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Posts posted by bonanova


You are wearing gloves while trying to retrieve the marbles contained in a bag. Because of the gloves, you're doing a pretty poor job of it. With each grab, you are only able to retrieve a random number of marbles, evenly distributed between 1 and n, the number of marbles currently in the bag. With 30 marbles initially in the bag, how many grabs do you expect it will take to retrieve them all?

Twenty coins lie on a table, with ten coins showing heads and the other ten showing tails. You are seated at the table, blindfolded and wearing gloves. You are tasked with creating two groups of coins, with each group showing the same numbers of heads (and tails) as the other group. You are only permitted to move or flip coins, and you are unable to determine their initial state. What's your plan?

For the purposes of this puzzle, consider our old friend Albert to have the shape of a rectangular paralellepiped (when he was born the doctor remarked to his mother, I don't explain them, ma'am I just deliver them) just meaning a solid having (six) rectangular sides. At a recent physical exam, Albert was found to be 2 meters tall, 1 meter wide and .20 meters thick (front to back.) He maintains his geometric rectitude by never leaning forward when he walks or runs.
So anyway, Albert, alas, has found himself caught in a rainstorm that has 1000 raindrops / cubic meter that are falling at a constant speed of 10 meters / second, and he is 100 meters from his house.
Just how fast should Albert run to his house so as to encounter as few raindrops as possible?

If you chose to answer this question completely at random, what is the probability you will be correct?
 25%
 50%
 0%
 25%

If
5+3+2 = 151022
9+2+4 = 183652
8+6+3 = 482466
5+4+5 = 202541Then
7+5+2 = ______ ?

12 hours ago, aiemdao said:I dont really understand the question, why there are seven red x, what is symbol ( in the middle of dividend.
It's an alternative representation to put the quotient to the right side.
Here is the more familiar placement. The green (overlined) digits repeat forever._________________
x x x x x x x x x x x x

x x x x x x / x x x x x x x
x x x x x x

x x x x x x x
x x x x x x

x x x x x x x
x x x x x x x

x x x x x x x
x x x x x x

x x x x x x x
x x x x x x x

x x x x x x x
x x x x x x x

x x x x x x x
x x x x x x x

x x x x x x x
x x x x x x x

x x x x x x
x x x x x x

x x x x x x 
Here are the placeholders for a long division, solvable, even with none of the digits filled in. The quotient has been placed to the side. It has a decimal point, not shown, and its last nine digits are repeating. Meaning, of course, the last row of X's replicates a previous row.
Can you piece together the dividend?
 _________________
x x x x x x / x x x x x x x ( x x x x x x x x x x x x
x x x x x x

x x x x x x x
x x x x x x

x x x x x x x
x x x x x x x

x x x x x x x
x x x x x x

x x x x x x x
x x x x x x x

x x x x x x x
x x x x x x x

x x x x x x x
x x x x x x x

x x x x x x x
x x x x x x x

x x x x x x
x x x x x x

x x x x x x 

Here are the placeholders for a long division, along with a single digit in the quotient.
Can you piece together the dividend?
x 7 x x x

x x x / x x x x x x x x
x x x x

x x x
x x x

x x x x
x x x

x x x x
x x x x

   
 
On 3/11/2018 at 12:44 PM, plasmid said:SMH, continuing from that last line of formula...Edit: Nevermind, I think you have to calculate the value over the range of integration for the numerator and denominator separately, I don't think you can just divide through like that. And after looking it up, those limits go to zero so I've got an indeterminate 0/0 still.
SpoilerFor Poisson, appropriately scaled, cumulative prob for (L<x) is F(x) = (1  e^{x}) and the pdf for (L=x) is f(x) = F'(x) = e^{x}. So f(x) says our interval is x, and F(x), for each neighbor, says that's longer than they are. So a large interval's expected length is int (0,inf) {x f(x) F(x) F(x) } dx. (The same integral without the x gives the probability that an interval is large. Turns out it's 1/3.) Integrate by parts using u=x and dv= { f F^{2} dx } = (1/3) d( F^{3}  1 ). Then the (uv) term {x (F^{3} 1) } is zero at both 0 and inf, leaving just int (0, inf) ( 1  F^{3} ) dx = int (0, inf) { 3e^{x}  3e^{2x} + e^{3x }} dx. Was your inf involved with the (uv) term?

