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Posts posted by bonanova


On 5/9/2018 at 7:10 AM, 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?
Yes.
SpoilerA selection strategy exists that gives us equal expected winnings, namely that we select H with probabilities p and q respectively, where
100pq + 25(1p)(1q) {my winnings} = 50[p(1q) + q(1p)] {your winnings}
This happens when p = q = 1/3 (Wolfram solution / contour plot); both sides of the equation evaluate to 200/9.
In fact, if either of us shows H with probability 1/3, the equation holds, regardless of opponent's strategy:
If p = 1/3 both sides evaluate to (50/3)(1 + q), and vv, due to symmetry.Knowing this, both of us can avoid a net expected loss by presenting H 1/3 of the time.
Additionally, no advantage can be gained by either of us by slightly increasing or decreasing that value. If we both increase or both decrease, I come out slightly ahead. If one increases while the other decreases, you come out ahead by the same amount: in the case of 1/3 +/ 0.1, that amount is 2.25.
And as already stated, if only one of us increases or decreases, no advantage or disadvantage is gained if the other stays at 1/3.
.

Doctoring the figure a bit (while I think about solving it.)
o  o  o  o  o
 /
o o o o
 /
o  o  oQuestion: You mention the number of nails (12) and also the number of nails in the perimeter (here that's 10.) So for this example, would we be given N=12 or N=10? I'm thinking the latter, but would like to confirm.

Welcome back TSLF. Nice puzzle!

Cute.
Six people are in the room. Two are alive.

Looks like
SpoilerThe angles now are
[  90^{o}/1  90^{o}/2 + 180^{o}/3 ] + [  90^{o}/3  90^{o}/4 + 180^{o}/5 ] + ... + [  90^{o}/(2n1)  90^{o}/2n + 180^{o}/(2n+1) ] + ....
= (90^{o}) Sum_{(n=1, inf)} {  1/(2n1)  1/2n + 2/(2n+1) }
= 90^{o} { ln(2)  2 } (Wolfram)
= 117.6167...^{o}
Twice as much as the first answer. Showing that rearrangement of terms in an infinite series that is not absolutely convergent (both answers were forms of alternating harmonic series) can change the result. If the terms of a convergent series are all positive, the order does not affect the result.

With that interpretation,
SpoilerThe successive angles are 180/3  90/1 + 180/5  90/2 + ... so that
Final angle = 180^{o} { 1/3  1/2 + 1/5  1/4 + 1/7  1/6 + ... }
= 180^{o} { (1/3  1/2) + (1/5  1/4) + (1/7  1/6) + ... }
= 180^{o} { 1/6  1/20  1/42  ...  (1/(2n+1)  1/2n)  .... }
= 90^{o} { 1/3 + 1/10 + 1/21 + ... + 1/(n(2n+1)) + .... }
= 90^{o} Sum_{(}_{n}_{=1, inf)} 1/(n(2n+1))
= 90^{o} { 2  2 ln(2) } (Wolfram)
= 55.2335 ...^{o }(CCW from yaxis) 
Assuming the above, we note that
SpoilerEach ngon bends the path uniquely. For odd n, the bend is given by 180^{o}/n.
For each 4pgon, where p=1 2 3 ..., the bend is 90^{o}/p.
Since the bends are all in the plane, they can be added in arbitrary order.
4pgons (square, octagon, etc., 90/p,  +  +  + ...)
Bend_{4}_{p} = 90^{o}, + 45^{o}, 22.5^{o} , = (90^{o}) Sum_{(}_{i}_{=1, inf)} (1)^{i}^{+1}(1/i) = (90) ln 2 = 62.383246250395077847550890931236^{o}
ngons (n odd, 180/n, + +   + +   ...)Bend_{n}_{odd} = 180^{o} { 1/3 + 1/5  1/7  1/9 + 1/11 + 1/13  1/15 + ... }
Off the top of my head, I don't know this sum. Here are some thoughts ...
Alternating series are generally easier to deal with. We can sum pairs of terms to do this in two ways. In the first approach, we sum adjacent terms of like sign:
Bend_{n}_{odd} = 180^{o} { (1/3+1/5)  (1/7+1/9) + (1/11+1/13)  (1/15+1/17) + ...}
= 180^{o} ( 8/15  16/63 + 24/143  32/255 + ... (2^{n}^{+1}/2^{2}^{n}1) ...
In the second approach, we sum adjacent odd terms and separately adjacent even terms to obtain two series:Bend_{n}_{odd1} = 180^{o} (( 1/3  1/7) + (1/11  1/15) + (1/19  1/23) ...)
= 720^{o} ( 1/21 + 1/165 + 1/437 + ... )And then the even terms pairwise to get
Bend_{n}_{odd2} = 180^{o} ( (1/5  1/9) + (1/13  1/17) + (1/21  1/25) ...)
= 720^{o} ( 1/45 + 1/221 + 1/525 + ... )Then add the two results.
Final angle that you're facing after doing this forever is { (90)^{o} ln 2 } + Bend_{n}_{odd} (CW is positive)
To be continued ...
I'm leaving now, to plow through some old texts on evaluating alternating series ... ugh.

