harey

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There are 100 (quite small) marbles in a (quite large) jar.
Pull out one marble, look at it, put it back.
You have done it 100 times. All marbles were white.Would you bet 5:1 that they are all white?
Bonus (and most interesting) question: Suppose you have 1520 seconds to decide.

The speed is a real number. As there is an infinity or real numbers, p(two bullets have the same speed)=zero, at least for a "small" n. The same way we do not suppose that 3 (or more) bullets can collide.
However, there is something else that gives me headaches.
It works when there are n bullets in line AFTER all n bullets were shot. What if the first two bullets collide and then n2 bullets are shot? More generally, what if there are collisions before all bullets are shot?

Yes, the "official" one.
But as I know it, it would be unfair to publish it. So I will leave it to someone else. We gave some hints now, did not we?

The list of given examples is not exhaustive nor limiting. You can tailor a property to any number and we can argue ad eternum whether this property is interesting or not. Just my personal opinion

Well, I know the "official" answer to this problem, but I do not agree with it. Human language is very imprecise, "interesting" is quite subject to interpretation.
I.e. my ex does not find 4 interesting at all. 2+2=2*2=2^2? So what?

I hope I got it:
General idea: As long as the first bullet in line is not the fastest, there will be a collision leading to a system of (n2) bullets. p(the first bullet in line is never the fastest) is not so hard to calculate.
More in detail: Let's suppose the bullets move from left to right and that the rightmost bullet is numbered bullet_1. When reading, think of a convenient n, something like 8 or 12.
There are two possibilities:
A1) bullet_1 is the fastest, it will not be reached by any other, p=1/n, all hope is lost.
B1) bullet_1 is not the fastest, p=(n1)/n
There WILL BE a collision. (It does not matter whether bullet_1 is involved or not, whether the slowest bullet is involved or not...) After the collision, we still have (n2) bullets. If necessary, the rightmost is renumbered as bullet_1.
Again, there are two possibilities:
A2) bullet_1 is the fastest, it will not be reached by any other, p=1/(n2). [There are (n2) bullets now.)]
B2) bullet_1 is not the fastest, p=(n3)/(n2). [There are (n2) bullets now.)]
There WILL BE a collision. (It does not matter whether bullet_1 is involved or not, whether the slowest bullet is involved or not...) After the collision, we still have (n4) bullets. If necessary, the rightmost is renumbered as bullet_1.
Again, there are two possibilities:
A3) ...
B3) bullet_1 is not the fastest, p=(n5)/(n4).
This will continue until B(n/2) and no more bullets left.
To annihilate all bullets, must occur B1 and B2 and B3 and and B(n/2).
p(annihilation)=p(B1) * p(B2) * p(B3)=
(n1)/n * (n3)/(n2) * (n5)/(n4) * * (1/2)Writing backwards, the conjectured formula: (1/2)*(3/4)*(5/6)* *(n1)/n.
Surprisingly, it does not matter whether the bullets are shut in regular intervals or not.
 1

On the right track but I have a different time.
I calculated very roughly. Interpolating an exponential is quite dangerous. If the temperature was rounded by 0.2 degrees and measured 23 minutes before/after the whole hour, you can get quite a different result.
Proof is left to the reader.

The body temperature decreases exponentially. This means 84 at 8:00, 92 at 7:00, 100 somewhere before 6:30.
As you do not study from books anymore but online, various logs will show active connection from my room.

Don't I still need to compute collision times to discover whether it's zilch or escape?
Not necessarily.
v3>v2>v1 and diff_1<>diff_2 assumed
diff_1=v2v1
diff_2=v3v2
if(diff_2>diff_1) then [3 reaches 2] else [2 reaches 1]I just cannot figure out p(diff_2>diff_1).

On the first roll, p(12)=1/36, p(twice 7)=0
On each consecutive roll, p(12)=1/36 and p(twice 7)=1/36
p(12 wins)=1/36 + x
p(twice 7 wins)=x
p(one of them wins)=1=(1/36 +x) + (x)=>x=35/72
=>p(12 wins)=37/72=51.39% which does not correspond to the simulation.Where am I wrong?

Scores:
0:0 (p=1), then
1:0 or 0:1 (p=0.5+0.5), then
2:0 or 1:1 or 0:2 (p=0.25+0.5+0.25)
...
7:0 6:1 5:2 4:3...Pascal's triangle, but I will not hazard myself into calculations.

@hhh3: Can you tell how?

What about a supplementary rule excluding Fred has 354?

