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Posts posted by bonanova


The King has decreed that his daughter the Princess shall marry the most wonderful Prince in all the land. One hundred suitors have been selected from their written applications, and on a certain day the King arranges for them, in turn, to interview the Princess. Each suitor must either be chosen or eliminated on the spot. If the Princess does not choose any of them, she will marry the last Prince to speak with her.
You have been chosen as the Royal Advisor to the Princess and tasked with implementing her best strategy to choose the Most Wonderful Prince of the realm. You devise an evaluation scheme by which the princess can assign a unique "wonder number" to each prince as she meets him. The strategy then is to have the Princess reject, but record the highest score of, the first N princes that she meets. The Princess will then choose the first Prince that she subsequently interviews whose score exceeds that recorded score.
That's it. The puzzle is basically solved. Except, of course, to decide on the optimal value of N. It requires some thought. If N to too high, the most wonderful prince is likely to be eliminated at the outset, and she ends up with the last guy. If N is too small, the Princess will likely settle for a fairly undistinguished prince.
What value of N optimally balances these two risks? What is the probability that the Most Wonderful Prince will be chosen?
Disclaimer: I recall this puzzle being posted before, with different flavor text. And it's somewhat of a classic. To give it a fair play here, I'll ask not to post any links and not to just give the answer if you know it, at least not without "showing your work."

There is survival in numbers.
SpoilerSay the colony numbered N squirrels, and they each dug up 2 holes.
They would create N piles of nuts whose average size is 101 and whose standard deviation is about 40.
So if they share among themselves, they will all get through the winter.
If each eats only what s/he finds, maybe only 2/3 of them will survive. 
12 hours ago, harey said:Though they are similar, I see at least one huge difference. In the traffic jam puzzle, you cut ANY part. In the marble problem, you cut the REMAINING part. Enough for different formulae.
The cars in front of the slowest car are the remaining cars. (Each slowest car captures those behind it.)

I think this puzzle is exactly the same as the traffic jam puzzle I posted recently, although that's not obvious at first glance.
@plainglazed found a solution in which he formed clusters of cars by recursively locating the slowest of a group of cars, assuming on average it was in the center of the remaining cars. This corresponds to "grabbing" on average onehalf of the remaining marbles from the bag. Picture the marbles in a line and, grabbing a random percentage of them starting from one end. This has to end up having the same number of marble grabs and car clusters. It leads to a logarithmic answer, but to the wrong base  it should be the natural log_{e}, not log_{2}, which gives too large an answer. Instead of decreasing the number by 1/2 each grab, the remaining number is decreased by 1/e each grab. Same must go for locating the slowest of the remaining cars.
@plasmid found a solution that leads to Sum { 1/k }, which as you point out is ln { n } + gamma, and is here confirmed by simulation. What I can't find is a corresponding analysis for the marbles problem (although plainglazed's approach is applicable to both problems) that will lead to that same sum, instead of yours, where both sums appear to be correct.
This has been fun to think about.
.

That agrees with simulations I ran, which gave, approximately,
Spoiler4, 10, 30, 84 and 244 (geometric series in e) as the number of marbles that require integer numbers 2, 3, 4, 5 and 6 grabs.
So if you multiply the number of marbles by e, it takes one more grab.exp { 1.4 2.4 3.4 4.4 5.4 } =~ { 4 10 30 84 224 }.
That's interesting.
Nice solve.

On 3/22/2018 at 10:27 AM, Molly Mae said:This is a thought exercise I've considered in the past. The answer that I've always come up with is that it doesn't matter when disregarding thickness. If the raindrops are constant and fill any cubic meter equally, he'll encounter the same number of raindrops on his "face" regardless of speed.
The .20m thickness, though, needs to be considered. We know that when he moves at 0m/s, he will encounter some amount of rain along that thickness and he will have made no progress. So to minimize the number of raindrops that lands on top of him, we just increase his speed infinitely. We also know that this can cause severe issues, however.
If you'd like some math to support this argument:
You're right. And it's not much of a puzzle after all.
You have to allow Albert to lean forward as he runs to make this puzzle at all interesting. But even then, the obvious solution is to have Albert lie horizontally and crawl at infinite speed. Only the top of his head gets wet then.
I ran into this puzzle a few years back and "solved" it, more interestingly but also more incorrectly, by multiplying his front and top areas respectively by sin theta and cos theta where theta was determined by his speed compared to the speed of the rain, and some other stuff. It was nonsense.

On 3/21/2018 at 1:18 AM, aiemdao said:??? there are 3 digits before green digits , but there are 4 subtraction different from all after
More clues...
Spoiler_________________
x x.x x x x x x x x x x <  Showing the decimal point in the quotient

x x x x x x / x x x x x x x
x x x x x x <+
 
x x x x x x 1 < clue 
x x x x x x 
 (These create the first three digits)
x x x x x x 0 < clue 
x x x x x x x 
 
x x x x x x 0 <+
x x x x x x 
 
x x x x x x 0 < etc. 
x x x x x x x 
 
x x x x x x 0 
x x x x x x x 
 (These create the repeating digits)
x x x x x x 0 
x x x x x x x 
 
x x x x x x 0 
x x x x x x x 
 
x x 0 0 0 0 
x x x x x x 
 
x x x x x x <+ 

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.
Squirrel
in New Logic/Math Puzzles
Posted · Report reply
One million simulations:
# holes % of squirrels that find 100+ nuts in their holes
1 hole 1.01%
2 holes 50.40%
3 holes 32.85%
4 holes 11.98%
5 holes 3.04%
6 holes 0.60%
7 holes 0.10%
8 holes 0.01%
9+ holes 0.00%
Every hole has a 1% chance of containing a single nut.
If you want to guarantee all squirrels survive, they must each dig up 100 holes