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 Member Title bonanova
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#327213 Sudoku Binary puzzle!
Posted by bonanova on 11 January 2013  08:22 AM
#326663 Philosophic humor
Posted by bonanova on 07 December 2012  09:04 AM
René Descartes was flying home from a conference when the flight
attendant asked, "Monseur Descartes, would you like a cocktail?"
To which the philosopher replied, "I think not," and promptly disappeared.
will immediately note that RD predated the Wright brothers by several centuries.
But ... could I reasonably have posited a stewardess inside his carriage?
Really?
 1
#326089 Blonde Jokes
Posted by bonanova on 17 October 2012  08:14 AM
So this brunette walks into the Doctor's office and says, Doc, something is terribly wrong  I hurt everywhere!
What do you mean? the Doctor asks.
Well I hurt here, she said as she touched her head, and here, touching her knee, and here, touching her shoulder, and here, touching her stomach, and here, touching her elbow, and ....
OK I understand, said the Doctor. I'm scheduling you for a comprehensive set of tests immediately.
Two hour pass, the results are in, and the woman is back in the Doctor's office.
The Doctor approaches her and asks, You're not really brunette, are you?
No, she admits, I'm blonde. How did you know?
You have a broken finger.
 2
#325282 Canned word
Posted by bonanova on 13 September 2012  08:58 AM
There was one duplicate letter.
Using them a target practice, he shot the cans in the following order: 5 6 3 4 2 9 1 8 7.
Amazingly, before and after each shot, the letters on the cans still on the fence spelled an English word.
The initial was was ... ?
 1
#323197 A merciless emperor
Posted by bonanova on 26 July 2012  02:06 PM
bonanova would be correct if there were 1000 bottles.
Being an engineer and not a mathematician, I built in some redundancy.
And next puzzle I solve, I'll wear my glasses.
 1
#321377 All the people like us are We, and everyone else is They.
Posted by bonanova on 06 July 2012  07:48 AM
Spoiler for My point
[1]
1.0000....
0.9999....
are equal to each other only when we goes until infinity.
[2]
there is always some difference between the above two numbers.
[1]
And so we have it ... two numbers written in decimal, differing in every decimal place, nevertheless equal.
Why? Because the "..." notation is a symbol that says, in your words, "go to infinity."
That's the answer to the OP.
[2]
There is never any difference between the above two numbers.
You, as jim, refer to 0.99... as if it were one of the terms of the sequence 0.9, 0.99, 0.999, 0.9999 ....
It is not one of the terms of that sequence; it is the limit of that sequence.
As such, it does not become something, it does not change, it is not approximately anything.
It is  they are both  precisely, exactly, and forever identical with, unity.
 1
#320792 Interesting one  A Cat and A mouse
Posted by bonanova on 02 July 2012  10:57 AM
The cat clearly travels faster than the mouse.
The mouse's path is a side of the square.
The cat's path length is longer than a diagonal but shorter than two sides.
So n lies between 1.414 and 2.
The slope seems to be somewhat proportional to the path length.
So I'm guessing a catenary. In which case,
n = 1.6161 times as fast as the mouse.
If the cat starts at the origin and the mouse goes from [1,0] to [1,1]
and if the curve is a catenary, the cat's path is given by
y = [cosh(nx)1]/n
 0
#320300 Which Chair you will choose ?
Posted by bonanova on 28 June 2012  07:26 AM
I think we differ only in one case: mmb.
See previous post for my analysis of that case.
See comment in red inside your spoiler for a question about yours.
Thanks.
Spoiler for now that I had time
Case 1: No Mixed
WBB (2)  Seat 1
BWB (2)  Seat 2
BBW (2)  Seat 3
(Person in the seat can see four black stamps, so both of his must be white)
Case 2: 1 Mixed
BWM (8)  Seat 3
MBW (8)  Seat 1 (on second turn)
BMW (8)  Seat 2 (on second turn)
(If his were White or Black one of the other would of know, since no other person spoke it must be mixed.)
Case 3: 2 Mixed
MBM (8)  Seat 1 (on second turn)
(If seat 1 were black or white, player three would have know that his was mixed, therefore seat 1 has to be mixed.)
BMM (8)  Seat 2 (on second turn)
MMB (8)  Seat 2 (on second turn)
(Seat 2 can see a person with black stamps, since the other person did not know, then the other person can't see two black or one black and one white, therefore seat 2 must be mixed.)
I guess that is the point I'm not getting.
How would seeing one black and one white disclose your color?
I think at this point Seat 2 knows only that she is not black.
What tells her that she also is not white?
Case 4: All Mixed
MMM (16)  Seat 2 (on second turn)
(For seat 2 the only other possiblilty would be MBM (or MWM), if that was the case then seat 1 would have known (see above), therefore seat 2 has to be mixed.)
I think Seat 2 can't know she is mixed in this case until her third turn.
The (see above) part of your analysis is what I asked about above.
Seat 1  18
Seat 2  42
Seat 3  10
So seat 2 would be the best to choice he would have 42/70 chance that the stamp you laid out in a combination that he could work out.
