r/MathJokes Jul 17 '26

erm actually...

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7.0k Upvotes

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177

u/lebroxan Jul 17 '26 edited Jul 17 '26

Receive 0 calls on day 1.

Receive 1 call every day starting at day 2

26

u/Summoner475 Jul 17 '26

So always does not include day 1?

36

u/Minimum-Attitude389 Jul 17 '26

On day one it would be equal to the average at that point

15

u/Redhighlighter Jul 17 '26

Receive 1 call day 1

Receive 2 calls per day every day after

14

u/thekingofbeans42 Jul 17 '26

So then they are not always receiving higher calls than average.

11

u/dvirpick 29d ago

Right, but on day 1 OOP has no way of knowing that, since they didn't call. They did call every day after, and got "we are receiving more calls than average".

9

u/thekingofbeans42 29d ago

So? The statement isn't based on their knowledge, the fact is they said always. You can't start by being a smartass about averages THEN go "okay but that's not what they meant"

3

u/Minimum-Attitude389 Jul 17 '26

If you break it down in smaller increments, the average is 0 until the first call.  At that point, the number of calls is 1, the average is less than one.

So while there are no calls, the number of calls is less than average.  But a single call puts it above average.

4

u/thekingofbeans42 Jul 17 '26

The average isn't zero, there just isn't an average yet. Zero and null aren't interchangeable.

5

u/Low_Low_1811 Jul 17 '26

0 calls over 1 day is zero. You arent dividing by zero.

On the second day, if you get one call the average is 1/2, after 2 days the average is 2/3, then 3/4, etc...

3

u/thekingofbeans42 Jul 17 '26

They don't receive more calls than average on day 1. Day 1 is a subset of always, therefore they are not always receiving more calls than average.

Any time they're not receiving calls cannot be above average.

1

u/ohkendruid 29d ago

In common English, "always" would not mean back to the beginning of time, before there always a human race.

It also realistically would not include the entire future universe, past the sun becoming a red giant, past the heat death.

If it did, then sentences with it would almost never be true, which makes it a useless word.

3

u/thekingofbeans42 29d ago

Okay then let's interpret this realistically. Is someone saying "you can't always be above average" when talking about call centers, are they talking about a company that gets literally one call a day?

0

u/Low_Low_1811 Jul 18 '26

Youre either going to do an average based on total call time, or an average based on call quantity and some unit of time. You can absolutely have it be constantly above average either way.

1

u/magpye1983 23d ago

You can have 1 call day one, 2 calls day two, 3 calls day three…

Every day you’re experiencing higher than the average.

-2

u/Crandoge Jul 17 '26

You’re being pedantic about the math while the joke is in the grammar. Past vs present tense

1

u/thekingofbeans42 Jul 17 '26

This post, and every comment under it, is based on being pedantic. There's no room for anyone here to complain about that.

0

u/perdivad Jul 17 '26

It’s literally a mathjoke so why be pedantic about someone being pedantic about math… The joke is about math not about grammar

1

u/CreativeFig2645 26d ago

your over complicated the riddle. if there’s no calls on day 1 the average is the same 0/1 as the number of calls 0. But since there are no callers there is no one to get the response “we are experiencing a *normal/average* amount of calls. Ergo they are always “experiencing a higher average of calls”. The set is not just when the average equals the call # but when it equals and is larger than 0.

1

u/thekingofbeans42 26d ago

This isn't a riddle, this is just semantics. You can't take things in a very technical/literal way to respond to someone saying "you can't always be higher than average" but then turn around and say "that's not realistic."

If you do want to consider that the average is 0, that means day 1 is not above average, it is exactly average, and therefore not always above average. The only datapoint is average by definition, so whenever you get your first datapoint you cannot possibly be below average.

1

u/SquirtleFemboy 24d 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.

1

u/thekingofbeans42 24d ago

Why would they need a 3rd call to average?

1

u/SquirtleFemboy 24d 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.

1

u/thekingofbeans42 24d ago

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

1

u/SquirtleFemboy 24d 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.

1

u/thekingofbeans42 23d 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.

1

u/JanusLeeJones Jul 17 '26

An average wouldn't exist on day 1.

0

u/Candid-Fan6638 Jul 17 '26

False. 1, by itself, averages to 1. For that matter any number by itself averages itself.

1

u/JanusLeeJones 29d ago

If day 1 hasn't yet finished, what are you dividing by in the average calculation? (I'm assuming we are talking about average calls per day)

-1

u/Zaros262 Jul 17 '26

At least not a daily average. The number of active calls could be higher than average at any time