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Chances of playing the exact same game of pool

  1. Jan 19, 2014 #1
    So I was thinking what would be the chances of two players on seperate pool tables playing the exact same game of pool? So player 1 breaks and the balls are scattered. What are the chances that player 2 breaks and the balls just so happen to scatter and come to rest in the same position as player 1?

    To make it less complicated obviously I don't want to calculate every particle, just that each ball lands in the same position. I literally have no idea how to go about working this out and I was just thinking could I say something like... The total number of individual places for a ball to come to rest is about 1097 based on a full size playing table and professional issue billiard balls. So a regular game of pool consists of 16 balls including the cue ball and since no ball can occupy the same space as another there is 1081 empty positions.

    I'm trying to figure out what the probability is that each ball from table 2 will come to rest in exactly the same positions as each ball from table 1.

    Each ball from either table is numbered and ball number 1 has to come to rest where ball number 1 from table 1 did ect. This is simply just a question of curiosity, it isn't homework or anything so if you know the correct math notion ect and how to work it that would be great.

    I understand that some balls might be potted upon the break ect but for the moment we will assume that no balls are potted.

    P.S obviously I know there are millions if unique positions but for simplicity a unique position is defined by a the balls diametre.
     
    Last edited: Jan 19, 2014
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  3. Jan 19, 2014 #2

    mathman

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    Where did you get the numbers 1097 and 1081 from? Wouldn't the number be infinite?
     
  4. Jan 19, 2014 #3
    I'm not sure it's infinite, as presumably there's Planck length type discretisation the might restrict it, though I suspect that if it is finite it's probably a little more than 1081...
     
  5. Jan 19, 2014 #4
    Well it was just an approximation. If you take 1097 billiard balls and put them on the table, there will be no places left for anymore balls. Just as a triangle doesn't have enough room for 16 balls and can only fit 15.

    I understand that there are many many unique positions for a pool ball but if you saw I said for simplicity assume that a unique position is defined by its diametre. I know you can place a ball on the table and call that position 1, then you can move the ball a nanometre to the left and that will be a new position, and again 1 nanometre to the left and that will be a new position so technically there are trillions of truely unique positions.

    But it will be impossible to calculate that? So I said just assume that a position is defined by the diametre of the ball. Attached picture shows 1097 balls
     

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    Last edited: Jan 19, 2014
  6. Jan 19, 2014 #5

    SteamKing

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    I think the fallacy in your argument is that you assume a given ball can only occupy one of the slots in your diagram. A real game of pool is much messier, since a ball can occupy almost any space on the table.
     
  7. Jan 19, 2014 #6
    Yes I understand that. OK scrap what I said originally then about individual spots. How can we calculate the number of unique spots any given ball can occupy? I understand that are too many variables in reality to calculate it accurately but how can we guestimate it?

    So to make this easier let me just test it with this value and you guys can tell me if it makes even the slightest bit of sense.

    Say there are ##16## balls and ##x## unique number of positions that a given ball can land in. ##(\frac{1}{x})^{16}## but we need to know or approximate how many unique positions there are. I said at the start to just use 1097 for simplicity and to get the ball rolling (pun intended) so ##(\frac{1}{1097})^{16}=\frac{1}{4398514633741029999818369522798225510625321083521}## but I don't know if this is correct or what the number actually represents. I think this is the probability of 16 balls from table 2 matching up with the positions of table 1 after the break. Assuming no balls were potted of course.

    Is this correct so far? Not only do the balls have to land in the same positions as table number 1 but the same balls have to land in the exact positions. Obviously to keep track of the balls positions during the break we would number them before hand.

    Please help :/
     
  8. Jan 19, 2014 #7

    AlephZero

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  9. Jan 19, 2014 #8
    FFS people, OP acknowledged in the first post that the "real" problem was significantly more complicated than his watered down version. Maybe PF needs an "Overly pedantic and completely unhelpful" button to go along with its "Thanks" button.

