Jump to content
BrainDen.com - Brain Teasers

Molly Mae

Members
  • Content Count

    3592
  • Joined

  • Last visited

  • Days Won

    9

Posts posted by Molly Mae


  1. 3 hours ago, plainglazed said:

    You and I each have a coin, select its orientation (heads or tails), and reveal to each other simultaneously.  If both are heads, you win a dollar.  If both are tails, you win a quarter.  But if both are different, I win fifty cents.  Fair enough?
     

    If we were choosing randomly with equal probability, I would take this bet and laugh my way to the bank.

    Since we're not choosing randomly, I won't be taking this bet.  Optimal play from you actually has you win significantly over time.  This is tricky, though, and I haven't found the optimal play for you yet, but it's somewhere around choosing with probability 1/3 heads, 2/3 tails.


  2. 1 hour ago, TimeSpaceLightForce said:

    Can you pick the misplaced color that belongs to the other row of colors? Which one is it?

     

    N2vd1.png

    Tough.  I might not even be on the right track, but I think it's the top right.

    The filename is N2vd1.png.12e9edc95937c6e48ebdfd71093cdf49.png

    We can break this down and remove the non-hexy bits and get 12e9edc95937c6e48ebdfd71093cdf49. 

    First, I tried just converting that to ASCII, but it was gibberish.  Better, probably was breaking it into RGB chunks and converting those to hex colours.  But the string is 32 hexy bits* long.

    So I tried a few methods:  first, I tried taking the first six, then the next six, and so on, dropping the last two.  That should give us five colours, which makes sense.  Then we could use the last 6 and go backward and drop the first two.  That seems pretty clever, since it would give us two groups of five colours each.  Sadly, it doesn't work.

    So next I went with the first 6, then I skipped the first digit, then the first two digits, and so on.  It looks....closer.  I'm sure there's a way to sample the colours given (probably in paint.net or GIMP) to get their hex value.  Comparing hex values is a lot easier for me than using MSPaint to copy part of the sample given and paste to compare.

    I'm sure, given more time and attention, I could say for sure which colour is out of place, as I'm pretty sure that appendage to the file name is key.

    *Note: Bits here means pieces, not bits as in binary bits.  Obviously.


  3. 2 hours ago, TimeSpaceLightForce said:

    What is the least number of illegal chess moves that would result in such position as shown?

     

     

    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.


  4. 7 hours ago, plasmid said:

    I'll call that a hit. My thinking was...

      Hide contents

    ...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.


  5. 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.


  6. 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 spume

    Alas, I find myself engaged
    By man with a wretched harpoon
    Assaulted thus, I shriek enraged
    With hellish retort for the goon

    Coffee in a thermos?


  7. On 4/1/2018 at 4:53 AM, bonanova said:

    One approach, among several, is to

      Hide contents

    consider the conditions under which a given number of passengers move

    and

      Hide contents

    whether it matters that Al was first to board.

     

    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).


  8. 29 minutes ago, plainglazed said:

    more musings

      Hide contents

    If you flip the fair coin and it happens to land all heads or all tails you'd travel along the x- or y-axis 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:
      Reveal hidden contents

    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


  9. 18 minutes ago, Molly Mae said:
      Hide contents

    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√2-2)/2 units from origin.

    Which means that I did some math.

    EDIT: And I'm dumb.  I calculated (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. 10 hours ago, bonanova said:

    You are both on the right track, but I realize now that I mis-stated 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 sort-of 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√2-2)/2 units from origin.

    Which means that I did some math.

    EDIT: And I'm dumb.  I calculated (2√2)as 8 instead of 4.  I'll recalculate.


  11. 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).


  12. EDIT:Ninja'd by PG.

    bona_gold_star.gif

    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 shot-put, 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.


  13. 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.

×
×
  • Create New...