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Posts posted by bonanova
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I'll see your Newton's Third Law and raise you a Solar Powered Truck!
Gas ........? What gas?
Good one.
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I'd love to hear your solution.
But consider this, regarding mine.
You're claiming the crow would be supported by increased pressure
[born by the truck - generalization of Newton's 3rd law] on the bottom of its wings.
But it's not Newton's 3rd law that supports a crow in flight. It's Bernoulli's principle.
The air speed is greater, and the pressure is less, on the top of the wing.
So normal atmospheric pressure exists beneath its wing, and no weight is added to the truck's axles.
Here's a thought experiment.
Attach the crow to the truck with a short rubber band.
Tell the crow to "press down" on the truck using maximum wing force.
Does the rubber band not stretch? Thereby decreasing the truck's force on the bridge?
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He lands lightly and tangentially, just touching the tips of his claws to the truck.
He then glides on a horizontal path, his outspread wings bearing all his weight.
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Further clarification.
This differs a little from the OP for clarity - it does not change the solution.
Jamie is blindfolded.
The four mugs are placed on the corners, randomly up or down.
Each move, Jamie touches only two mugs [not all of them].
Here is what constitutes one move:
Davey turns the lazy susan an arbitrary multiple of 90 degrees.
Touching only the lazy susan, Jamie chooses any two adjacent corners or any two diagonally opposite corners.
He picks up those two mugs.
He puts them back on the same corners, individually flipped or unflipped, i.e. individually up or down, as he chooses.
Jamie gets 5 moves.
At the end of any move: if the mugs on the lazy susan are all up or all down, Alex rings the bell and Jamie wins.
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Clarifying.
Jamie is blindfolded.
The mugs are placed on the four corners, randomly up or down.
Move #1 - Jamie knows where the four corners of the lazy susan are and can pick up
[1] two adjacent mugs - or -
[2] two diagonal mugs
Then he can feel whether they are up or down. He can flip either, neither, or both and replace them.
Davey randomly turns the lazy susan to a new position. Jamie doesn't know what Davey did.
Move #2 - Jamie [still blindfolded] does the same thing again.
Davey turns the lazy susan again.
Move #3 ... same.
Move #4 ... same.
Move #5 ... same.
At this point [or before] all the mugs must be up or all must be down.
At any time, if all the mugs are up or all the mugs are down, Alex rings the bell, and Jamie wins.
Can Jamie ensure a win - without relying on luck?
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It seems, said Alex the other night, that no one likes questions about odds.
They're either too simple to care about or too subtle to believe.
Ian, Davey and Jamie all agreed, and the four of them downed their ale.
Which brings me to tonight's challenge, said Alex.
It has nothing to do with odds -- just this little lazy susan thing.
And he laid it on the center of the table, giving it a spin to show how it worked.
Now here's the deal, boys.
Each of us puts his mug on one corner of this lazy susan thing, upside down or not, doesn't matter.
Then, one of you has to turn them all upside down or all right side up.
There's only three simple rules:
[1] You pick up any two mugs, feel them, and then put them back, whichever way you like - up or down - they don't have to be the same.
[2] Then someone else turns the lazy susan to a new position each time, after the mugs are touched.
[3] And ... you have to do it blindfolded.
Whenever you get them all up or all down, I'll ring this bell here, and you win.
But ... you must do it in 5 moves or fewer.
Davey rolled his eyes and declined; so did Ian.
Jamie thought for a long time and then agreed to try.
Is there any way Jamie can ensure he will win the bet, without just being lucky?
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Is it safe to assume that if he jumps from 2X hight, that it will take him 2Y minutes to touch ground?
Wind resistance eventually decreases his acceleration to zero. [terminal velocity]
Starting great distances from the earth, gravity is initially weaker.
Both influences taken into account, however, his speed never decreases.
Since he starts with zero [downward] velocity, the first half of his fall will take longer than the second half does.
I think, in all cases, is it will take less than 2Y minutes to fall 2X distance.
Two caveats:
Closer to earth, [1] wind resistance increases, decreasing the terminal velocity, and [2] gravitational pull increases, tending to increase terminal velocity. Because the thickness of the atmospheric layer is a small fraction of the earth's radius, the first effect is stronger.
His terminal velocity I believe actually does decrease.
But not enough to make up for zero initial velocity, I think.
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If you reason that because the only cards that can show a B face [the BB and BW cards]There are only 2 cards with a black side on them. AND, of those two cards, only 1 is black on the other side.... whatever. I guess I'll just agree to disagree.are drawn with equal likelihood the answer must be 50%, then try the experiment.