Just to be clear, n = 3 4 5 7 8 9 11 12 13 ..., and CW and CCW alternate untethered (my new favorite word) to parity?

6 hours ago, kaspar said:Rocdocmac is close. You get one day for the 5 minutes, one day for crossing the international date line, and one day for a leap year.
So if you are east of the international date line at 11:59 on the 28th of February, then cross the line and wait a minute or so, it becomes the 2nd of March. On a subsequent leap year these dates could correspond to Sunday and Wednesday.
Nice solve. And the "few years back" is actually two years ago, when Feb 28 was Sunday and March 2 was Wednesday.

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?

A few years back while visiting friends, we celebrated the birthdays of their identical twin daughters Joan and Jane, born just 5 minutes apart. Joan had her party on Sunday, and Jane had hers on Wednesday.
Explanation?

This was my take.
SpoilerBert enters a plane with a random empty seat. Chances are 1/100 that it's his, when no one has to move. Else, chances are 1/99 the empty seat belongs to the guy in Bert's seat, so probability that 1 passenger moves is (99/100) (1/99). Surprise!!! That's also 1/100. Chances that two passengers move is (99/100)(98/99)(1/98) = OMG also 1/100. And so on.
What does that mean? There are 100 equally likely cases (1/100 for each) that { 0 1 2 3 4 ... 48 49, 50 51 ... 97 98 99 } passengers have to move. The 99 candidate passengers who move, or not, are indistinguishable (being first to board does not make Al special in any way.) The chances that any passenger will be asked to move varies uniformly from 0/99 to 99/99, with expected value of 49.5/99 = 50%.
Al is one of those passengers.

8 hours ago, plasmid said:When Al gets on the plane, he will do one of three things: (1) he will sit in his own seat, (2) he will sit in Bert's seat, or (3) he will sit in someone else's seat. The odds that he will sit in his own seat (and be guaranteed to not have to move) or in Bert's seat (and be guaranteed to have to move) are equal.
If Al sits in someone else's seat, call that someone else Charlie. Al will have to move if and only if Charlie has to move and claim his assigned seat which Al took. And Charlie will either (1) sit in Al's seat so Al won't have to move, (2) sit in Bert's seat so Al will have to move (note that the odds of picking Al's seat or Bert's seat are again equal), or (3) sit in someone else's seat so that Al and Charlie will both have to move if and only if that someone else has to claim their seat which Charlie took.
This keeps repeating until someone sits in Al's seat (so Al doesn't have to move) or sits in Bert's seat (so Al has to move), and at each step in the process the person who's taking a seat has equal probability of choosing Al's seat or choosing Bert's seat. So the odds that Al has to move are 50/50.
Nice.
One thing I liked about this puzzle is that it's open to clear thinking. Even tho at first it seems too complex.