17 hours ago, plainglazed said:Flipping a coin eleven times results in _{11}C_{8}+_{11}C_{9}+_{11}C_{10}+_{11}C_{11 }or 232 trials that have 8 or more heads. There are _{11}C_{7} or 330 trials that would have exactly seven heads, half of which would have eight heads with one more flip:
(_{11}C_{7}/2+_{11}C_{8}+_{11}C_{9}+_{11}C_{10}+_{11}C_{11})/2^{11}=.193848
SpoilerEquates with the 3x speed version of ants on a checkerboard:
probability of reaching either of the ends of the five meeting points = 794 / 4096 
Looks like ...
Spoilerdy/dx = 1
Spoilerb = x^{a} = e^{a ln x} = e^{ln ln x} = ln x
y = e^{b} = e^{ln x} = x 
8 hours ago, CaptainEd said:That was the firstcut answer in the antchessboard problem.
SpoilerThe value is too low. It counts only once the (multiple) cases where 8 heads is reached before the "full complement" of 4 tails occurs.
The solution is easier to find if we (reasonably) assume that
Spoilerthe final toss will be Heads.

SpoilerSpoilerBut seriously, working on it.

1 hour ago, plainglazed said:Agree.
SpoilerI think you have the solution at 6.
I've been doing this in my head for two days. I was only able to eliminate 5. 
Just now, Molly Mae said:Signature updated!
I'm humbled.
btw I sharpened (with humor) the flavor text and posted this on another favorite site. It will be interesting to see how quickly it's solved there.

1 hour ago, Molly Mae said:@Molly Mae gets the coveted bonanova gold star for finding an algorithm and getting the answer in less than one day.
So now that the cat (algorithm) is out of the bag, so to speak, it should be trivial to find whether an integer can triple, (or quadruple, or quintuple) under the same manipulation.
It's also interesting that his solution forms a ring that shows the doubling process also occurs for four other ending digits (not 1, so his is the smallest.) But which ones? Incidentally, this also shows that all such integers have 18 digits.

Fun one.
SpoilerThe pickle solids weigh, whatever. They're now much easier to carry.

Things are not going well at the Acme Company.
Executive talent is hard to come by, and it is not cheap. Folks at the water cooler have no ideas, and the coffeebreakers can't imagine how to improve things either. But those who party around the teapot, they came up with something. They suggested to the Board that Acme promote the newest hire in the mail room and make him the CEO! We need to shake things up, but good. Qualifications, job experience, brains, judgment, integrity, these are all things of the past.
Some were not so sure. Doesn't make sense at all, the old timers said. Almost like appointing some guy with orange complexion to be President. That's exactly the idea, said the teapeople. Turn things on end, let the bull loose in the china shop, and see what happens. Hey  how could it be worse than what we have now? Not surprisingly, the debate was long and heated.
Such a risk merited proof of possible gain, so the old guard posed a challenge: produce a concrete example of where the idea had been tried with incontrovertible benefit. In fact, make it mathematical. You know, something that might make a good BrainDen puzzle.
We'll promote the mail room guy, they said, if you can show us an integer that doubles in value when its least significant digit is promoted to its mostsignificant position.
That is, give us a number { some digits } q that has half the value of q { same digits }.
Spoiler102 doesn't quite work, since 210 does not quite equal 204
That all happened last week, and now we're looking for the mail room guy. Was he promoted? Did the tea people find such a number? Is there one?
We need a number or a proof that one does not exist.
T.L.D.R.
What number doubles in value by by moving its last digit to the first position (if there is one)?

Al, Bert and Charlie are on the ballot for secretary of the local yacht club, and they finish in a threeway tie. To break it, they solicit the members' second choices, and again it's a threeway tie. Al notes that since the number of members is odd, a series of twoway votes would be decisive. As a reward for finding a way past the impasse, Al receives the bye and will take on the winner of a Bert vs Charlie vote to decide who gets the position as secretary. Assuming voter preferences do not change, what are the winning probabilities for the three gentlemen?

Alternate approach to the antcheckerboard problem asks: What is the probability of flipping 8 heads before 5 tails?