Do precisely what Alex did, ignoring the WW card for simplicity.
Make two cards: BB and BW. Place one from your pocket onto a table.
[1] If it shows B, add 1 to your possible outcomes total.
[2] If the reverse side is B, add 1 to your favorable outcomes total.
Repeat until you have 30 possible outcomes.
Now, are the favorables closer to 15? or to 20?
If it's 17 or 18, do another 30. And maybe another 30.
1/2 or 2/3 will eventually come into sharp focus.
And when it does, keep in mind that you drew the BB card 1/2 of the time!
If I were a betting man, and I'm not, I'd bet an entire donut on 2/3. Why?
Because of the equal likelihood requirement.
You see a B face 100% of the time when the BB card is drawn
but only 50% of the time the BW card is drawn.
2 cards x 2 faces = 4 equally likely events.
3 events show a B face.
2 events are favorable.
Basically, you count the BB card twice - because it can show a B face
two ways; and count the BW card once - it can show a B face only one way.
Does that make sense?
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[1] take a key from the piano and unlock the door.
[2] Use the saw to cut a hole in the wall.
[3] ?
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No.So lets say Will jumps from an airplane that is at X hight. It took him Y minutes to reach the ground.Is it safe to assume that if he jumps from 2X hight, that it will take him 2Y minutes to touch ground?
How about 5X and 5Y?
Or 100X and 100Y?
And I hope he brought a parachute.
His speed the first X [feet, let's say] is less than his speed the 2nd X feet,
so it will take him less than 2Y minutes jumping from 2X height.
In general, ignoring wind resistance, Y increases as the square root of X.
For 2X, the time is Y x sqrt[2] = about 1.414Y.
For 5X, the time is Y x sqrt[5] = about 2.236Y.
For 100X, the time is 10Y.
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You make a valid point. I hope that doesn't sound condescending ...I think the part I always have a problem with is the part leading up to that. I don't think you can count the sides independently of each other. For instance. He pulled the card at random from his pocket and there was at least ONE black side. There were only TWO cards in his pocket that had at least ONE black side, so there's a 50% chance that it will be the black&white card and a 50% chance that it will be the black&black card. Since the ONLY SCENARIO in which there could be 2 black sides would be if he chose the black&black card, there is a 50% chance that the other side will be black because there is a 50% chance that he chose that card.The answer depends on the premises of the set-up.
The way you describe the set-up I agree the answer is 1/2.
But look back at the story. I created a narrative that presupposed
nothing except that a card was placed on the table.
And then it was observed that the top face was black.
It wasn't presupposed to be black.
And if it doesn't matter which color it would have been, a card needn'tI pull one at random from my pocket, and place it on a table,like this. There. As you can see, the side facing up is black.
Now what are the odds that the other side is black, also?
even be drawn; only to ask the question: if I pull a card at random
from my pocket and place it on the table ... what are the odds
that the other side will match?
Where the answer is clearly 2/3.
Presupposing the color that's visible [as in your analysis] makes the answer 1/2.
Part of the fun I get in constructing these stories is providing enough clues for an unambiguous answer.
Sometimes I succeed , Sometimes I get it wrong.
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Get Alex, Jamie and Davie to remove 2 or 3 glasses at a time,So a pint weighs about a pound and a tenth. If you assume a force of 3 pints will upset the balance enough to spill all the drinks, how do you get the pints down to drink them?keeping things balanced at each step.
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Intuitive probability fails when it compares favorable to possible outcomes that are not equally likely.
I buy a lottery ticket.
There are two outcomes - it's a winning ticket or it's not.
One outcome is favorable.
My odds of winning the lottery are 1/2. Ooops ... reality check!
The times that this is really fun is when the faulty result is feasible.
With the black and white card problem, you can easily make the cards
and do the experiment say 30 times.
A valid intuitive solution notes that 1/3 of the cards have opposite color on the other side.
The probability of opposite color is 1/3 and same color [black or white doesn't matter] is 2/3.
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That's it, exactly.As long as there are no indications of which side is which (such as the mixed card having a 1 on the black side and a 2 on the white side and the other cards numbering their sides as well) then it seems there is a 50% chance that the other side is black. If you have a card with one side being black laying on the table, then you either have 1 of 2 cards.... the mixed or the all black which would mean that you have a 50% chance.EDIT:
I think you're considering each side as a separate entity ...What if it was side 2 that was black?
Draw the numbers 1 and 2 on the faces of the black card.
Draw the numbers 3 and 4 on the faces of the white card.
Draw the numbers 5 and 6 on the faces of the mixed card - say 5 is on the black face, and 6 is on the white face.
