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#332917 Probability of a rope cut exactly in half, in a single slice
Posted by bonanova on 11 June 2013  05:09 AM
#332708 A stand up Logician
Posted by bonanova on 30 May 2013  05:30 AM
Spoiler for Slight change in the process does itHave them stand in a straight line.Instead of requesting a show of hands, ask them to take a step forward. Liars will step forward at one or the other request. Truthtellers from out of town will take two steps.Truthtelling locals wont' move.The onesteppers comprise all of the liars.
Truthtellers (T) and Liars (L) who are Residents ( R) or from Out of town (O).
"All from out of town take a step forward." (TO and LR step forward one step.)
"All who stepped forward take a step forward." (TO and LO step forward one step.)
Those who took one step are LR and LO: only liars and all the liars.
You have thus used the same method to distinguish the Ls from the Ts.
Q.E.D.
 1
#332606 A stand up Logician
Posted by bonanova on 27 May 2013  03:53 PM
Instead of requesting a show of hands, ask them to take a step forward.
Liars will step forward at one or the other request.
Truthtellers from out of town will take two steps.
Truthtelling locals wont' move.
The onesteppers comprise all of the liars.
 0
#332604 Rolling a coin
Posted by bonanova on 27 May 2013  03:20 PM
No need to spoiler.
If it has made a complete revolution, (and it has) it's facing the same "way" after as before.
Maybe I'm missing something.
 0
#332571 The AgeOld Lie
Posted by bonanova on 26 May 2013  01:07 PM
Pat is 27 and told the truth. See analysis below.
Chris then lied.
Let's assume he ONLY lied about his present age and was accurate otherwise.
Without that restriction we can't conclude very much.
Chris used an age of xy.
He then calculated an age plus age next year times 5 to be xy5.
Adding last digit of birth year gave 229,
so he claimed to be 22 in 1993 with a birth year ending in 4.
That is impossible.
Chris' age actually is 9, 19, 29, 39, ...
If he's about the same age as his friend Pat, he's 29.
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#332441 Coin Triplets
Posted by bonanova on 22 May 2013  12:00 PM
Player 1 picks a triplet T_{1}.
Player 2 picks a triplet T_{2}.
Coin is tossed until T_{1} or T_{2} appears and wins.
Every triplet T_{1} has an evil twin T_{2} that will beat it on average.
Suppose T_{1} = TTT.
If TTT is not the result of the first three flips, it will be preceded by H and lose to HTT.
TTT is the result of the first three flips 1/8 of the time.
Thus TTT loses to HTT with 7:1 odds.
Other pairwise triplet probabilities can be deduced similarly.
The table gives P_{2}'s winning probability for every triplet combination T_{1}, T_{2}.
The bottom two rows give the best and average winning probability for each T_{1}
What is the optimal strategy for each player?
P_{1} should pick T_{1}= HTH, HTT, THH or THT.
The best that P2 can achieve then is a winning probability of .667,
That is the smallest best winning probability among the possible T_{1} choices.
P_{2} should look in the column headed by T_{1} and choose the T_{2} that has the highest winning probability.
With best play, who is most likely to win?
P_{2} can always gain favorable odds, ranging from 2:1 to 7:1 depending on T_{1}
Suppose the triplets were chosen in secret?
I interpret this to mean P2 chooses T_{2} without knowing T_{1}.
Nothing different for P_{1}. He should still choose HTH, HTT, THH or THT.
P_{2} does not know T_{1}, so he should seek the T_{2} with best average winning probability.
These numbers are given in the last column of the table.
The choices are either HTT or THH with average winning probability of .507.
HTT loses only to HHT; draws against itself, HTH, THH and THT; wins against HHH, TTH and TTT.
THH loses only to TTH; draws against itself, THT, HTT and HTH; wins against TTT, HHT and HHH.
What would be the optimal strategy against a randomly selected triplet?
I take this to ask P_{1}'s optimal strategy in the case that T2 will be chosen at random.
P_{1} should look at the bottom row of the table to see P_{2}'s winning probability averaged over all the T2s.
P_{1} should thus choose HTT or THH.
This gives T_{2} the lowest average winning probability: .368.
P2\P1 HHH HHT HTH HTT THH THT TTH TTT  MAX  AVG 
\+++++++++++
HHH   .500 .400 .400 .125 .417 .300 .500 .500 .330 
HHT .500  .667 .667 .250 .625 .500 .700 .700 .489 
HTH .600 .333  .500 .500 .500 .375 .583 .600 .424 
HTT .600 .333 .500  .500 .500 .750 .875 .875 .507 
THH .875 .750 .500 .500  .500 .333 .600 .875 .507 
THT .583 .375 .500 .500 .500  .333 .600 .600 .424 
TTH .700 .500 .625 .250 .667 .667  .500 .700 .489 
TTT .500 .300 .417 .125 .400 .400 .500  .500 .330 
+++++++++++
MAX .875 .750 .667 .667 .667 .667 .750 .875 
+++++++++
AVG .545 .386 .451 .368 .368 .451 .386 .545 
+++++++++
 0
#332193 Driving man's delight  2 a harder puzzle
Posted by bonanova on 15 May 2013  08:16 PM
I owe it to the Den to post at least one of these without error.
Possibly this one does that.
Give the longest route (sequence of city numbers) that visits all the cities
(a) not returning to starting city (sum of 7 distances  starting point matters)
(b) returning to starting city (sum of 8 distances  starting point does not matter)
This puzzle has more choices than the first one.
Cities lie clockwise on the perimeter of a 6x6 square:
6OO+O
3 4 5
 
