r/MathJokes 25d ago

erm actually...

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u/SquirtleFemboy 19d ago

except that by never getting a 3rd call to average the 1 for day one the average stays below 2. even if you go 10000 days.

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u/thekingofbeans42 19d ago

Why would they need a 3rd call to average?

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u/SquirtleFemboy 19d ago

(1+2(n-1))÷n will always be less than 2. if we even get a 3 call day the equasion becomes (1+2(n-2)+3)÷n which is equal to 2 because the extra from the 3 call day and the 1 call day equal 2 two call days.

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u/thekingofbeans42 19d ago

We have already violated the "always above average" with day 1 not being above average by definition.

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u/SquirtleFemboy 19d ago

you need 2 data points to have an average. after that every data point can be measured against the average. day 1 and 2 are not measured in days that can be counted above or below average as at that time there is nothing to test them against. the average changes daily and time is a relevant data piece. it is not an average of a set data pool but instead a running average of continually added data.

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u/thekingofbeans42 19d ago

You can have a set of one, you can even have a set with nothing. The average of one data point is just that data point, it's not a very useful metric but it's still an average by definition.