Frequency test Kaskada
Does the distribution of hits for all numbers match the model?
- Game
- Kaskada
- Number pool
- 24
- Selected range
- 50 most recent
- Draws in the sample
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- From
- —
- To
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- Latest draw in the sample
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- Data version (calculated)
- 2026-09-16T17:11:17+02:00
What do these numbers show?
- Frequency-test statistic for 50 draws: T = 16.100 at df = 23. Method: chi-square with the (m−1)/m correction for drawing 12 of 24 numbers without replacement.
- Descriptive standardization (T − df)/√(2·df) = -1.02. A positive value means T is above the expected df; a negative value means it is below. This is descriptive, not a test result.
- Asymptotic p-value: 0.850955 (from a chi-square distribution with df degrees of freedom). This is not the probability that the random model is true.
- Largest standardized deviation in the range: max z = 1.414.
These are historical data. Each draw is independent of previous draws, so these numbers do not tell us what will be drawn next.
How do I use this page?
- Choose the number of recent draws or a year range.
- Compare T with the df degrees of freedom: under the random model T is usually close to df.
- Check the asymptotic p-value and calibrated percentile; they show how unusual T is under the random model.
- Use the contribution table to see which numbers differ most from expected hits.
What this page does not do: does not point to numbers to play and does not increase the chance of winning.
Number frequency test
Each of the 24 numbers should occur equally often under the random model: expected hits E = N·k/m for k = 12. T sums squared standardized deviations for all numbers with the (m−1)/m correction, because numbers in one draw are not independent.
Statistic T (N = 50 draws)
16,100
degrees of freedom df = 23 · (T − df)/√(2·df) = -1,02
Asymptotic p-value
0,850955
A value of 0.000000 means p is below 0.0000005 after rounding to six decimal places — not proof that such a T is impossible.
Range too short for a reliable test: p-value and calibration disabled.
Calibration by model simulation
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threshold c95 = — · c99 = — · max z = 1,414
Numbers with the largest contribution to T
z = (hits − E)/SD, where SD = √(N·p·(1−p)), p = k/m. Contribution = z²·(m−1)/m, calculated from raw hits. Percent is the share of the sum of all numbers' contributions.
| number | hits | E | z | contribution | % T |
|---|
The contribution table loads from a separate data file when the page opens. It is not available without JavaScript; the measures above come from the analysis file.
Methodology
For each number n ∈ {1…m}: p = k/m, E = N·p, SD = √(N·p·(1−p)), z = (hits − E)/SD. Statistic T = ((m−1)/m)·Σz², degrees of freedom df = m−1. The (m−1)/m correction results from drawing k numbers out of m without replacement.
The p-value is asymptotic: calculated from a chi-squared distribution with df degrees of freedom, which approximates the distribution of T for large N. The calibrated percentile and the c95/c99 thresholds come from a simulation of the random model. The p-value is not the probability that the random model is true — it tells how often the model gives a T at least this large.
For a range of years, T, df, z, and contributions are calculated from summed annual counters; the p-value, percentile, and thresholds are not available for such a combination. In short ranges, we show the facts and T, but disable the p-value and calibration.