 
4+ +
 
 
 
2O2 6O
 
 
1 8 7
0O+OO
   
0 2 4 6
++++
City x  y  Distances:
++++ 8.485 15 37
 1  0  0  7.211 25 36 38 47
 2  0  2  6.325 14 16 27 48 58
 3  0  6  6.000 13 17 26 35 57
 4  2  6  5.656 46
 5  6  6  4.472 24 28
 6  6  2  4.000 18 23 45 56
 7  6  0  2.828 68
 8  4  0  2.000 12 34 67 78
++++
 0
#332190 Walking man's delight
Posted by bonanova on 15 May 2013  06:00 PM
on the graphic, 7 and 8 are inverted, also it seems the 6.083 distance whould have been 6.325
Spoiler for my take on itnot returning: 1 5 2 6 3 8 4 7 for 42.63
returning: 1 5 2 7 3 8 4 6 1 for 47.10
Xavier is correct on both counts. I've modified the OP. Square root of 40 is 6.325.
Edit: both of these paths are just shy of optimal.
 0
#332163 Walking man's delight
Posted by bonanova on 15 May 2013  01:26 AM
Enough with traveling salesman who hate long road trips.
Mario Andretti has joined the team, and he loves to drive!
His route today comprises eight towns, placed at the corners of an octagon.
Here are their coordinates:
Edit: 7 and 8 are labeled incorrectly. They should be switched.
++++
City x y
++++
1 0 2
2 0 4
3 2 6
4 4 6
5 6 4
6 6 2
7 4 0
8 2 0
++++
The intercity distances, to save some calculations, are NS and EW distances of 6, along with four different diagonal distances of 2.828, 4.472, 5.657 and 6.083 6.325
If Mario begins driving at city 1 (coordinates 0 2) and drives until he has visited them all, how far will he be able to drive, and what city will he visit last?
If his objective is to return to his hotel in City 1 after visiting the other cities, how long of a trip could he accomplish?
 0
#332092 Slicing a pizza a whole different way
Posted by bonanova on 14 May 2013  10:58 AM
#332026 Sunny skies over Titan?
Posted by bonanova on 12 May 2013  09:06 AM
The exciting part of this puzzle is that in 2004 you were given a large supply of beef jerky, a warm parka, and the job of being Titan's chief inperson, feet on the ground, up close and personal, weather observer. Upon your recent return, you reported your findings on Titan's rain activity. Let's call the days on Titan that it did not rain "sunny" days, even with the constant nitrogenous smog. (You grew up in Los Angeles.) You found that sunny days on Titan were followed by another sunny Titan day 29 days out of 30, (p_{ss} = 29/30), while rainy days were followed by another rainy day with probability p_{rr} = 0.7.
Like many heavenly satellites, Titan's rotation is tidally locked to its orbital period (16 Earth days.) If you were on Titan for say 9 Earth years, on about how many Titan days did you observe rain?
 0
#331852 Three Big Wheels
Posted by bonanova on 09 May 2013  09:14 AM
by three non overlapping circles completely contained by the triangle?
 0
#331745 Which liar done it?
Posted by bonanova on 07 May 2013  09:21 AM
Sergeant McGuffy remarked to his lieutenant one day after it was all over. "The way I reconstruct the crime, the innocent people were so excited when they first talked to us, they each got one fact wrong. But, the killer wanted to confuse our trail  and so coolly lied from beginning to end." Let's recap what we know ...
Egmont VanDorn has been found dead in his apartment. From the beginning, it is pretty clear to the police that this was no accident. Before they can be separated, the four young people who found the body  Arnold, Betsy, Charles and Daisy  eagerly begin to tell their story.