But actually ...
SpoilerIf Rectangle (a,b) and Circle (r=1) have the same area, then ab = pi.
If they also have the same perimeter, a+b = pi.
Substituting a=pi/b into the second equation leads tob^{2}  pi b + pi = 0
which has no real solutions.
So the premise of the OP is not true.That's because circles maximize area for a given perimeter.
An equalperimeter rectangle must have less area.The most efficient rectangle is a square, with side pi/2 and area pi^{2}/4.
That's (pi/4), or about 78%, of the circle's area.Good one. You had me going for a few days.


Do any of the clues imply that the indicated needles are adjacent?
I would take "on the left side of" to imply that "pain" is adjacent to "seek," while "to the left of" simply means "the other pain" is not to the right of "interesting little creation." But I'd like to be certain of that. Also, is "interesting little creation" one of the needles? If so, you have described nine, not eight, needles.
Thanks.

You have 10 sets of ten coins. One set of the ten is counterfeit, the others are genuine. The genuine coins weigh exactly 0.10 ounces. The counterfeit coins are exactly a 0.01 ounces off, making the entire set of ten coins 0.10 ounces off. You may use an extremely accurate digital scale only once. How do you determine which set is counterfeit?

SpoilerCircle: Area = pi r^{2} Circumference = 2 pi r
Rectangle: Area = ab = pi r^{2} Perimeter = 2 (a + b) = 2 pi r Diagonal = d = sqrt (a^{2} + b^{2})^{1/2} Width = b.2 (a + b) = 2 pi r => (a + b) = pi r => (a + b)^{2} = pi^{2 }r^{2}.
[1] (a^{2} + 2ab + b^{2}) = pi^{2 }r^{2}.
ab = pi r^{2} =>
[2] 2ab = 2 pi r^{2}
[2]  [1] (a^{2} + b^{2}) = pi^{2} r^{2}  2 pi r^{2} = pi r^{2} (pi  2) = ab (pi  2) =>
[3] a^{2}  (pi  2) b a + b^{2} = 0.
a = { (pi  2) b +/ [ (pi  2)^{2} b^{2}  4b^{4} ]^{1/2} } / 2 = f (b)
Ratio of circumference to diagonal = g.
Write g in terms of rectangle width b.
g = 2 pi r / d = 2 (a + b) / ( a^{2} + b^{2} )^{1/2}
g(b) = 2 ( f (b) + b ) / ( f^{2} (b) + b^{2} )^{1/2}

Clue.
SpoilerWhat is the probability that an arbitrary passenger must move? What does that imply?

@ThunderCloud you're homing in on it, but now you're a little high.
@BMAD It's certainly true that if the FIRST child was a boy born on a tuesday, then it's just the prob that the second child is a boy. But ... the OP does not tell you that. That is, "one is a boy" does not imply "my oldest child is a boy." So your "second" child simply means the "other" child.

Not that I know of. Tables of powers, or spreadsheet where different abc values can be simply typed in, or inspired guesswork?
It's not my fav type of puzzle, but some like this type. 

On 4/11/2018 at 1:14 AM, plasmid said:Is there an easier way to reach the solution? If I try to handle the case of an infinite number of princes to choose from, and I deal with the exact formula I was working with initially instead of the simplified approximation that I ended up using to be able to actually calculate an answer, then I get
So this guy Thomas Bruss solved the general stopping problem.
Spoiler(Check out section 2 on page 1388 of the pdf file available in the above link.)
Bruss assigns 1 or 0 to each event in case it's a best case so far. Then the problem reduces to finding the final 1 in the sequence. The probability p_{k} of say the k^{th} event being a 1 is 1/k, making the odds r_{k} = p_{k}/(1p_{k}) = 1/(k1). He then summed the odds starting with the last event and working backward, stopping when that sum first reaches (or exceeds) unity: (Edit): R_{s} = 1/(n1) + 1/(n2) + ... + 1/(s1) >=1. He proved that s is the optimal stopping point. And doing the math he showed that s/n = 1/e and that ((s1)/n)R_{s} (the probability of success) approaches 1/e, as well as n > infinity.
It's really a beautiful result.

