• Nollij@sopuli.xyz
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    1 day ago

    There’s a similar and related math problem for this:

    How many people do you need in a room before 2 of them share a birthday?

    The answer is around 50, which is way less than most people expect.

      • otp@sh.itjust.works
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        1 day ago

        I think the question is usually frames as “how many people does it take to make it at least 50% likely that two people will share a birthday”, or more likely than not etc.

        A guarantee would need 366 people. But most people are satisfied with “more likely than not”, “90% chance”, or “99% chance”.

        EDIT: I meant 367, not 366!

        • frongt@lemmy.zip
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          1 day ago

          366 would not guarantee it. That’s not how probability works. You cannot guarantee a shared birthday without selecting people. And not to mention, birthdays aren’t evenly distributed.

            • JasonDJ@lemmy.zip
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              1 day ago

              366 people wouldnt guarantee no shared birthdays though. There could still be one leap year baby in that bunch. But what are the odds in that?

              2.6 • 10^-158 , if anyone is curious.

              • clif@lemmy.world
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                1 day ago

                That sad experiment where 366 people in a room all have the exact same birthday.

                Statisticly unlikely, but definitely possible.

            • frongt@lemmy.zip
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              1 day ago

              I misunderstood the scenario. For some reason I was thinking that if you randomly selected people and had a duplicate birthday that’s what you didn’t want.

    • howl2@lemmy.zip
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      1 day ago

      If you assume one mass shooting every three days for the last 15 years, and there being 1700 people “present” for each (within earshot, not necessarily immediately in danger), there are now over 3 million people who have now been present for shootings.