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Posts posted by harey

  1. @bmad: yes, "non-mathematical" 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.)

  2. 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 15-20 seconds to decide.

  3. 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 n-2 bullets are shot? More generally, what if there are collisions before all bullets are shot?

  4. 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 (n-2) 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=(n-1)/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 (n-2) 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/(n-2). [There are (n-2) bullets now.)]

    B2) bullet_1 is not the fastest, p=(n-3)/(n-2). [There are (n-2) 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 (n-4) bullets. If necessary, the rightmost is renumbered as bullet_1.

    Again, there are two possibilities:

    A3) ...

    B3) bullet_1 is not the fastest, p=(n-5)/(n-4).

    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)=
    (n-1)/n  *  (n-3)/(n-2)  *  (n-5)/(n-4)  *    * (1/2)

    Writing backwards, the conjectured formula: (1/2)*(3/4)*(5/6)*  *(n-1)/n.

    Surprisingly, it does not matter whether the bullets are shut in regular intervals or not.

    • Upvote 1
  5. 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 2-3 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.


  7. 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

    if(diff_2>diff_1) then [3 reaches 2] else [2 reaches 1]

    I just cannot figure out p(diff_2>diff_1).

  8. 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)

    =>p(12 wins)=37/72=51.39% which does not correspond to the simulation.

    Where am I wrong?

  9. 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)

    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: prime

    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:

    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:

    This excludes 453 for Bob (6), leaving for Bob:

    a) Let's assume Fred has 354:
    1st digit: 4:    7-fred (1 andy, 2 bob,  3 fred, 5 tim)
    2nd digit: 1, 3: 4-bob  (2 andy, 4 tim,  5 bob)
    3d digit:  1, 3: 4-bob  (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)

    • Upvote 1
  10. Set aside the sets calculated by Bonanova:

    As 600.60*5=1501.50*2=3003=3*1001, remain TVs for 231, 273, 429, 1001 (multiple solutions reserved).

    -> all divisible by 3

    9+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.

  11. 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.

  12. 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.


  13. 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.

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