---
title: "Poker test, 1000 four-digit numbers (2073)"
subject: "Simulation and Modeling"
chapter: 6
key: "semester_8/simulation_and_modeling:poker-test-numerical-2073-magh"
source: https://www.bctengineer.com/q/semester_8/simulation_and_modeling/poker-test-numerical-2073-magh
site: BCT Engineering
---

# Poker test, 1000 four-digit numbers (2073)

Simulation and Modeling · Chapter 6: Random Number

- Asked in: 2073 Magh, 8 marks
- Video: [Poker Test (4 digits) | Random Number Independency | Simulation & Modelling | in Nepali](https://www.youtube.com/watch?v=BNIt43dxQaY) — Monkey D. Luffy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/simulation_and_modeling/poker-test-numerical-2073-magh
- Whole chapter: https://www.bctengineer.com/md/semester_8/simulation_and_modeling/chapter-6

## Question

A sequence of 1000 four digit numbers has been generated and an analysis indicates the following combination and frequencies:

| Combination (i) | Observed Frequency (Oi) |
|---|---|
| Four different digit | 520 |
| One pair | 390 |
| Two pair | 55 |
| Three like digit | 34 |
| Four like digit | 1 |
|  | 1000 |

## Short answer

Using probabilities for four‑digit numbers (all different 0.504, one pair 0.432, two pairs 0.027, three like 0.036, four like 0.001), χ0² = 33.74. With df=4, critical value at α=0.05 is 9.488. Since 33.74 > 9.488, independence is rejected.

## Answer

**Setup.** H0: the numbers are independent. H1: they are not. N = 1000, five patterns for four-digit numbers, so df = 5 − 1 = 4. The question gives no α; take α = 0.05, so χ²0.05,4 = 9.488.

For four-digit numbers the probabilities are: four different 0.504, one pair 0.432, two pairs 0.027, three of a kind 0.036, four of a kind 0.001.

| Pattern | Probability | Oi | Ei | (Oi − Ei)²/Ei |
|---|---|---|---|---|
| Four different | 0.504 | 520 | 504 | 0.508 |
| One pair | 0.432 | 390 | 432 | 4.083 |
| Two pairs | 0.027 | 55 | 27 | 29.037 |
| Three of a kind | 0.036 | 34 | 36 | 0.111 |
| Four of a kind | 0.001 | 1 | 1 | 0.000 |

**χ0² = 33.74**

**Conclusion.** Since 33.74 > 9.488, H0 is **rejected**; the numbers are not independent. The main cause is "two pairs": 55 observed against 27 expected (contribution 29.04). Merging the last two classes (Ei = 37, df = 3, critical 7.815) gives χ0² = 33.74 and the same decision.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which are the four-digit poker probabilities (four different, one pair, two pairs, three, four alike)?
   A. 0.504, 0.432, 0.027, 0.036, 0.001
   B. 0.5, 0.3, 0.1, 0.05, 0.05
   C. 0.432, 0.504, 0.036, 0.027, 0.001
   D. 0.2 each
   Answer: A. These come from counting 10^4 equally likely digit strings. (Key point: Probabilities)

2. With five patterns, df and critical value at alpha = 0.05 are:
   A. df = 5, 11.07
   B. df = 4, 9.488
   C. df = 3, 7.815
   D. df = 4, 7.78
   Answer: B. df = 5 - 1 = 4, chi-square 0.05,4 = 9.488. (Key point: Degrees of freedom)

3. Two pairs: observed 55, expected 27. What is its contribution (Oi - Ei)^2/Ei?
   A. 29.04
   B. 1.04
   C. 28.0
   D. 2.04
   Answer: A. (55 - 27)^2 / 27 = 784/27 = 29.04. (Key point: Two-pair contribution)

4. chi-square0 = 33.74. What is concluded?
   A. Independent, since 33.74 is a valid value
   B. Not independent, since 33.74 > 9.488
   C. Independent, since 33.74 > 9.488
   D. Cannot be decided
   Answer: B. The statistic exceeds the critical value, so H0 is rejected. (Key point: Decision)

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