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Posts posted by Molly Mae


10. Sounds like tape.

7 hours ago, plasmid said:
I'll call that a hit. My thinking was...
...a fireplace, but after thinking about it I can't tell the difference between a fireplace and a wood burning oven anyway.
For the wave, I was thinking roof of a house (which is why the wave that the fireplace/chimney is mostly under is singular, although potentially in a sea of waves in a neighborhood). The oven's aroma is certainly as characteristic a spume as smoke from a chimney. The clue about harpooning is a reference to a fire poker which will rile the fire up, but it's apparently no longer commonplace enough to be widely recognized by everyone since people rarely have wood burning fireplaces any more.
Oh, man, I do like that answer. It does totally make sense. I was just too close to food, I guess.
At one point, I was putting my hands in the shape of a "peaking wave" (and it totally is the shape of a roof) and wondering to myself what has that shape and would be above something insulated. And I landed on oven hood.

1 hour ago, plasmid said:
Getting warmer, but I'll definitely cause you (the listener of the riddle) more calamity if my insides get out and have my own "peaked wave" in the sea to sit under (mostly).
An oven? Thinking the peaked wave might be an oven hood.
EDIT: Or a frying pan, thinking about bacon and the exploding spray of oil.

A crockpot/slow cooker?

A septic tank? I think it neatly fits.
EDIT: Actually....I think I like freezer better.

2 hours ago, plasmid said:
I feast on meals of mindless fare
With blubber to keep it all in
For what’s inside should stay right there
Or you’ll face a calamitous end‘Neath peaking wave I spend the days
While mostly conceal’d, I presume
Unless my presence breath betrays
As characteristic’s my spumeAlas, I find myself engaged
By man with a wretched harpoon
Assaulted thus, I shriek enraged
With hellish retort for the goonCoffee in a thermos?

6 hours ago, bonanova said:
The professor writes a problem on the whiteboard, thus:
25  55 + (85 + 65) = ?
He then inexplicably states that, even though you might disagree, the correct answer is actually 5!Explanation?
Classy.
25  55 + (85 + 65) = 120 = 5! = 5*4*3*2

On 4/1/2018 at 4:53 AM, bonanova said:
Given that the first 99 board randomly, it doesn't matter when Al boards, as long as he does so before Bert. He's just as likely to get any one seat if he boards 1st as he is 99th (that is, the possible arrangements of passengers remains the same, regardless of when they board).

29 minutes ago, plainglazed said:
more musings
If you flip the fair coin and it happens to land all heads or all tails you'd travel along the x or yaxis a distance of 2sqrt(2). The line segment connecting these two points, (0,2sqrt(2)) and (2sqrt(2),0), describes the endpoints of all random walks. The centroid of this segment is (sqrt(2),sqrt(2)) which makes sense because after the first step of sqrt(2), all subsequent steps combine for a distance of sqrt(2) thus making the distance from the origin to the centroid the minimum distance. All points northwest of the segment's centroid and all points southeast are the inverse of one another so either half of the segment can be considered in search of the expected distance travelled.
So we can look along the segment from (sqrt(2),sqrt(2)) (which is the minimum distance from (0,0) and is equal to 2) to (2sqrt(2),0) (which is a maximum); to find the expected distance travelled. The arithmetic average of these two numbers is sqrt(2)+1. But that does not jive with the midpoint of this segment which is a distance of sqrt(5) from the origin. So the distribution along the segment is not linear? And my explanation for my third guess is foolishness and is somewhat debunked above.
I did a simulation which for me means a spreadsheet. Only sixteen paces (the last was still pretty small at .00004316) and only 100,000 trials and came up with an expected distance of 2.33668. For me that number leaves no clue as how one might back out an explanation. Having difficulty understanding how repeated irrational distances could clump at all but I have fractally little maths. Have found this very interesting and hope someone with better maths can explain. Thanks bn
Oh, derp. I see that I subtracted the 2 to find the centroid of the triangle when I shouldn't have, since that 2 is definitely part of the distance. I agree that sqrt(2)+1 is the centroid of that triangle. If you run more trials, does the expected difference come closer to 2.4142?
But...I still like an expected end coordinate of (√2, √2) and a total distance of 2. For any "walk" that chooses directions using X or Y, there's an inverse that swaps Xs with Ys and Ys with Xs. So I'd expect √2 in one direction and √2 in the other as the average.
10 minutes ago, Molly Mae said:Oh, derp. I see that I subtracted the 2 to find the centroid of the triangle when I shouldn't have, since that 2 is definitely part of the distance. I agree that sqrt(2)+1 is the centroid of that triangle. If you run more trials, does the expected difference come closer to 2.4142?
But...I still like an expected end coordinate of (√2, √2) and a total distance of 2. For any "walk" that chooses directions using X or Y, there's an inverse that swaps Xs with Ys and Ys with Xs. So I'd expect √2 in one direction and √2 in the other as the average.
Missed edit time. This makes the assumption, though, that the average of two paths which are inverse of each other is
√2, √2 
18 minutes ago, Molly Mae said:
I feel like my newest answer is wrong, since I believe no walk will ever be less than two. The current plot point I have is (√.5, √.5). This is almost certainly wrong.
But I think I need to calculate a different triangle, specifically the one that involves the point that is 2√2 distance along the bisector of the original triangle.
Without actually drawing it out, I'm pretty confident I can say the centroid of this new triangle is (2√22)/2 units from origin.
Which means that I did some math.
EDIT: And I'm dumb. I calculated (2√2)^{2 }as 8 instead of 4. I'll recalculate.
And maybe I'm not dumb. I think this evaluates correctly.
I'll stand by my answer.
Except that I'll simplify it to (√2)1

