[The answer should be plain to see]

Incidentally, the title was a hint too: plain in color, and also plain -->plane, since a piece of paper is flat.

[The answer should be plain to see]

My intention was like this:

As pure in shade as mother's blessing true, (so same color as milk)

And yet in form I'm still possessing, too; (so not liquid... The other way the line could be taken would be about origami, but that was just a secondary thought)

Partner both in math and drinking (math for math papers, drinking as in paper cups)

I was born of man's quick thinking (because truth be told paper kinda blows my mind still)

[The answer should be plain to see]

Thalia: a solid guess, but no cigar. Hint #2:

Spoiler

Partner in drinking refers to what a cup is made of.

 

[The answer should be plain to see]

A hint:

Spoiler

Mother's blessing is milk. What's the same color as milk?

 

[The answer should be plain to see]

Nope. Look more at the first line: what's the "mother's blessing"? I think Molly understood that.

[The answer should be plain to see]

Nope. Bit more common than that.

[The answer should be plain to see]

As pure in shade as mother's blessing true,

And yet in form I'm still possessing, too;

Partner both in math and drinking

I was born of man's quick thinking

[Destroying a checkerboard]

Spoiler

I can do six, I think. cut board in half, stack halves, cut in half again (2x8 chunks now), stack, cut again, stack (1x8), cut in half other way, stack (1x4), cut in half, stack (1x2), cut.

 

[I’m not a gardener]

Something that should be odd, but in this case is even, which makes it a symbol of good luck. hmm...

Spoiler

Something involving the number thirteen would be my guess... maybe floor thirteen renamed to fourteen to avoid bad luck? red and pink are elevator lights, taken up by an elevator...

 

[3 Quickies]

#3:

Spoiler

Out of pure optimism, I'll say 1.

#1:

Spoiler

Well there's a 3/4*2/3=1/2 chance that I didn't draw the white disk the first go around.

In the scenario that i did draw the guaranteed white counter, then the chances of me drawing another white one are 1/2.

In the scenario that I didn't draw the guaranteed white counter, there's a 3/4 chance that I draw a white counter (1/2 to draw the guaranteed one, 1/2 to draw the coin flip, and for that coin flip 1/2 chance that it's white).

So, I would say that there is a 5/8 chance of drawing a white counter.

At least to me, it wouldn't make sense for the chances of drawing a white counter to be less than 50%.

 

[I’m not a gardener]

Spoiler

it burns itself (or so I gather from "taken up from flame we stoke")

serendipity: i.e. luck, good fortune.

Honestly I'm kinda with Molly's guess. Carbon seems like a fair fit. The only issue I can see are the pink and white roses.

 

[Random triangles in a circle]

A thought:

Spoiler

The first two points (call them A and B) don't matter, since if you are able to choose the third one, it's always possible to choose one such that the center is within the triangle. (draw a line from the center of the line segment AB to the center, and then place the third point on the line at any point after it intersects the center.)

Hence, the question is only really about the third point.

More thoughts:

Spoiler

The farther you go on that line, the more tolerance you have for moving off it (i.e. when you've only gone one unit on the line, you might only be able to deviate .1 units off it, whereas if you go ten units you might be able to deviate by a full unit). Hence, the region where the third point needs to be placed is always within a certain sector of the circle.

 

[I'm Back]

On 1/10/2018 at 6:29 AM, Cavenglok said:

Hello. I was here, um, must be seven years ago. I used to run a series called "Brian Dennis". It was an interactive detective story, sort of an RP situation. It was sorta popular at first, but because I just impulsively rushed into making these with little to no planning, the adventures very quickly started making no sense. (At one point there was an actual half-tiger/half-woman in what was supposed to be a noir setting... as well as a fencing match on top of the Devils Tower in Wyoming) But I was thirteen then. I'm a little older now, and I think I can do better. (For starters I'm actually planning the course.)

I don't know if there are any members left from seven years ago but as I'm planning it I'd appreciate some feedback? Also, if anyone is interested, just post here, so I can gauge whether this is a good idea.

I'm very much interested. I recognize the name, too! I expect great things, my friend.

[Digging Probabilities]

Very nice! Well done!

[Digging Probabilities]

18 hours ago, bonanova said:

My understanding of the puzzle

  Hide contents

In general:

• Each dig reveals a prize -- there are no empty holes.|
• Game ends when you get health.

In Method 1,

• Each dig takes 10 seconds.
• Gravedigger continues digging until health upgrade is found, then stops.
• You won't get more than (8, 4, 2) of the (green, blue, red) stones
• Best / Worst cases are 1 / 15 digs (10 / 150 seconds) to get health.

In Method 2,

• Each dig takes 5 seconds
• You reset and go to the next round after every dig
• You can end up with any number of (green, blue, red) stones
• Best case is 5 seconds. No worst case.

 

If that's all true, then

Q1:

  Hide contents

Probability to take maximum digs (15) to get health.