    I think one thing your model is lacking is a way to discern a "legitimate" layout of billiard balls from one that is not, where an illegitimate layout would involve some overlap of of the balls. For instance, you can't have two balls occupying the same grid slot in your setup. This is something that your ##(1/x)^{16}## computation doesn't take into consideration. That'll make the actual computation a little more complicated, but not by much.

    If I could offer a little advice on maybe rephrasing your problem in a way that is a bit more realistic (and perhaps invites a bit less ***hattery), you could ask what is the probability that two games of pool appear to be the same. You are then permitted to define appear however you choose; for example two layouts appear to be the same if the position of each ball on the second table is within ##x## of the position of its counterpart on the first table, where ##x## is some unit of distance.
     
  10. Jan 20, 2014 #9
    Surely Heisenberg uncertainty would cause issues with replicating the exact game?
     
  11. Jan 21, 2014 #10
    You are using sampling with replacement, when you had ought to be using sampling without replacement.

    1907 positions, sixteen balls. The first ball can go in any of 1907 positions, the second into any of 1906, etc. So there are 1907!/(1907-16)! possible such combinations.
     
  12. Jan 22, 2014 #11
    Whoops. That would be correct for identical balls, but the balls are numbered so there are more combinations. I don't recall how to do it right now.
     
  13. Jan 22, 2014 #12
    So this answer that we have based this off is on the assumption that there are only 1097 unique spots. We can keep refining this value to get a more accurate answer. Assume that each unique position is determind by 1mm. We work out how many millimetres a table consists on and then simply use that figure instead of 1097?

    I know one step at a time and in reality there probably isn't a limit once we go quantum but for now I'm happy to reduce it gradually. So it would be [tex]\frac{3757910!}{(1907-3757910)!}[/tex] but again this doesn't take into account the numbered balls. I think it's safe to say the number just grows insanely larger as the length decreases.

    Again there is no point behind this post it was merely for curiosity :P
     
  14. Jan 22, 2014 #13
    [tex]\frac{3757910!}{(3757910-16)!}[/tex]

    To take care of the numbering of the balls I think it is necessary to multiply by 16!. It's the number of physical layouts times the number of permutations of the order of the balls. So

    [tex]\frac{3757910!16!}{(3757910-16)!}[/tex]
     
  15. Jan 22, 2014 #14
    [tex]\frac{3757910!}{(3757910-16)!}[/tex]

    You'd have to account for each ball covering more than one lattice point somehow. I don't feel like doing it.

    To take care of the numbering of the balls I think it is necessary to multiply by 16!. It's the number of physical layouts times the number of permutations of the order of the balls. So

    [tex]\frac{3757910!16!}{(3757910-16)!}[/tex]
     
  16. Jan 28, 2014 #15

    FactChecker

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    I think you were right the first time. Ball 1 can be in any of the 1907 positions, ball two in any of the remaining 1906 positions ... So that gives the final answer of 1907!/(1907-16)! = 2.87215784575433e+052. If the balls were identical and could be swapped around without changing the answer, you would divide the answer by 16!
     
  17. Jan 28, 2014 #16

    WWGD

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    Maybe you can use some results from the ball-packing literature --yes, it's true, but let the jokes roll-in. Maybe you can see the paper" ball-packing in fock spaces of tight curvature" ( if you can find it, since I just made it up :) )
     
  18. Jan 29, 2014 #17
    Think of it this way. First find the answer for unnumbered, identical balls. In other words, count the number of subsets of size 16 of the set of locations. That's 1907!/(1907-16)!. Then multiply by the number of permutations of balls for each subset of locations. That's 16!
     
  19. Jan 29, 2014 #18

    FactChecker

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    It's the reverse of that. The number of permutations of 1907 objects taken 16 at a time is 1907!/(1907-16)!.
    That is where order counts and the balls are numbered and considered different. If all the balls are considered the same, the number of combinations is 1907!/( (1907-16)! * 16!)
     