If we place four matches in the form of a square, they form 4 right angles.
If we place them like a hashtag (#) they form 16 right angles.
If someone removes one match, can we still form 12 right angles?
(No bending or breaking of the matches is allowed.) 
On 3/3/2018 at 10:52 AM, plasmid said:Suppose you have one guy that’s responsible for putting the bags on the conveyor belt, and the amount of time it takes him to put each subsequent bag on the belt is randomly drawn from a uniform distribution from 0 to 1 second. Also say the conveyor belt is moving at 1 meter per second to make the math easier. So you can work with a string of distances: x_{1}, x_{2}, x_{3}, … and we’re looking for the ratio of (the sum of all terms x_{n} where x_{n} > x_{n1} and x_{n} > x_{n+1}) divided by (the sum of all terms x_{n} regardless of whether it's a nonnearest neighbor segment).
For any given value of x_{n}, the probability that x_{n} > x_{n1} is simply the value of x_{n} (since we’re dealing with a uniform distribution of values over a range from 0 to 1), and similarly the probability that x_{n} > x_{n+1} is x_{n} (since these numbers are randomly generated independently of each other). So the probability that x_{n} is a nonnearest neighbor segment is the product of those two, or x_{n}^{2}.
Since the length of each segment is generated from a uniform distribution from 0 to 1, we can do an easy integral. To calculate the fraction of the conveyor belt that’s covered by nonnearest neighbor segments, I think we would want to take the integral of [length of the segment] · [probability that it’s a nonnearest neighbor segment] (which is x · x^{2} = x^{3}) and divide that by the integral of [length of the segment].
Int_{[x=0 to 1]} (x^{3} dx) / Int_{[x=0 to 1]} (x dx)
_{[x=0 to 1]} (x^{4}/4) / _{[x=0 to 1]} (x^{2}/2)
(1/4) / (1/2) = 1/2That answer would very likely change if you used a different method to generate random distances between suitcases on the conveyor belt.
The answer probably does change with the nature of the distances.
SpoilerYour answer seems lower than expected.
SpoilerI wonder what you'd calculate assuming a Poisson distribution, where the probability that any interval is smaller than say a distance x can be taken, with appropriate scaling, to be 1  e^{}^{x}. Equivalently, the probability density function for an interval of distance x would just be e^{}^{x}.

Regarding the 5 probabilities after 12 ant moves:
SpoilerEnumerating paths and treating them as equals gives { C(12,8) C(12,7) C(12,6) C(12,5) C(12,4) } / (3498) multiplied by { 1 4 6 4 1 } / 16 and summing for a result of .22995. But Al (the faster ant)'s paths involve from 8 to 12 choices, so they are not equally likely. Probabilities derived by dividing each path by the sum of all paths (3498 in this case) will be incorrect. What is the right approach?

@plasmid augments paths that touch an intermediate border point, where a choice is eliminated, by multiplying by 2 at each point. The effect is to create virtual paths that increase the count to each end point by 299, making 4096 = 2^{12} total paths and leads to the correct probability of .20551.
I have three other approaches. The first is the one I thought would be found here. Another is quick and dirty. The last one occurred to me as I'm writing this.

This one seems evidently correct but is the most involved of the three.
Call the Final points { F_{1} F_{2} F_{3} F_{4} F_{5} } with probabilities of { f_{1 }f_{2 }f_{3 }f_{4 }f_{5} }.
Start with a result from the first problem, the probabilities { d_{1 }d_{2 }d_{3} d_{4 }d_{5 }d_{6 }d_{7 }d_{8 }d_{9} } of reaching the 9 diagonal points. Then (by inspection) determine the probabilities of reaching each F_{i} from the onetofive possible starting diagonal points Dj. Call those p(Dj, Fi). There are 25, but really only 15 due to symmetry, and some are trivial. Others take just a little thought. Clearly p(D1 ,F1)=1. Then, since p(D2, F2) (4 E moves) = 1/16, we have p(D2, F1) = 15/16, and so on. Then we have
f_{i} = sum(j) { dj x p(Dj, Fi) }

This one involves the least work.
Paths to { F2 F3 F4 } don't hit an edge. So { f2 f3 f4 } = { C(12,7) C(12,6) C(12,5) } / 2^{12}
Then f_{1} = f_{5} = 1/2 ( 1  sum { f_{2} f_{3 }f_{4} } )

This one recognizes an equivalent calculation.
What is the probability of flipping 8 heads before flipping 5 tails? I don't know whether this is an easy question or not, but the answer gives f_{1} and f_{5}.

@plasmid augments paths that touch an intermediate border point, where a choice is eliminated, by multiplying by 2 at each point. The effect is to create virtual paths that increase the count to each end point by 299, making 4096 = 2^{12} total paths and leads to the correct probability of .20551.
Same numbers of heads
in New Logic/Math Puzzles
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@Thalia You are so right. Thanks.
Locking this thread.