Where is Bushindo when you need him? Mr. Bayes.
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Posted by bushindo on 22 November 2013 - 08:55 AM
Posted by bushindo on 15 May 2013 - 12:52 AM
Anna (A), BIll (B), Cindy ©, and Dante (D) work on a project.
Together, A, B, and C can complete it in 10 days.
Together, B, C, and D can complete it in 11 days.
Together, C, D, and A can complete it in 12 days.
Together, D, A, and B can complete it in 13 days.
Who is the best performer? Prove your answer.
In each scenario, 1 worker is excluded and everyone else contributes.
Therefore the best worker is the one without whom the work takes the longest. That would mean C is the best worker since without him it takes 13 days to complete the task.
Posted by bushindo on 11 February 2013 - 03:13 AM
Seven International Hostages are lined up in a row .Some are standing or sitting ,facing backward or forward.All were blindfolded with different colored duct tapesand ordered not to move. None has any information about the other hostages.The terrorist leader announced to the nationals before him on the speakers that..they shall all be released if they all can guess their blindfold's color correctlywhich are in the order as ff : Red Orange Yellow Green Blue Violet WhiteChi has 2 hostages facing backward on his right, 3 sitting down on his right.Jap has 2 hostages facing forward on his left, 3 standing up on his leftFra has 1 hostage facing backward on his right, 3 standing up on his left.Rus has 3 hostages facing backward on his left, 2 sitting down on his left.Eng has 2 hostages facing forward on his right, 3 sitting down on his right.Ger has 2 hostages facing forward on his left, 1 standing up on his right.Spa has 2 hostages facing backward on his left, 2 standing up on his leftIn 60 seconds 1 hostage should raise both his tied hands then will be pulledbehind the row and will be shot in the head right on the spot in case he failsto shout the correct color of his blindfold .The other nationals will share thesame fate. If he is right, all the others should not fail or else everyone is goingto die. If none of them will guess in a minute all will remain hostages..Who would first raise his hands?
The two possible arrangements are
Posted by bushindo on 07 February 2013 - 10:49 PM
Prove that you cannot
cover a 10 x 10 chessboard with 25 figures
(Problem from Russian Math Olympiads. 6-th grade, 1964.)
(1) B W B B (2) W B W W
So, let A be the number of pieces of type (1), and B be the number of pieces of type (2). The following two equations would have to be true if we can fill a 10x10 board
A + B = 25
3*A + B = 50
But obviously, there are no integer solutions for the above, so we can't fill the board with this Tetris shape.
Posted by bushindo on 03 February 2013 - 12:29 AM
Posted by bushindo on 30 January 2013 - 06:46 PM
You glue two small square pyramids together at their bases to form an eight-sided object that becomes a fair die. You mark the opposite pairs of faces with 0 1 2 and 3.
You begin the game with $1. You roll the die; the number that shows is your new stake. That is, with equal probability you lose you dollar, you keep your dollar, you double your dollar or you triple your dollar.
A minute later, you bet each of your dollars, if any remain after the first roll, with another roll of the die; one roll for each dollar that you have, and collect your winnings, if any. After another minute passes, each of your dollars, if any, suffers the fate of another roll. To be clear, in all cases each dollar is wagered individually. As the minutes turn into years, you eventually become rich, or you go bust.
What is the probability that you go bust?
I like the conditions that Prime gave in his last game. You can simulate the game if you like. But a submitted solution must comprise an answer and a method, both of which are correct. Enjoy!
Here's an approach
P(1) = (1/4) * P(0 ) + (1/4) * P(1) + (1/4) * P(2 ) + (1/4) P(3) P(2) = 0.0625 * P(0 ) + 0.1250* P(1 ) + 0.1875* P(2 ) + 0.2500* P(3 ) + 0.1875* P(4 ) + 0.1250* P(5 ) + 0.0625 * P(6 ) P(3) = 0.015625* P(0 ) + 0.046875* P(2 ) + 0.093750* P(2 ) + 0.156250* P(3 ) + 0.187500* P(4 ) + 0.187500* P(5 ) + 0.156250* P(6 ) + 0.093750* P(7 ) + 0.046875* P(8 ) + 0.015625* P(9 )
The key now is that we will assume P(N) = 0 for N >= 30. This is a reasonable assumption because a rough order-of-magnitude guess for P(30) = (1/4)^(30) = 8e-19, so that's close enough to zero.
So, we we construct 30 equations P(N) for 0 <= N <= 29 as described above. This system of 30 equations can be rewritten as a matrix equation
A * p = p
where A is a matrix constructed about of the coefficients an, and p is the 30-dimensional vector (P(0), P(1), P(2), … , P(29 ) ).
p is obviously an eigenvector of A corresponding to the eigenvalue 1. It is then straightforward to find p using eigen decomposition. The value for P(1) is .4142136. The bust probabilities for all starting dollar values between 1 and 29 are
 4.142136e-01 1.715729e-01 7.106781e-02 2.943725e-02 1.219331e-02  5.050634e-03 2.092041e-03 8.665518e-04 3.589375e-04 1.486768e-04  6.158394e-05 2.550890e-05 1.056613e-05 4.376635e-06 1.812861e-06  7.509115e-07 3.110374e-07 1.288356e-07 5.336514e-08 2.210427e-08  9.155624e-09 3.792164e-09 1.570594e-09 6.504336e-10 2.693280e-10  1.114981e-10 4.614419e-11 1.908862e-11 7.891683e-12
Posted by bushindo on 14 January 2013 - 09:25 PM
If I repeatedly throw a fair 6-sided die, what is the probability that the running total will at one point equal 137?