I have got 3 solutions, but I do not really understand the point (9)
Andy: 1AB (4)  the smallest number
1A5 (4)  divisible by 5
125 (11)  A=1 or 2, 1 excluded (2)Tim:
1st digit: at least 3 (8)
2nd digit: 4 (11)
3rd digit: 2 (7), Bob has at least 2
342/2=171: not prime
542/2=271: primeBob:
1st digit: 2 3 4 (1 excluded, (Andy); 5 excluded (6))
2nd digit: 1 3 5 (10)
3rd digit: 1 3 (10) (5 excluded (Andy))
w/o repetition (1) and divisible by 3 (5) leaves:
213
231
351
453Fred:
1st digit: 3 4 (1 excluded, (Andy); 2 excluded (6); 5 excluded (Tim)
2nd digit: 3 5 (10) (1 is not considered as prime)
3rd digit: 1 2 3 4 (5 excluded (Andy))
w/o repetition (1) and divisible by 3 (5) leaves:
351
354
453This excludes 453 for Bob (6), leaving for Bob:
213
231a) Let's assume Fred has 354:
Greg:
1st digit: 4: 7fred (1 andy, 2 bob, 3 fred, 5 tim)
2nd digit: 1, 3: 4bob (2 andy, 4 tim, 5 bob)
3d digit: 1, 3: 4bob (2 tim, 4,greg 5 andy)
> 413 compatible with Bob (1st solution)
> 431 compatible with bob (2nd solution)a) Let's assume Fred has NOT 354:
fred: 351 or 453
bob: 213 or 231
> greg's 4th digit=4
> fred 453 (someone has to have the first digit 4)
> bob 231 (213 excluced)
> andy 125
> tim 542
> greg 314 (3rd solution) 1

I calculated with 660.60 and the price is 600.60.
Just a small typo as I often do...
 1

Set aside the sets calculated by Bonanova:
13,519.90(4*660.60)(1*1501.50)=9,376As 600.60*5=1501.50*2=3003=3*1001, remain TVs for 231, 273, 429, 1001 (multiple solutions reserved).
231=3*7*11
273=3*7*13
429=3*11*13
> all divisible by 39+3+7+6=25; so 9,376 is not divisible by 3;
> there are at most 8 TVs priced 1,001 and only 7,374 and 1,368 fits.The same way substract multiples of 231 and consider multiples of 39 (3*13, common divisor of 273 and 429).
Alternatively, I think the problem could be solved by Diophantine equations, but I never studied them. 
What about:
(12*231)+(9*273)+(5*429)+(4*660.60)+(1*1501.50)+(2*1001)=13519.90

I do not follow here. Worst case: the cards are sorted.
Bonanova wrote:
What is the expected number of cards in the left pile after all N cards have been drawn?
OK, that answer is not difficult to determine, since the definitions of the piles are symmetrical.

100% loss
if the bin contains even numbers starting at 4

by ignoring travel perpendicular to the direction of the wind, where it has no effect on the plane's speed
Even here, it increases the travel time.
A nonzero component of the plane's speed is used to compensate the wind.

This assumes that a point is (at least partially) illuminated if a straight line can be drawn uninterrupted between it and a light source.

Make the sides as small as possible and you approach a circle.

These thoughts suggest what I think is an interesting question:
for each N, find an example of a convex solid that can be illuminated with the fewest lights.
That's the start to the original question
What bothers me with points and lines:
We do not illuminate an object as such, we illuminate it's surfaces. (Already a problem with a sphere and alike: in this case, we may not consider the sphere has one surface, but infinity.) If a surface is illuminated, does it imply that it's edges and vertexes are illuminated, too? One answer would be that this question does not have a sense, but it seems a little week as argument.

Not so quickly...
If we need (in 3D) four lights for a sphere, a cylinder can be illuminated by three lights only. (I spent a long time to find a way to show that the edges can be smoothened and still illuminated until I consulted http://en.wikipedia.org/wiki/Convex_body and realized that a cylinder is a convex solid).
For a cone, two lights are enough.
What bothers me at most: Consider a unit cube in the xyz coordinates and a line (0;0;0)  (1;1;1). If we need 0 lights to illuminate a point or a line, the whole cube would be illuminated by just two lights placed on this line just slightly outside the cube.
100 white marbles
in New Logic/Math Puzzles
Posted
@bmad: yes, "nonmathematical" equivalence of "the jar is quite large" and "the marbles are quite small"
@jasen: In overwhelming majority of cases, I base my decision on probability, very exceptionally on luck.
@bonanova: The answer I expected. I would reason something like that: if there is one black marble, p(black)=1/100. Repeating 100 times, I would estimate p(all white) about 95% and would be very surprised if I were told it is less than 90%.
Knowing the solution, I recall the problem of n people having birthday the same day.
However, your formula is missing something. If you complete it, you will get another surprise. (Please hurry, I do not want to be faced with the problem of two best answers again.)