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.
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.
1. How often does this happen? This is a question about expected value, and the answer could be anything zero or above.
2. What are the chances that this will happen within a year? This is a question about probability, and the answer must lie between zero and one. There is no such thing as "a 775% chance it would happen yearly".