- Asked 4/10
- 2072 Ashwin · 8 marks
Poker test, 10,000 five-digit numbers (2072)
Short answer
For five‑digit numbers, expected frequencies from probabilities give χ0² = 318.37. With 6 degrees of freedom and α=0.05 critical value 12.592, since 318.37 > 12.592, the null hypothesis of independence is rejected.
A sequence of 10,000 five digit numbers has been generated and analysis indicates the following combinations and frequencies.
| Combinations (i) | Observed Frequency (Oi) |
|---|---|
| All different | 3450 |
| One pair | 4510 |
| Two pair | 1150 |
| Three of a kind | 750 |
| Full house | 85 |
| Four of a kind | 40 |
| Five of a kind | 15 |
| Total | 10,000 |
Based on Poker Test check whether the numbers are independent. Use α = 0.05 and N = 6 is 12.592
Setup. H0: the numbers are independent. H1: they are not. N = 10,000, 7 classes, so df = 6 and χ²0.05,6 = 12.592.
For five-digit numbers, counting the 10⁵ equally likely strings gives the probabilities: all different 0.3024, one pair 0.5040, two pairs 0.1080, three of a kind 0.0720, full house 0.0090, four of a kind 0.0045, five of a kind 0.0001 (they sum to 1).
Expected frequency Ei = 10,000 × probability.
| Pattern | Probability | Oi | Ei | (Oi − Ei)²/Ei |
|---|---|---|---|---|
| All different | 0.3024 | 3450 | 3024 | 60.012 |
| One pair | 0.504 | 4510 | 5040 | 55.734 |
| Two pair | 0.108 | 1150 | 1080 | 4.537 |
| Three of a kind | 0.072 | 750 | 720 | 1.250 |
| Full house | 0.009 | 85 | 90 | 0.278 |
| Four of a kind | 0.0045 | 40 | 45 | 0.556 |
| Five of a kind | 0.0001 | 15 | 1 | 196.000 |
χ0² = 318.37
Conclusion. Since 318.37 > 12.592, H0 is rejected. The numbers are not independent. The large deviations are in "all different" (3450 against 3024 expected), "one pair" (4510 against 5040), and "five of a kind" (15 against 1). Even after merging the two smallest classes, the statistic stays far above the critical value.
✓ Answer checked by Claude Opus 5.5 (high effort)
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