2 hours ago, tojo928 said:Think I have the ideal solution here
A (0,0); B (1,0); C(1,1); D (0,1); E (cos(30),sin(30)); F (cos(60),sin(60))
This solution assumes if you are excatly the same distance from points you get to choose. Very slight tweaks to the points can force the longest path
Start at A. B,D,E&F are all 1 mile away so pick E (+1 mile)
At E. B and F are both same distance away (2 sin(15)~.5176) so pick B (+.5176 mile)
At B. A,E,&F are all 1 mile away so pick F (+1 mile)
At F, D is closest (2 sin(15)) (+.5176 mile)
At D, C is closest (+1 mile)
At C, return to A (2^.5~1.414
Total distance walked= 3*(1)+2*(2sin(15))+2^.5~5.449
SpoilerExtremely close to optimal. So close I'm declaring it solved.
SpoilerThe best solution I know of has length of L = (5+√2+√5+√6) / 2 = 5.549+ by moving one of your points (E) to the center of side AB and starting by going from that point to B. Check it out.
The triangle puzzle
in New Logic/Math Puzzles
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You've probably seen this puzzle. There are 15 holes in triangular array. (See sketch below.) The game begins with pegs in 14 of the holes. The play is to jump pegs over adjacent pegs, removing the "jumped" pegs afterward, as in checkers. The jump is made in a straight line. To make a jump, you need a contiguous group consisting of { peg1, peg2, hole } in a straight line. Peg1 ends up in the hole, and peg2 is removed. The object is to make 13 legal jumps and end up with a single peg.
This happens about 6% of the time. That is, about 94% of the time you get a configuration, with more than one peg remaining, that permits no further legal jumps. In some games the peg must end up in the original empty hole, and that happens only about 3% of the time. So, it's not a trivial puzzle.
This puzzle asks for something different, and easier: Lose as badly as possible. That is, select a location for the empty hole, and then find a sequence of moves that leaves the greatest number of pegs on the board where there are no more legal jumps.
It's simple enough to play, even without the game, by marking hole locations on a sheet of paper and using pennies.
Hint: you might first want to sketch patterns of pegs that have no further legal jumps. Then decide where to place the initial empty hole and how to make the jumps to get to the desired configuration.
As already stated, there are 15 holes. There are also 36 possible jumps. For convenience in writing sequences of jumps, they can be numbered, as follows:
Number the jumps like this: and the holes
> o  1
So Jump #1 means the / \ like this:
peg in hole #1 jumps 1 2 > 2 3
over the peg in hole
#2 into the empty 4 5 6
hole #4. o o
/ \ / \ 7 8 9 10
Jump #18 is peg 7 3 4 5 6
over peg 8, into 7 13 11 12 13 14 15
hole 9. / \
o8 o 14o
Holes 4, 6, 13 / \ / \ / \
begin 4 jumps; 9 10 11 12 15 16
the others 17 19 21 23
begin two. / / \ \
o18 o20 22o 24o
There are
36 jumps.
25 27 29 30 33 35
/ / \ / \ \
o26 o28 31o32 34o 36o
With symmetries taken into account, the holes have four equivalence classes:
This means that there are just four distinct places for the empty hole to start a game: { 1 2 4 5 }. All other holes are symmetrically equivalent to one of these.
Just to be sure the numbering above is understood, here is a winning game of the normal type. Start with pegs in every hole except #1. (The top hole is empty.) Then make these jumps: { 7 14 2 17 23 27 34 26 30 6 35 14 7 }. If done correctly, the original hole #1 contains the final peg.
Enjoy.