After 14 digs we must have {X} = {BBBBBBBB GGGG RR} in some order.
No matter what, (with complete certainty) the next dig gives you H.

If we say we have to dig exactly 8 B's, 4 G's and 2 R's the probability would be

p = .4^8 x .3^4 x .2^2 = .00065536 x .0081 x .04 = 2.1233664 x10-7

But that's unnecessarily restrictive. We could also get {X} by digging 14 B's:

p = .4^14 = 2.68435456 x 10-6 (more than 10 x as likely)

But there are many more paths to get to {X}:

1. First, dig anything but H (p=.9) until you have dug or made two R's.
2. Then, dig anything but H or R (p=.7) until you have dug or made four G's.
3. Then, dig only B's (p=.4) until you have 14 rupies.

That admits a slew of cases, which we'd have to enumerate and compute a weighted average. Or, we might compute the expected number of steps (1) and (2). Too complex for my taste. Instead, here's a conservative estimate that involves doing (1) twice, (2) four times and (3) eight times. Now it's seen to be much more likely.

p = .9^2 x .7^4 x .4^8 = .0049787136  (about .5%)

Including the cases we omitted above for simplicity increases the result. If steps 1 and 2 are done only a few more times, it easily becomes a few percent.

Q2:

  Hide contents

Average number of digs to get health upgrade.
What are the most likely ways to get H?

• H   p=.1  <digs> = 10  (100 sec)
• RRR   p = .008  <digs> = 125  (1250 sec)
• GGGG(G or R)RR   p = .0081 x .5 x .04 = .000163   <digs> = 6172.8  (61728 sec)

The likelihood of getting H (in the first 10 digs) without just digging it is very small.

Answer: 10 digs. (100 seconds)

Q3:

  Hide contents

Compare Method 1 and Method 2 for speed of win.

In Method 2, there are no "upgrades."
You can't get H without actually digging it, with p = .1.
<Digs> is still 10, but now that's only 50 seconds.

Method 2 wins.

 

 

Question 1:

Spoiler

Everything you've said so far is true. Can you give me a better approximation than "a few percent"?

Questions 2 and 3 are correct! Followup question: What are the chances I get the health from the 14th dig versus the 15th dig? Do they differ significantly?

[Digging Probabilities]

So the other day I was watching a speedrun of the Legend of Zelda: Ocarina of Time (a speedrun is a playthrough of a game with the intent of beating the game as fast as possible). In one particular part of the game, the player is forced to follow around a gravedigger as he digs up various holes. There is one particular outcome that is desired (the "jackpot" of the game, essentially): a permanent health upgrade. There are also three undesirable outcomes that only give out money: a green rupee (the least valuable prize, pretty much $1), a blue rupee (a fairly desirable prize, say about $5), and a red rupee (a very desirable prize, say $20).

Here are the rules:

The chances of digging up a green rupee is 40%, a blue rupee 30%, a red rupee 20%, and the health upgrade 10%.

If the gravedigger has dug up eight green rupees already, and he would dig up a ninth green rupee this time around, he will instead dig up the health upgrade.

If the gravedigger has dug up four blue rupees already and he would dig up a fifth this time around, he will instead dig up a green rupee.

If the gravedigger has dug up two red rupees already and he would dig up a third this time around, he will instead dig up a blue rupee.

As a result, the maximum number of attempts to dig up the health upgrade is fifteen.

Example: four blue rupees and two red rupees have been dug up. If the gravedigger hit the 20% chance to dig up a red rupee on his seventh total attempt, he would instead dig up a blue rupee. However, four blue rupees have already been dug up, so he would actually dig up a green rupee.

After digging up the health upgrade, the game is over.

 

Three questions:

One: what is the probability that it will take a player the maximum number of tries to dig up the most desirable outcome (the health upgrade)?

Two: What is the expected average number of tries for the health upgrade?

Three: There exists two methods to play this minigame. First is the method described above. Second is that after the first dig, the player exits and reenters the area. This resets the green/blue/red rupee counter, but also allows the player to try another dig far faster. This introduces the potential for an infinite number of tries (the world record currently sits at about 100, which is pretty unlucky to say the least). Say it takes a player ten seconds between digs using the first method, and five seconds between digs using the second method. On average, which one will get you the health upgrade the fastest?

 

Disclaimer: I've got an answer to the first question and maybe the second, but I've got no clue for the third.

[Jelly beans join the clean plate club]

6 minutes ago, flamebirde said:

But no matter what you'll always have at least one pair whose average is even. Try it: pick any three numbers such that the average of any two is odd. I'm fairly sure it's impossible, since between a+b, b+c, and a+c at least one is guaranteed to be even. Do you have another counterexample of two numbers whose average is even but can't be reached via doubling?

Especially when you consider as well that at any point you've got an extra plate either to transfer to or to transfer from (although that does complicate things a bit, it also ensures that pairs such as {4,8} are solvable).

11 minutes ago, CaptainEd said:

Here’s a tiny observation about what the next to last step looks like.