10 hours ago, bonanova said:
You are both on the right track, but I realize now that I misstated the OP. I didn't ask for what I wanted.
What I wanted to get at was the average location, that is the average of all the possible ending location coordinates, more precisely, their centroid, and its distance from the origin.
That's not the same as the expected distance of the ending points  which does take sortof serious math. My bad.
I edited the OP.
I feel like my newest answer is wrong, since I believe no walk will ever be less than two. The current plot point I have is (√.5, √.5). This is almost certainly wrong.
But I think I need to calculate a different triangle, specifically the one that involves the point that is 2√2 distance along the bisector of the original triangle.
Without actually drawing it out, I'm pretty confident I can say the centroid of this new triangle is (2√22)/2 units from origin.
Which means that I did some math.
EDIT: And I'm dumb. I calculated (2√2)^{2 }as 8 instead of 4. I'll recalculate.

In that case...maybe...
I just need to cut my answer in half and say 1. My distance was where the bisector intersected the hypotenuse of the original triangle with legs of 2
√2. The center of this isosceles right triangle should be halfway along the bisector (which creates 2 isosceles right triangles). 
STAR WARS?
I think I'm missing a few clues, but I can make an argument for a few and weak arguments for a few others.

EDIT:Ninja'd by PG.
A little bit of trial and error based on some observations:
The total points scored is 40, so if we assume integer points per place, we know that each event gives a total number of points that's a factor of 40 AND that the total number of events is also a factor of 40 AND that those two numbers multiplied by each other equals 40. A little jumping around and I came up with these values:
1st Place is worth 5 points, 2nd is worth 2 points, and 3rd is worth 1 point. Total points given per event = 8, so there must be 5 events.
We know Bert has 5 points for one event, so he must have 4 third places
Bert: 1 / 0 / 4
Al is easy to calculate. There's only one configuration that works to get to 22 points in 5 events:
Al: 4 / 1 / 0
So we can fill in the rest based on places that weren't taken:
Charlie: 0 / 4 / 1
These equal the correct point totals.
Since Bert got 1st in the shotput, we know that his 1st place didn't come from the javelin. So Al must have taken 1st in the javelin. Bert didn't get a second place at all, so Charlie must have gotten 2nd in the javelin.

Clarification questions:
Are points awarded in integers?
I'll also assume that the points are 1st>2nd>3rd>0, since this is obvious.

Well I did a little bit of math...
...which means I'm probably wrong.
So one thing I can logically deduce is that the maximum distance from origin is 2√2. So I made a right triangle with sides 2√2 and solved for the hypotenuse (4). Then I bisected the right angle of the triangle to create two more triangles (yay, triangles!). I know one side is half the length of the hypotenuse of the triangle I cut in half. I also know that the 90 degree angle I cut in half is 45 degrees, so I have two sides that are length 2. I don't need to solve for the hypotenuse of this triangle, because that bisector is length 2, and that's what I was looking for.
So my answer is 2.