  20. Jan 29, 2014 #19
    Interesting idea.

    So far I see discussion of ball positions after each shot, and how to "resolve" those to the table surface.

    What about ball rotation orientation? If the two ball ends up in the same position but in one case the label "2" points one way and in the other case it points another direction, is that considered the same or different?

    Likewise, the same final resting position of the two ball may have come from simply rolling straight a few feet, or as a result of contact with other balls, or it may have bounced off the bumpers to get there, or a combination of these; are those the same?

    Since we're just thinking out loud, the way I would approach this question would be to try to take the phrase "exact same game" very seriously to include not just each post shot configuration, but all the motions, velocities, accelerations, vibrations, contacts with other balls and bumpers, slides, spins, final resting orientations, etc..., as well as the sequence order of the pocketed balls, etc...

    That sound like a big order, but to me it boils down to two big issues:

    How to gauge the physical resolution.
    How to represent each possible momentary configuration (before, during, and after each shot).

    To gauge the physical resolution, I would make a variable "z" that could be changed to "zoom in" indefinitely close.

    To represent the configurations (including motions), I would use a "three dimensional picture" like a phase space comprised of pixels, the pixel size based on the "zoom" variable "z".

    The picture approach might be done like this:

    Taking a single xy plane as an example, a screen is conceived to represent that plane, and the boundary of that screen matches the aspect ratio of the table.
    Each pixel within that boundary may be assigned a value:

    0=just empty table space unoccupied by a ball
    1-16=occupied by part of a ball (number being that ball's number and calling the que ball "16")

    Two basic steps take place:

    1] Assign a resolution based on "z" to the screen

    2] Compute every possible screen picture at that resolution and give it a number by:

    a] Starting with all pixels set to 0

    b] Treat the total string of pixels as representing a number and incrementing
    (so the first screen's number, for example, is 00000 from top left, then next line 000..., then next line 000..., down to bottom right 000) so first screen's number is all 0's. 00000...0000

    c] increment to get 00000...0001 That is the next picture
    Repeat to get 00000...0002 That is the next picture
    The increments are base 17 (0-16)
    Eventually you will finish up with the second to the last picture number being
    17 17 17 17 17...17 17 17 16
    and the last one will be
    17 17 17 17 17...17 17 17 17.

    Now, the total number of unique distinct pictures is going to depend on the resolution variable z chosen for the screen.
    If z provides that there will be 1000 pixels in this plane, then the number of pictures will be
    17^1000
    If z provides 10,000 pixels, then no. of pictures is 17^10,000

    This is the number of unique pictures only for that plane, all three dimensions need to be represented, and the z variable will determine the number of planes in each dimension, so what you end up with is a large set of 3d representations of every possible 3D representation, within the limits of the resolution variable z. Most of these will not represent a physically realizable possibility in a proper game.

    The next step is to eliminate all the static representations that are physically impossible or unrealizable in a proper game with a set of rules that perform tests on each 3d representation:

    1] Only whole balls correctly represented, no overlapping, or partial balls, etc...
    2] Only 16 or less unique balls (with que ball if no scratch)
    3] etc...

    So now you have a set of representations of all possible static moments in all possible proper games, to the limits of the z resolution variable.

    Then compose all possible sequence orders of all "good static" representations and use classical physics rules to filter out the dynamically impossible ones with respect to time:

    1] Start with all balls and cue ball
    2] All balls slide, roll, contact, bounce, and move at correct angles and speeds
    3] Ball don't disappear, reappear, break apart, float, change size or number label, or otherwise misbehave, etc...
    4] Continuity and consistency with classical physics, etc.

    The result will be a set of 3d representations of all possible dynamically physically correct proper pool games achievable at the resolution of z.

    Now start playing with z...

    Assuming this is modeled with a computer, you may run trials with increasingly fine z resolutions until the total number of unique dynamically physically correct proper games looks like it is converging or appearing to break down approaching the quantum levels... my sense is that it would likely converge to a very large value long before the quantum level.
     
    Last edited: Jan 29, 2014
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