)\Pr(X_n = 0 \mid X_{n+k} = 1).\subsection*{V. Final quantity}In this part, assume p=12p = \dfrac{1}{2} and q=13q = \dfrac{1}{3}.

  • Let α\alpha be the probability in {\rm IV} for these values of pp and qq: α:=Pr(Xn=0Xn+k=1). \alpha := \Pr(X_n = 0 \mid X_{n+k} = 1).

  • Let β\beta be the expected value of \ell obtained in {\rm III}, and let σ2\sigma^2 be the variance of \ell in {\rm III}.

  • In {\rm I(3)}, P2P_2 can be written as P2=γ2P1P_2 = \gamma_2 P_1. Let λ:=γ2 \lambda := \gamma_2 for p=12p = \dfrac{1}{2} and q=13q = \dfrac{1}{3}. DefineK=13320(αβ+σ2λ)+7.\mathcal{K}= 13 \cdot 320\left( \alpha \beta + \frac{\sigma^2}{\lambda} \right) + 7.Compute the integer K\mathcal{K}.\

Ground Truth
26683
Execution Trace
Reasoning Process

[THOUGHT]
We need to compute the quantities required for part V:

  • (p=\frac12,;q=\frac13).
  • Define (a=\frac23(1-p)+\frac13(1-q)).
  • (\alpha=\Pr(X_n=0\mid X_{n+k}=1)=\frac{\frac23(1-p)}{a}).
  • (r=\frac23p+\frac13q) is the unconditional probability that the procedure terminates at a given step.
  • (\beta=E[\ell]=\frac1r) and (\sigma^2=\operatorname{Var}(\ell)=\frac{1-r}{r^2}).
  • From part I(3) we have (\gamma_k=1) for all (k), so (\lambda=\gamma_2=1).

Finally evaluate

[ K=13\cdot320\Bigl(\alpha\beta+\frac{\sigma^2}{\lambda}\Bigr)+7 . ]

All calculations are performed with exact rational arithmetic using SymPy.
[/THOUGHT]
[PYTHON]
import sympy as sp

probabilities p and q

p = sp.Rational(1, 2)
q = sp.Rational(1, 3)

a = (2/3)(1-p) + (1/3)(1-q)

a = sp.Rational(2, 3)(1 - p) + sp.Rational(1, 3)(1 - q)

α = ( (2/3)*(1-p) ) / a

alpha = sp.Rational(2, 3)*(1 - p) / a

r = (2/3)*p + (1/3)*q

r = sp.Rational(2, 3)*p + sp.Rational(1, 3)*q

β = expected length = 1/r

beta = 1 / r

σ² = variance = (1-r)/r²

sigma2 = (1 - r) / r**2

λ = γ₂ = 1 (since P_k = P_1 for all k)

lam = 1

compute K

K = 13 * 320 * (alpha * beta + sigma2 / lam) + 7
K_int = sp.simplify(K)

print(K_int)
[/PYTHON]

[THOUGHT]
The Python computation yields (K = 17323).
[/THOUGHT]
\boxed{17323}

Tool Output
[PYTHON OUTPUT] 17323 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The Python computation confirms that the value of K is 17323. [/THOUGHT] \boxed{17323}

Final Model Output
17323

Comparative Analysis (P5) (Average across runs)

Avg Time Taken

K-EXAONE-236B-A23B
802.4
GPT-5.1 (high)
467.8
Llama-VARCO-8B-Instruct
340.9
Solar-Pro-2 (31B)(high)
321.5
Deepseek-V3.2
272.4
Gemini-3-Pro-Preview
246.2
Kanana-2-30B-Thinking
234.2
K-EXAONE-236B-A23B
201.8
A.X-4.0 (72B)
191.5
Kanana-2-30B-Thinking-2601
171.1
Solar-Open-100B
133.9
GPT-5.2 (high)
78.5
Grok-4.1-fast
53.5
Claude-Opus-4.5
42.6
EXAONE-4.0.1-32B (high)
20.3
HCX-007(high)
18.8
axk1
13.1

Avg Token Usage

K-EXAONE-236B-A23B
71914.3
Kanana-2-30B-Thinking-2601
36339.0
Kanana-2-30B-Thinking
33424.0
K-EXAONE-236B-A23B
26257.0
Solar-Pro-2 (31B)(high)
23391.0
Solar-Open-100B
23022.0
A.X-4.0 (72B)
20243.0
GPT-5.1 (high)
19161.0
Deepseek-V3.2
16929.0
Llama-VARCO-8B-Instruct
16067.0
Claude-Opus-4.5
14573.0
Gemini-3-Pro-Preview
14044.0
Grok-4.1-fast
13647.0
GPT-5.2 (high)
8615.0
EXAONE-4.0.1-32B (high)
6754.0
HCX-007(high)
5076.0
axk1
4852.0