The numbers 1, 2, 3, 4, 5, 6 have equal likelihood of being visible.
Nothing favors one of these numbers over any of the others.
If 1, 2 or 5, is visible, you see black.
If 1 or 2 is visible, the opposite side is black.
The odds are 2/3.
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kewl.
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If you're referring to the God I know, the answer is this.
God cannot fail.
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We know side 1 is black, so there are only 2 possible states:
Side down is black
Side down is white.
Odds are 50/50.
What if it was side 2 that was black?
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A teenage boy is driving in a car with his father, when they are in a terrible car accident.
The father is killed instantly, and the son is seriously injured.
The boy is rushed to the hospital, and imediately gets sent into emergency surgery.
The surgeon takes one look at him, and says "I can't operate on this boy, he is my son".
The boy was never adopted and his biological parents are still married.
How can this be?
It can't be.
Does the phrase "Til death do us part" ring a bell?
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Ian waved to Jamie and approached the bar. What's wrong
with Alex? he said. He's sitting in the corner fumbling in his
pockets and talking to himself.
I think he's had a few too many, Jamie replied. Plus, a friend
gave him a birthday present - a puzzle - and it has him stumped.
You remember the friend, the one with two children, one
being a boy, who kept asking the probability the other's a girl?
Anyway, he's been there a while. Let's see if we can help.
So I've got three cards, Alex explained...
* a black card - black on both sides,
* a white card - white on both sides, and
* a mixed card - black on one side and white on the other.
I pull one at random from my pocket, and place it on a table,
like this. There. As you can see, the side facing up is black.
Now what are the odds that the other side is black, also?
After thinking a moment - and downing a couple O'Doul's,
Ian smiled and whispered his answer to Jamie.
What would you have said?
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Explanation [i think this post was lost over the weekend]
The title - return from infinity
Alex's promise to stay away from infinity
How far from infinity can you get?
Alex standing on his head before going to infinity tonight
That's inversion. What is 1/[infinity]?
improved descriptions for 2 of the 16 numbers
grid with the correct 16 numbers
8 2 8 8 2 1 1 1 4 2 8 4 1 4 2 8
16 = 4 + 4 + 4 + 4 [make four groups of four numbers]
8 2 8 8
2 1 1 1
4 2 8 4
1 4 2 8
do something mathematically simple to the numbers in each group [use only one arithmetic function]
Add them. => 26 5 18 15
then one final step.
Count into the alphabet.
answer is a single-digit number
Z E R O
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A+B+C+D+E+F+G must = 1/2 of 8J+2(H+I). = 4J + (H+I)
IF this is true, then A+B+C+D and E+F+G must both be fractions Right?
They're not half of an odd number if If H and I are both odd or both even.
Then 4J + [H+I] is even, and (A+B+C+D) and (E+F+G) are each half of an even number.
Now put [H and I are both odd or both even] together with 3H=2I and you're well on your way.
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Happy Birthday to me ... HBTM ... etc sang Alex as he strolled into Morty's
last night and then announced, Boys, I've got a real treat for ya tonight!
First, take a look at this. And he hung from the ceiling a complicated,
interconnected set of five scales, whose 10 balance pans he'd labeled
A, B, C, ... H, I, J.
Now the barkeep gave me one request, bein' it's my birthday on Saturday, and all.
He's gonna draw out fifty-five cold pints. Then, if we can place a different number
of pints into each of these balance pans, we'll have a real party, cuz the drinks
will be on the house!
Every pan gets a pint to start with, and all five scales have to balance.
Alright boys, let's get at it ... we have until midnight to get this done.
Otherwise we pay for the drinks.
Feel free to help Alex, Jamie, Davey and the boys celebrate!
Edit for clarity
The marks on the bars are equally spaced.
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Barkeep,
A pint for the smiley one here!
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You are the director for the annual Open city-wide Men's Singles Tournament.
So you're in charge of drawing up the brackets and seeding the entries.
Your job will be easiest if the entrants number a power of two.
As luck would have it, 97 locals sign up. But at least they all have distinct
rankings, so the seeding is straightforward; and, since it's a single-elimination
format, there is no losers bracket. But since there's an odd number of entrants,
you'll need some play-in matches to start, and maybe some byes for the
higher seeds. Eventually you can reduce the number of surviving players
to a power of two, and the rest of the pairings are straightforward.
You spend a sleepless night working on it and finally come up with a
set of brackets that work.
How many matches are in the brackets that you drew up?
96.
Play-ins, seedings, byes don't change this.
Each match eliminates an entrant and you need one survivor.
How do you make it?
in New Logic/Math Puzzles
Posted
Amends ... have one on me.
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