South Africa's lottery probed as 5, 6, 7, 8, 9 and 10 drawn

Re:South Africa's lottery probed as 5, 6, 7, 8, 9 and 10 drawn by smallnamespace on Wednesday December 02, 2020 @10:01PM (#25284811)

Those two questions are closely related here by a very simple transformation: if the expected number of occurrences is N over many independent tries, then probability of 0 occurrences is approximately 1-e^(-N), or 99.96% if N=7.75.

Note that for N close to 0, 1-N is also a good approximation to 1-e^(-N).

For large N, it's generally more convenient to talk about the expectation rather than the probability of 0 hits—I'm sure many readers implicitly converted 775% to the expectation in their heads.

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Re:South Africa's lottery probed as 5, 6, 7, 8, 9 and 10 drawn by thaumasiotes on Thursday December 03, 2020 @12:48AM (#25285836)

> I'm sure many readers implicitly converted 775% to the expectation in their heads.

Most people cannot do this correctly; the most obvious interpretation of a "775% chance" is that it represents a 25% chance of seven occurrences and a 75% chance of eight occurrences, with no other possibilities.

The problem gets even worse when you have expectations less than one. If the expected number of occurrences is 80%, what are the odds of getting any occurrences at all? They're less than 80% as long as it's possible to have more than one occurrence.

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