With some recursive code, I get
Posted by bushindo on 08 January 2013 - 08:27 PM
Posted by bushindo on 08 January 2013 - 04:43 AM
Here's an approach,
The general strategy to determine a number N beforehand. During the game, the statistician should to look at the first N girls' cards and note the maximum dowry value among the N cards. We shall denote this maximum dowry value as M.
Having done this, the statistician should then look at the remaining (100-N) girls, and choose the first girl that has a dowry value larger than M.
The probability of winning using this strategy is
(100 - N)*N/[ 100 * 99]
Using derivatives, it is straightforward to see that the optimal choice for N is 50. So the optimal winning chance is (1/2)*(50/99)
Forgot about some cases in computation of winning chance. Back to the drawing board.
Posted by bushindo on 17 October 2012 - 05:08 PM
Posted by bushindo on 27 July 2012 - 06:22 AM
Spoiler for looks like
Four extra prisoners can correct a single faulty bit for up to 2048 bottles.
So long as they're hamming it up a bit as they drink.
Posted by bushindo on 30 June 2012 - 07:37 AM
I just thought of a positive integer less than 4
said Alex to his friends Davie and Jaimie last
night at Morty's while idly tossing darts in the
back corner of the room.
It was a slow night, and Alex loved to keep things
interesting, as best he could. Now if I were to
give you two questions of the standard Yes/No
variety, I have no doubt that even you two geniuses
could deduce what my number is. So, I'll raise
the challenge a bit, and buy the next round of
beer, if you can tell me my number - with absolute
certainty - after asking me a single Yes/No question.
Davie didn't even bother to stroke his beard.
Can't be done, pure and simple. A waste of time,
If you ask me, agreed Jaimie. Neither of them
noticed the twinkle in Alex's eye as he continued
to reel them in. All right, I'll turn it around. You
pick the number and I'll ask the question. And if
you can't answer it, you'll have to say so. But it
will cost you to play. If I get it right, with certainty,
you buy. If I get it wrong, I'll buy. Either of you
up to that?
Jaimie Just walked away, muttering something
quietly about complexity. This time Davey did
stroke his beard and after a minute said, OK,
Who bought the next round?
Posted by bushindo on 27 June 2012 - 07:44 PM
Personally I find it bizarre. I've noted that the "signal to noise ratio" (if signal means correct/true/good reasoning/answer/process, noise means bad/incorrect/false) on brainden is worse than most ad hoc forums or messages, even for the trivially easy puzzles. One of these forms is where people give answers that don't mean a thing to anyone else; the words and logic is practically gibberish in English. How does this happen? Even for most people to acknowledge a correct logical answer seems to be impossible here. If this place lacks common sense, you have to wonder what "communal" value there is given the objectives of contributing here.
In this case, the fact that we have completely different solutions means most of us are completely wrong, and yet believe we are all on the right track.
I have no problem accepting I could be wrong, but I'm the only one who has given a proof, and I have implicitly disproved all other answers. But in my experience, proofs and disproofs mean nothing to most people.
Spoiler for Analysis of the 70 distributions of stamps
- Exactly two students have mixed stamps.  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.
Spoiler for 'possible error
The possible error is highlighted in red. I find that there is indeed a chair preference here in that the third 3rd chair can not win if there are two students with mixed stamps. Consider the two states bmm and mmb, my calculations show that they both give will the win to the second chair.
Also, I believe that if there are 3 students with mixed stamps, then the third chair would win.
Spoiler for solution'
Third chair: 24 left.
So I think there are 2 or 3 mixed pairs. Second chair would have won if one of 1st or 3rd pair was not mixed, therefore they must both be mixed.
I believe possibilities left are MMMMMM or MMXXMM. This would mewan third must have XY no matter what.
Can check this by seeing how many possibilities MMMMMM and MMXXMM form:
MMXXMM: 2x for XY or YX on third pair, 2x for XY/YX on 1st pair, 2x for identify of X, 1x for the rest must be YY. 8 ways.
MMMMMM: 8x for three pairs XY/YX. Remaining is also a pair, 2x for that. 16 ways.
Totals 24 so correct.Spoiler for 'possible error
The possible error is highlighted in green. If the states are MMXXMM, then the 3rd chair would say 'No' on turn 3, allowing the 1st chair to rule out the state XX and YY for his own state. The 1st chair would then win on turn 4.
Posted by bushindo on 14 May 2012 - 05:33 PM
Each of eight logicians has a unique positive integer less than 100 stapled to their foreheads. They stand in a circle facing one another so every logician can see everyone's number, save for their own. Each logician must discern their number and raise their hand. After all logicians have raised a hand, they must then all declare their number in unison. What strategy (within the assumed spirit of these types of puzzles) might they employ such that all are correct?
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