 "If, as the police say, Egmont was injured between four and five," said Daisy, "that must let us all out. We were all together having dinner for at least two hours before we came over here."
 "But we all arrived at seven," Charles pointed out.
 "Arnold said that it was six o'clock at the time, just before we opened the door. Didn't you Arnold?" said Daisy.
 "I said it was just seven on the nose," said Arnold. "Sorry, honey."
 "I have no idea what he said or what time we got here," declared Betsy. "The thing I remember is the gas in the hall. We rang the doorbell, and no one came and no one came, and then I smelled the gas. My heart turned over, I thought to myself, he's dead. I just know he's dead."
 "Don't dramatize yourself," said Daisy coolly. "There is no way that you could have smelled the gas before we opened the door. The place was locked up, and sealed too, tighter than a drum. We'd still be in the hall if Eggy hadn't given me a key last week."
 "There was gas in the hall, all right," said Charles. "I smelled it before we opened the door. You seemed to take forever getting your key out. When you finally got the door opened, the gas just streamed out."
 "That was a pretty dangerous thing you did, Charles," said Arnold, "turning on the lights the way you did. Didn't it occur to you that a spark at the light switch could have blown us all up."
 "The lights were already on, Arnold," Charles replied.
 "For my part, I'm sorry about pulling him out of the oven  tampering with the evidence and all that," said Arnold. "Murder never crossed my mind. Locked room, you know. All I could think of was that maybe he was alive and we could still save him."
 "I don't believe this," said Daisy. "You didn't pull him out of the oven. I did. You ran and opened the window. Very good move, too, I thought at the time."
 "I opened the window!" cried Betsy. "I was dizzy from the fumes, and I knew I needed to do something fast."
 "The only things you opened were the door to the liquor cabinet and a bottle of Scotch, Betsy." Charles laughed at her. "And I thought they were good moves."
 "That was Arnold who got out the Scotch," said Betsy. "Don't you remember our sitting there after you called the police, and Arnold passing out the glasses?"
 "It couldn't have been me. Must have been you," said Arnold. "I've never been here before, I didn't even know where he kept the stuff. Charles, what did you do?"
 "Do you know what? I don't think I did anything. I remember quite clearly, walking over here with you after that long dinner we had together, and then seeing poor Eggy's feet through the door. But after that, I don't think I did a thing, except stand there gasping."
OK it's time to make an arrest.
Which liar done it?
 1
#331531 Ice slice
Posted by bonanova on 03 May 2013  09:04 AM
But I checked, and its side is just 1/2 sqrt (2) = .7071 for a unit cube making an area of six equilateral triangles with that side length, or 1.29.. square units.
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#331239 Particle detectors
Posted by bonanova on 25 April 2013  03:11 AM
ok, why is a circular detector bigger than the square detector ?
Circles aren't bigger or smaller than squares until constraints are added.
 If the constraint is a given perimeter, circles are bigger (area wise)
 If the constraint is maximum and minimum values of x and y, then squares are bigger.
The present constraints are of the second type.
 0
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