Advertisement
Why is one ordinary interval insufficient for 49 numbers?
A simultaneous band controls false-alarm risk for the full family of numbers. It is wider than treating 49 separate intervals as if each were the only comparison.
The multiple-number problem
For one number, a 95% interval has a 5% probability of missing the true value in repeated sampling. Across 49 separate checks, the chance of at least one random crossing can be much larger.
Correction
A simple Bonferroni correction uses for m comparisons. With α = 0.05 and m = 49, this is about 0.00102 per number.
Example
A number may cross its ordinary 95% interval while remaining inside the simultaneous band. It is unusual alone but not necessarily unusual after accounting for 49 opportunities to deviate.
Reading the chart
A band crossing flags a number for examination in that sample. If no points cross the band, all frequency estimates remain within the chosen simultaneous range.
What the band does not say
It neither proves mechanism fairness nor predicts another result. It controls one defined kind of deviation over one defined sample.
Limitations
Results depend on the band method, comparison count and model. A window chosen after inspecting data needs extra caution.
Frequently asked questions
- Is a simultaneous band always wider?
- Usually, because it controls joint risk across many numbers.
- Does one crossing prove a draw error?
- No. It is a statistical signal that requires checks of the data and model assumptions, plus confirmation that the signal persists in new data.
Check it on real draw data
Choose a draw range in Lotto Lab PRO and compare the observed result with the random model.
Related Lotto Lab PRO guides
Last updated: