dannnex

AGHHHHHHH I think I finally got it. I haven’t been able to think about literally ANYTHING else for the past 24 hours because this seems like it should be a really easy question to answer, but it was stumping me heres the equation for the coin flip example P=1-0.5^x plug in the number of flips for x and you get the probability of having at least one heads by that many flips. and theoretically you should be able to replace the 0.5 with whatever the starting probability is if you did it once. So instead of 1/2 for the coin flip you’d do 1/365 for the birthday thing. EXCEPT THAT DOESNT WORK AND I HAVE NO IDEA WHY AT ALL waitwaitwaitwaitwait this is starting to sound kinda like the uhh whatsitcalled...Compund Interest formula or no, what’s the other one Continuous interest? update: I AM TEACHING MYSELF ABOUT RECURSIVE SEQUENCES VIA AN OLD MATH TEXTBOOK IM SO CLOSE update: AAAAAGGGHHH I GOT IT Its a recursive sequence. P{n}=0.5^n+0.5{n-1} with brackets as subscript and you can successfully put in any original chance of something happening where the 0.5 is. calculator notation is u(n)=0.5^(n)+u(n-1) So. To finally answer OP’s question..... yeah idk. I guess it’s just guess and check by plugging in numbers to the calculator over and over until you get as close to 1 as possible, but my TI-84 Plus CE won’t even handle numbers that big. but I think I can answer the revised question, sorta. If you have 30 people, you can be 0.274% sure than one shares your birthday. Which isn’t the same as just, the odds of someone having your birthday, that would be 1/365 times 30, or about 8.2% .......I really hope that’s all correct, otherwise I’m actually going to go insane. *deep breath* its not. *internal shrieking* i did further testing with an event that has a 0.25 chance of occurring instead of the horizontal asymptote being at 1, it’s at 0.333 repeating. So according to my calculator, with an event that has a 25% chance of happening, even if you run the event infinite times, you still can only be 33% sure that it actually happened. And that’s obviously not right. and for the 1/365 it’s at like 0.0027 something. but it works for the lower numbers! I think? If you roll a 4 sided die twice, you have a 0.3125 chance of being sure that a certain thing was rolled. Right? Maybe all my base assumptions about this are wrong. How do you calculate the probability of an certain event occurring by a certain number of trials?? As opposed to just the probability of the event happening in general. That’s the whole question here. And I thought I was right but that means the numbers should all be approaching but never reaching 1, not 0.33 or 0.0027. Wagwhahfwhahwhffahwhwhfgagaaaaaa can someone who’s actually smart come save me please

No, but you’re thinking along the wrong lines so thats kinda a misleading answer.

7/10, will be 10/10 if I can have a sandwich.

Yes, other versions of this thing are common on Earth.

I’ll join! Still kinda think this is all going over my head. But I’ll do my best to not be dumb and ruin it! Uh no RP character for now.

Radriarc walked over and laid down on the bed, too tired to wonder how Chris apparently had electric machinery decades beyond the rest of Scadrial. “Thanks” he said, his voice muffled by the pillow. He was asleep within seconds.

Well it’s made of atoms, and atoms are made of those, so I’d say yes?

No

No

Tau = 2pi and it makes everything so much better Why have circumference be 2*pi*r when you cold just have Tau*R but more importantly it makes radians and trig SOOO much easier because one whole circle equals one Tau so the numbers actually go where you think they should go

Pi is bad Tau forever. As for the actual question, I think it is as simple as it sounds. If you choose one random date out of 365 possible dates, and then choose one truly random person, there is a 1/365 chance that they’d have that birthday. So if you gathered 365 people, you’d have a 1/365 * 365 chance. Which would be equal to 1, or 100%. I think some people were referencing the famous birthday paradox scenario, which isn’t actually a paradox, just counterintuitive. It asks a similar question: How many people would you need to gather to be certain that any two of them shared a birthday. It’s a surprisingly low number. I don’t remeber the exact math, but it has to do with you not counting the number of people, you need to count the number of pairs. And the required number of pairs can be reached with a low number of people. Next part is me trying to figure it out from memory: I think it’s 27. If you have 27 people, there is something like a 99% chance that 2 of them share a birthday. If you have 27 people, there are 351 possible pairings (I think) 351/365 equals 96% hmmmm wait I just thought about the original question some more and I’m less sure. What if we simplify the question to a coin flip. Same principle really, just different numbers. How many times would you have to flip a coin to be reasonably certain it has landed on heads at least once? Two times? That doesn’t sound right. so flip #1, there’s a 50% chance it lands on heads. flip #2 is the same, but I don’t think we can add the probabilities together, you have to multiply them, and then add that to the original probability. So I’m flip #2 you have a 75% chance of having landed a heads at least once. Flip number three would add 12.5% so 87.5 total chance of having at least one heads. Yeah this is sounding better. What would that be as an equation... P=0.5^n + 0.5^(n-1) + 0.5^(n-2)... where n is the number of flips? hmmm that’s not quite the notation, what happens when n is less than 0? I guess you could just add a clause “where n is not less than 0”. But I think there should be a better way to do that and I think there has to be a way to notate that without just doing ‘...’ maybe something to do with limits. ooooh yeah, limits are definitely the answer, but it’s 3:22 AM and I’m definitely overthinking this in the wrong direction so ima sleep and come back to this.

Well, congrats Ghanderflaffle, you are the first person to ever technically win this game! Even with infinite questions nobody could figure it out. Time for round four! I’M THINKING OF SOMETHING.

7/10, very true, except for when you forget an “r”

Wait what. I feel like I’m being tricked.

Oh I made one and put it in the FotT discord a while ago. I dunno where the link went but it’s back there in the discord somewhere. ya know, I think it should theoretically be possible to just have one Universal Quiz, and everyone would just compare their answers with each other. It would be less fun though as the questions couldn’t be personalized to the creator, they’d have to be more vague, probably all written answer too. I think I’ll give a shot at making that.

Sure, why not We are now house Iwgahftreejyw or house Wafreightejyw or house ightejywafre