This Forum and Equations
- Thread starter Guy Manges
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Yes
No need to define a position of the base of the ball as much. If a ball is in a space, it displaces the other.
We all know the breaks aren't completely random. In every game , balls tend to go certain direction/speed to end up in predictable locations. We see the patterns and often run very similar racks.
Guy Manges
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Many balls to all, Thank you all... Guy
Guy Manges
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Does this make pocket billiards a general game and not a precise game ?
Guy Manges
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What we need to do on the forum is let this whole world know that not to get discouraged because this pool can be a very Hard games... Guy
NP no P no NP WELL MAYBE P NO MP O WELL...
what is a ball configuration?
The answer is infinite unless there’s enough evidence that the center of mass of a ball always lands dead center between the threads on a worsted cloth. Any individual ball being off by 1mm, 0.1mm, 0.01mm……0.1x10^30 mm are all valid places for a ball to land.
Even if that assumption is true (the ball center of mass lands dead between the threads of the cloth), that error is probably well within the manufacturing tolerance of the balls which adds more options and shifts the balls actual position by fractions of a hair.
Even if that assumption is true (the ball center of mass lands dead between the threads of the cloth), that error is probably well within the manufacturing tolerance of the balls which adds more options and shifts the balls actual position by fractions of a hair.
a fraction of a hair is not a unit of measure,,,,,,, but the question is impossible to answer without a clarification as to what exactly a ball configuration is. I asked and he thought it was a funny question, which doesn't provide much hope for an answer.
I think the "ball configuration" is simply however the balls are arranged on the table.
pj
chgo
Then I think it's still impossible to answer.
Its not impossible to answer. The answer is infinite with no cloth and perfectly round balls and perfectly flat table. With cloth the solution can be solved with an upper and lower bound (meaning the real solution is in between a range, but the exact solution is unknown) and over time the upper and lower bound can be tightened as better approximations are found.
So at this time according to this thread there is an upper-lower bound of:
upper: infinite
lower: (958 choose 10) * 10!
And it would not be hard to up the lower bound...You just need to change from 1 inch spots to place balls to 1/2 inch or whatever you want...
Due to the symmetry of a table you are overcounting. And if the order of balls doesnt matter there is more over counting.
The following is a bit too low::------------------------------------------
The formula would be 1,258 times the next 9 descending whole numbers for each less spot on the table taken by the next ball....so
1,258 x 1,257 x 1,256, x 1,255 x 1,254 x 1,253 x 1,252 x 1,251 x 1,250 x 1,249 equals
9,577,054,998,202,898,152,157,349,600,000
That's 9.577 quintillion
---------------------------------------------------------------------------
The following is a bit too high:-----------------------------------------
That is about
1526 00000 00000 00000 00000
00000 00000 00000 00000
00000 00000 00000 00000
00000 00000 00000 00000
00000 00000 00000 00000
00000 00000 00000 00000
-----------------------------------------------------------------------------
The first underestimates the number of unique initial positions;
The later overestimates the unique combinations by allowing 2 balls to overlap.
The later is closer to the correct answer.
The formula would be 1,258 times the next 9 descending whole numbers for each less spot on the table taken by the next ball....so
1,258 x 1,257 x 1,256, x 1,255 x 1,254 x 1,253 x 1,252 x 1,251 x 1,250 x 1,249 equals
9,577,054,998,202,898,152,157,349,600,000
That's 9.577 quintillion
---------------------------------------------------------------------------
The following is a bit too high:-----------------------------------------
That is about
1526 00000 00000 00000 00000
00000 00000 00000 00000
00000 00000 00000 00000
00000 00000 00000 00000
00000 00000 00000 00000
00000 00000 00000 00000
-----------------------------------------------------------------------------
The first underestimates the number of unique initial positions;
The later overestimates the unique combinations by allowing 2 balls to overlap.
The later is closer to the correct answer.
I agree due to symmetry there is definitely an over-count when assigning ball locations. I think how much of an over-count is somewhat a matter of opinion depending upon if the shooter is ambidextrous or not. I suppose the solution to this is to divide (968 choose 10) by 4?
Furthermore, I agree when assigning the actual balls there is also a possible symmetry over-count. It seems to me this issue is harder (but still possible) to take into account than just a simple division by 4 because its more conditional upon the symmetry of the chosen positions than just the symmetry of the table. Maybe you see an exact solution. I don't off the top of my head.
I think the (simple) solution to these over count issues is to further define the problem to include a foot and head side of the table.
Either way good catch of this problem.
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If you know what you're doing...2
Respect the pool grid
40 area sections
all 10 in each section, 40 ways
(further study reduce for mirror images)
Notation: 10 balls in 1 section only 10 ways.
2 balls in 1 section or 2 balls in 2 sections
9 in 1 and 1 in 9, 81 ways
8 in 1 and 2 in 1 or 2 in 2,
7 in 1 and ... 3 in 1 or 3 in 2 or 3 in 3
...
If you finish the calculation you can say we collaborated. Its a math problem solved by student T.
40 area sections
all 10 in each section, 40 ways
(further study reduce for mirror images)
Notation: 10 balls in 1 section only 10 ways.
2 balls in 1 section or 2 balls in 2 sections
9 in 1 and 1 in 9, 81 ways
8 in 1 and 2 in 1 or 2 in 2,
7 in 1 and ... 3 in 1 or 3 in 2 or 3 in 3
...
If you finish the calculation you can say we collaborated. Its a math problem solved by student T.
