plasmid

Would that be equivalent to the following?

Not a jeweler. If I had been, I would probably have counted Molly's answer as close enough.

Here's what I came up with Edit: an aside Edit2: even further aside

I got a different answer for question #2...

Molly's answer isn't what I had in mind. Combining it with plainglazed's comment does make it make more sense to me (although not for all of the clues, particularly the second doublet), and is getting warmer.

In some ways I could see slate fitting, especially with clues about being woven tight and oddly even if multiple different sizes are patchworked together. I’m not sure if it comes in pink and white, but I can’t think of halos for them and there’s a reason for the number fourteen (although I admit it’s not the first number that someone would think of with the thing I have in mind.)

Now Izzy is making me hungry. The cake is a lie, unfortunately, or at least not the answer to this one. Teeth and braces seems along the right track, but there are a lot of tiny clues scattered around that make something else a clearly better fit.

My band of fourteen cultivates Roses pink and white The oddly even adulates When not dispensing slight Woven tight, mayhaps invoke Serendipity Taken up by flame we stoke In halo fittingly

Wait, did bonanova just say there’s a way to do it with a nasty integral?!?

Ah, this is probably what you have in mind.

The program isn't a proof, it's just an application of a greedy algorithm that starts from states with an empty plate and works backwards to see whether every state could eventually reach one with an empty plate. It covered every possible state for up to 500 jelly beans, but it doesn't prove that a plate can always be cleared if you have something like 8x1012 jelly beans.

I wrote a perl program to do what I said earlier...

@Molly Mae maybe this'll help. Q: How should you arrange all those plates of jelly beans? A: Put them on a table.

Not fully calculated because the math gets extremely messy and complicated pretty fast, but a description of how to go about it Although a better answer might be that if there are 10 people playing a game with at least $10 entry fee and $100 payout with a non-zero chance that no one will win and the house will keep everything if everyone ends up picking a number in common with at least one other player, then your best move is to not play.

My answer would be Edit: addendum It might also be worth saying how big of an effect that would be.

Do we get any information about how your sister pulls out the peas -- like should we act as if the number of peas that gets pulled out is a number drawn from a mean (perfect number of peas to fit in her hand) +/- standard deviation, a Poisson distribution, or something else -- or is that left for us to decide?

Sort of combining the two solutions already given

You are or more seriously

From this screenshot, I can say that Questions that might help work toward a solution if it’s neither of those: Can you say what format the door ID is: Number of characters? All alphabetical? All numeric? Both alpha and numeric? Colors? Coordinates? Is there a reasonably high chance to expect that this should end up spelling out a phrase in plain text when we find the solution? Any previous puzzles in the game building up to this one that we should know about in order to have a sense of how we’re supposed to approach it?

I'm not very smart, but could you find it in your heart to post the question for me anyway?

I think gavinksong is right. I sat down and did the math for the case where you're given that one focus is at (0, 0), and two points on the ellipse are at (x, 1) and (-x, 1) so the second focus must be somewhere on the Y-axis (since that math is reasonably tractable). As you move the second focus around and calculate the point at the top and/or bottom of the resulting ellipse, there are no repeats. Meaning that there is no point on the Y-axis where you could put a third point and have any ambiguity about the ellipse. My original argument was thinking in terms of a circle around the third point that contacts the curve defined by the first two points at multiple sites, but that doesn't happen if the third point is on the curve (at least in this case) -- if you think in terms of the picture from my previous post and gradually changing the value of R so you have two potential locations for the second focus along the curve which move outward along the curve as R increases, then apparently the circle around the third point grows fast enough that it "outruns" the point moving along the curve and doesn't intersect it again. If you were to move the third point farther away from the curve (when I did the math for the case above, I was covering cases where the third point sitting on the curve of potential focus locations), then the further away it gets from the curve the more distance along the curve will be covered as a circle around the third point increases in radius and the less likely you are to see any second intersection. So since it doesn't happen for points on the Y-axis, it seems unlikely to happen anywhere. But I can't definitively prove that three points will be always be enough, especially for cases where the first two given points are different distances from (0, 0) and the curve of potential spots for the second focus is not linear. (Well, there's the trivial case where all of the points including (0, 0) are co-linear so the "ellipse" is a line segment, but I'm not counting that case.)

I don't think it's possible with three points on the ellipse even if you know that one focus is at (0, 0).

That was pretty unusual for a mafia... it's rare that at endgame the players can openly conspire to secure a win, and even rarer that everyone actually gets their act together and actually cooperates with such a plan *cough sparrowhawk cough*. Most of the time there are maybe one or two pretty cleared players but the rest comes down to how you read everyone else's play, and such open planning can usually be countered. But life is tough for baddies when the RID kill is known to be gone. Thanks for hosting, araver.

Correction: Nana needs to not attempt to save anyone. Since she can't save self, if she attempts to save someone else and it fails then she would end up going into a 2 vs 1 or 1 vs 1 scenario without a vote, and that could make the goodies lose. I doubt it matters though because I'm pretty sure Nana or Bonanova is the baddie.

How not to mafia How to mafia If it goes into a final day phase with 1 goodie vs 1 baddie (without a vote manip) so they both vote each other and die in the lynch does that count as a tie? Other than that, anyone see problems? Is everyone OK with that plan and please confirm if you send those actions? Mine is sent.