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0 pointsgrayclhn4y ago0 comments

To be super pedantic, for one million ops it's closer to a 63% chance every day:

Pr[something happens across 1_000_000 events]

= 1 - Pr[nothing happens across 1_000_000 events]

= 1 - Pr[nothing happens once]^1_000_000 ## assuming independence

= 1 - (1 - Pr[something happens once])^1_000_000

= 1 - (1 - 1/1_000_000)^1_000_000

≈ 1 - 0.378

= 0.632

It's still below 99% for 4 million ops.

0 comments

Dylan168074y ago

> for one million ops

Okay, but they said millions.

> It's still below 99% for 4 million ops.

I find this misleading, because it's... 98%.

Which completely undermines your argument. If something happens 98% of days, it's fine to call that "every day".

grayclhnOP4y ago

Dude, it was meant to be mildly educational, not an "argument."

Dylan168074y ago

Then present it as a fun fact rather than as a correction?

And I'm not trying to be mean but I think the way you phrased your last line is accidentally anti-educational. Your last line treats 1 million ops and 4 million ops as nearly equivalent, when the truth is that 1 million ops is far from "every day" while 4 million ops can easily be called "every day".

And if you dislike the word "argument" pretend I said "point"? I think you're reading connotations into that word that I didn't intend.

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