 1
#320169 Which Chair you will choose ?
Posted by bonanova on 26 June 2012  05:27 AM
As noted, there are 70 equally probable distributions of the stamps.
They fall into four cases. Analysis for each case follows.
 No student has mixed stamps. [6] For example bwb, [where chair 2 would win.]
The student who differs in color from the other two wins: he sees four stamps of one color.
There is no chair preference for this case. Two for each chair.
 Exactly one student has mixed stamps. [24] For example bwm, [where chair 3 would win.]
The mixedstamp student thinks it might be case 1. But after the other two pass, he knows otherwise, and wins.
There is no chair preference for this case. Eight for each chair
 Exactly two students have mixed stamps. [24] For example, bmm [wherechair 2 would win.]
Here the chair that would win case 2 can't know he is mixed [he sees another mixed student] so he passes.
That second mixed student thus knows it's not case 2. So she know she is mixed, and wins.
There is no chair preference for this case. Eight for each chair.
 All three students are mixed. [16] Only example is mmm, [where chair 2 will always win.]
Everyone thinks this might be case 3, but after chair 1 passes the third time, it's not.
They all know it's case 4, and the chair that has first chance to say so is chair 2, who wins.
This is the only case with a chair preference: chair 2 always wins. Sixteen for chair 2.
Winning probabilities are thus .257 / .486 / .257
Interestingly, only in case 1 does the winning chair have samecolor stamps.
91.4% of the time the winner is mixedcolor.
 1
#320146 Which Chair you will choose ?
Posted by bonanova on 25 June 2012  09:17 PM
Second seat wins most frequently, but I get different odds from previous solvers.
I'll enumerate the cases so that if I'm wrong at some point, it will be clear where.
Let b [black], m [mixed], w [white] represent BB, (BW or WB), and WW respectively.
There are nineteen permissible b,m,w triplets, some, like bbb, bbm, etc. are not permissible.
If six of the eight stamps are distributed at random to the students, we can determine the relative frequency of their occurrence.
We can then group them in terms of how many students receive stamps of mixed color.
Case 1: No mixed students: 6 triplets, each occurring 1 time
Case 2: One mixed student: 6 triplets, each occurring 4 times
Case 3: Two mixed students: 6 triplets, each occurring 4 times
Case 4: Three mixed students: 1 triplet, occurring 16 times.
Case 1. The student wins who differs in color from the other two: If bww, Seat 1 wins.
Case 2. The student wins who has the mixed stamps: e.g. If bwm, Seat 3 wins.
Case 3. The student wins who follows the nonmixed student: If bmm, Seat 2 wins.
The first three cases have no chair preference, thus comprise 18 wins for each student.
Case 4. Seat 2 wins. There are 16 such occurrencess.
This case tips the odds to favor Seat 2.
Enumerating the nineteen cases:
Seat 1 wins wbb[1], bww[1], mbw[4], mwb[4], mmb[4], mmw[4] = 18 times
Seat 2 wins bwb[1], wbw[1], bmw[4], wmb[4], bmm[4], wmm[4], mmm[16] = 34 times
Seat 3 wins bbw[1], wwb[1], bwm[4], wbm[4], mbm[4], mwm[4] = 18 times
So the relative chances of winning seem to be 9, 17, 9.
 1
#320042 The Unsolvable Riddle
Posted by bonanova on 24 June 2012  09:26 PM
I expect full and appropriate credit for the crackling insight evident therein.
 2
#319756 The Letter Exchange IV
Posted by bonanova on 22 June 2012  06:07 AM
Enjoy!
1. goes with cotton or vermouth 3
2. close by 4
3. talk like a horse 5
4. swinging joint 6
5. done in from a platform 6
6. beware, Will Robinson! 6
7. made mad 7
8. made mad 7 [oops. Pure anagram here, no letters change.]
9. make amicable 6
10. mini bomb
11. eats sauce also 6
12. homes to radishes and roses 6
13. caught unawares 6
14. dweebs 5
15. lairs of lions, foxes and iniquity 4
16. finis 3
 1
#319133 A simple gold link puzzle
Posted by bonanova on 19 June 2012  09:13 AM
You can do it with two cuts: links 4 and 11.
You'd have lengths of 1, 1, 3, 6 and 12.
 1
#317029 Very Hard Grid Puzzle
Posted by bonanova on 31 May 2012  10:34 PM
Gerda Brownies . 12 $ 60
Klaus Face . . . 20 $100
Lydia Caricature 16 $ 80
Nancy Sand . . . 14 $ 50
Patch Weight . . 18 $ 70
Information you might overlook when solving.
Clue 1. The same person [turns out to be Gerda] both raised more money and served fewer customers than Nancy.
Clue 3. It was a woman who sold the brownies, eliminating Klaus and Patch as the brownie seller.
Clue 3. You find out early that 20 pairs with $100, and 16 pairs with $80. Later, you find that only one other 5x pair is possible.
 2
#316212 Star and cross dissections
Posted by bonanova on 21 May 2012  08:09 PM
Is that the minimum?
It is, unless you can show a dissection with fewer.
 1
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