 

  Hide contents

 

Any one of the following three patterns results in the x,x,y pattern that leads to an empty plate.

(Z, 3z, anything)

(Half of the total, anything, anything)

(Z, 2z, z+ anything)

 

 

Spoiler

Potentially the second one of these seems really useful... if we can ever get 1/2(a+b+c) on a plate then we can solve the entire thing. Although this won't always be possible if the sum of all the numbers is odd... hmm. I'm a little stumped, to be honest.

 

[Jelly beans join the clean plate club]

21 hours ago, Molly Mae said:

I can show you by counterexamples that they don't always land at the average (if, for example, their average is odd [say 4, 6], you'll never be able to double to the average).  That's why I considered evaluating the difference between two as being divisible by 4 (instead of 2).  Perhaps you'll succeed where I have failed.

But no matter what you'll always have at least one pair whose average is even. Try it: pick any three numbers such that the average of any two is odd. I'm fairly sure it's impossible, since between a+b, b+c, and a+c at least one is guaranteed to be even. Do you have another counterexample of two numbers whose average is even but can't be reached via doubling?

[Jelly beans join the clean plate club]

I think I've gotten closer.

Spoiler

If a+b is even, then probably continuously shifting beans between the two will eventually result in two identical bins, solving the question.

If b+c is even, then the case is identical to the one above.

If a+b is odd, and b+c is odd, then a+c is even, which means that the solution is identical to the one given above.

now the question is: can we prove that the function of doubling one bin by taking beans from the other converges to the average of the two? If so, then the question is solved.

 

[Jelly beans join the clean plate club]

Aw man, I was so confident I'd finally solved one of Bonanova's legendary puzzles.

A start on thinking:

Spoiler

I get this far:

a     b     c

2a     b-a    c

2a     2b-2a     c-b+a

but then I'm stumped, since it's not possible to definitively say that any one of these three values is larger than another. I would have to examine every possible case (where 2a is now the biggest, where 2b-2a is the biggest, and so on, for every branch) in order to determine whether or not beans can be moved, and even then my calculations would have to go to infinity in order to figure it all out. Thoughts?

 

[Largest unit sphere]

19 hours ago, rocdocmac said:

  Hide contents

The general formula for the volume of an even-dimensional n-sphere is ...

After the maximum unit volume in D7 is reached, the ratio starts dropping since grows much faster than .

So by the time you reach n=1000 (or higher towards infinity), the value for the numerator cannot compete with that of the denominator at all and therefore a second maximum cannot occur.

 

Nice, thanks!

[Whodunit?]

2 minutes ago, Molly Mae said:

I think if we're getting into the analysis of the logical AND, we could say that any of them are telling falsehoods.  The trees, for example, could have been pear trees instead, so the entire statement would evaluate false.  I doubt that's the intention of the puzzle.

Spoiler

But if we've proved that Al, Bert, and Chuck are all correct, then it must be Dick, yes?

 

[Largest unit sphere]

Question: the process shown demonstrates a local maximum for the function, but is there a way to prove that (for instance) at n=1000 it doesn't suddenly reach a new maximum? (I guess what I'm trying to say is, is there a proof, or is the solution presented here just process of elimination?)

[Jelly beans join the clean plate club]

23 hours ago, bonanova said:

On a table are three plates, containing a, b and c jelly beans, in some order, where a < b < c. At any time you may double the number of jelly beans on a plate, by transferring beans to it from a fuller plate. After one such move, for example, the plates could have 2a, b-a,  and c beans. Using a series of these moves, Is it possible to remove all the jelly beans from one of the plates?

 

Spoiler

No.

In order to get a plate with no beans on it, one must transfer all the beans on it to a different plate. In order to do that, the number of beans on that plate must be greater than the number of beans on the other. Since you can only transfer from "fuller" to "less full", you can't transfer beans if the amount of beans from one equals the amount of beans on the other. (This also has the side effect of potentially "locking" the plates if all three have the same # of beans.)

The only way to get rid of all the beans on a plate with n beans would be to transfer n beans to a different plate.  However, this violates the transfer condition, since this means another plate has exactly n beans on it as well (since the other plate must double in beans), which means neither plate is "fuller" when compared with the other. Hence, there is no way to get an empty plate.

Alternative solution: let me eat all of them.

 

[Whodunit?]

On 1/14/2018 at 7:31 AM, Molly Mae said:

  Reveal hidden contents

I considered this, but since Dick said another was lying, all of them can't be telling the truth.  I take "is lying" to mean "is lying today" and not "is lying right at this moment."  I also considered plays on the word "lying" (as in, lying down), but the correct form would be "laying" and absolute truth-tellers also don't make mistakes.

Spoiler

That's true. At least one of the brothers must be lying. I think it must be Dick. Since we have solutions/ proofs of all of the other three, Dick's statement must be false by default (the first half doesn't really matter, since the sentence is an "and" which means both parts must be true to evaluate true).

But why say "At least"? Man, that's gonna drive me up a wall.