Ah.
The odds of Al moving seem to be 50%. If we test it against a smaller number of passengers, we can see how often these closed loops form and how often Al is in his own assigned seat. It seems as the likelihood of the latter decreases, the likelihood of the former increases appropriately.
Here I have the layout for 3 and 4 passengers. 1 is Al and the last number is Bert (3 for 3 passengers, 4 for 4 passengers). Any time 3 can take the third spot, Al stays put. Any time Al is already in his assigned seat, Al can stay put. Any time the displaced person finds their seat, Al can stay put. Al staying put is denoted with a 'y'. Al moving gets an 'n'.
Al stays put 3/6 times with 3 Passengers:
123  y
132  y
213  y
231  n
312  n
321  nAl stays put 12/24 times with 4 passengers:
1234  y
1243  y
1324  y
1342  y
1423  y
1432  y
2134  y
2143  y
2314  y
2341  n
2413  n
2431  n
3124  y
3142  n
3214  y
3241  n
3412  y
3421  n
4123  n
4132  n
4213  n
4231  n
4312  n
4321  n 
A start:
Less than 99%. My first instinct is that Al only has a 1% chance of randomly choosing his assigned seat (and that's true). But Al doesn't have to move if the current person displacing another passenger finds their assigned seat empty.
My intuition now says that the chance Al has to move is actually rather low (probably below 5%). Basically, Al has to be in either (a) his assigned seat, or (b) a closed loop of passengers who have each others' seats. I can't even think of how to go about solving that mathematically.

10 hours ago, flamebirde said:
I hate to bring this one back up again, but what's the rationale behind #3?
It's just a letter substitution. 1 = A, 2 = B, 3 = C, etc.

On 3/20/2018 at 10:05 PM, bonanova said:
For the purposes of this puzzle, consider our old friend Albert to have the shape of a rectangular paralellepiped (when he was born the doctor remarked to his mother, I don't explain them, ma'am I just deliver them) just meaning a solid having (six) rectangular sides. At a recent physical exam, Albert was found to be 2 meters tall, 1 meter wide and .20 meters thick (front to back.) He maintains his geometric rectitude by never leaning forward when he walks or runs.
So anyway, Albert, alas, has found himself caught in a rainstorm that has 1000 raindrops / cubic meter that are falling at a constant speed of 10 meters / second, and he is 100 meters from his house.
Just how fast should Albert run to his house so as to encounter as few raindrops as possible?
This is a thought exercise I've considered in the past. The answer that I've always come up with is that it doesn't matter when disregarding thickness. If the raindrops are constant and fill any cubic meter equally, he'll encounter the same number of raindrops on his "face" regardless of speed.
The .20m thickness, though, needs to be considered. We know that when he moves at 0m/s, he will encounter some amount of rain along that thickness and he will have made no progress. So to minimize the number of raindrops that lands on top of him, we just increase his speed infinitely. We also know that this can cause severe issues, however.
If you'd like some math to support this argument:
...you should probably ask someone else.


Named for the first wherever you go
Hundreds of cycles in column and row.
Grouped into teams oft coloured the same.
Hardly one in each will ever earn fame.
The race will begin when ball touches ground
Millions of cheers can be heard all around.
You expect it to start when northern half cools,
But location may often vary the rules.
Too fast or too slow, not so much a race,
As all are required to maintain one pace. 
6 hours ago, BMAD said:
There is a machine with 20 pieces of candy. Five of those candies are butterscotch. If you put in a 25 cents, one candy is provided at random. If you put in 75 cents, two candies are dropped at random but you may give the machine back one candy in exchange for a 25 cents. And if you put in $1.50 you receive 5 pieces of candy at random but are guaranteed at least one butterscotch.
How much should I expect to spend to get all of the butterscotch?
Clarification question:
If I put in 75p, can I return a candy I received from a previous purchase for the 25p rebate?

You could represent the bazillion dollar coin as a binary number using the first 10 coins. That covers the 2
^{10} possibilities. You can make it more efficient by using the 11th bit to define whether heads is 0 or 1. This saves flips in the event that you need to flip more than half of the coins to make the correct binary number. Depending on the number of coins, you could extend or reduce the amount of bits you use by using the fewest bits that can represent the highest possible number.
16 Pawns puzzle
in New Logic/Math Puzzles
Posted · Edited by Molly Mae
Assuming a "move" is moving a single piece and not "swapping two pieces":
Eight. Eight of the pawns can make it to their current square without illegal moves. The other 8 must illegally hop or teleport to another space. I guess I could notate it out to make sure the answer isn't nine (where the other illegal move would be passing). But 8. Actually, 8 for sure, since the king on either side can triangulate to lose a tempo, if necessary.
0 if we're playing bughouse.