---
title: "Poker test, 10,000 four-digit values (2070)"
subject: "Simulation and Modeling"
chapter: 6
key: "semester_8/simulation_and_modeling:poker-test-numerical-2070-bhadra"
source: https://www.bctengineer.com/q/semester_8/simulation_and_modeling/poker-test-numerical-2070-bhadra
site: BCT Engineering
---

# Poker test, 10,000 four-digit values (2070)

Simulation and Modeling · Chapter 6: Random Number

- Asked in: 2070 Bhadra, 7 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-2070-bhadra
- Whole chapter: https://www.bctengineer.com/md/semester_8/simulation_and_modeling/chapter-6

## Question

A set of 10,000 4-digit random values have been generated. An observation shows that 5065 values have all different digits, 2000 have 2 of a kind digits, 760 have 3 of a kind, 1500 have 2 pairs and 675 have all same digits. Test these values for randomness using Poker test (Use α = 0.05).

## Short answer

From four‑digit probabilities (all different 0.504, one pair 0.432, two pairs 0.027, three of a kind 0.036, four of a kind 0.001), χ0² ≈ 51,516. With df=4 and critical value 9.488 at α=0.05, randomness is strongly rejected.

## Answer

**Setup.** H0: the numbers are random (independent). H1: they are not. N = 10,000. The counts given are: all different 5065, 2 of a kind (one pair) 2000, 3 of a kind 760, two pairs 1500, all same 675. They add up to 10,000. Five classes, df = 4, α = 0.05, so χ²0.05,4 = 9.488.

Probabilities: all different 0.504, one pair 0.432, two pairs 0.027, three of a kind 0.036, four of a kind 0.001. Ei = 10,000 × probability.

| Pattern | Probability | Oi | Ei | (Oi − Ei)²/Ei |
|---|---|---|---|---|
| All different | 0.504 | 5065 | 5040 | 0.124 |
| One pair (2 of a kind) | 0.432 | 2000 | 4320 | 1245.926 |
| Two pairs | 0.027 | 1500 | 270 | 5603.333 |
| Three of a kind | 0.036 | 760 | 360 | 444.444 |
| All same (4 of a kind) | 0.001 | 675 | 10 | 44222.500 |

**χ0² = 51,516**

**Conclusion.** Since χ0² ≈ 51,516 is far greater than 9.488, H0 is **rejected**. The values are not random. For example, 675 numbers with all four digits the same against only 10 expected, and 1500 two-pair numbers against 270 expected, show a strong departure from a random sequence.

## Check yourself

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

1. In this question, '2 of a kind', '3 of a kind', '2 pairs', 'all same' correspond to which expected probabilities (respectively)?
   A. 0.432, 0.036, 0.027, 0.001
   B. 0.504, 0.432, 0.036, 0.001
   C. 0.027, 0.036, 0.432, 0.001
   D. 0.001, 0.027, 0.036, 0.432
   Answer: A. One pair 0.432, three of a kind 0.036, two pairs 0.027, four alike 0.001. (Key point: Class identification)

2. For N = 10,000, what is the expected count of two pairs?
   A. 27
   B. 270
   C. 2700
   D. 1500
   Answer: B. 10,000 x 0.027 = 270. (Key point: Expected count)

3. Which class shows the strongest departure from expectation?
   A. All different
   B. One pair
   C. All same: 675 observed vs 10 expected
   D. Three of a kind
   Answer: C. 675 against 10 gives a contribution of about 44,222. (Key point: Largest departure)

4. chi-square0 is about 51,516 vs critical 9.488. Conclusion?
   A. Values are random
   B. Values are not random (reject H0)
   C. Accept since the total is 10,000
   D. Df should be 5
   Answer: B. The statistic far exceeds 9.488, so H0 is rejected. (Key point: Decision)

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