This Forum and Equations

vapoolplayer said:
Just for curiosity sake, wouldn’t it be more than 1258?

For example, pretend the pockets are corners and not pockets. You place the cue ball tight to a corner. Now move it a microscopic distance from the side rail, but still on the short rail. Then continue that microscopic movement all the way down the rail until you touch the other side.

Those are all different positions that ball could land in and be touching the rail.

Now you have to move it away from the short/end rail a microscopic distance. And then start working backwards.


Wouldn’t that equate to a ridiculously high number that’s not just as simple as 1258 positions?
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Wouldn't yield a significant difference

Bigger consideration is the additional space balls take up when they are less than a balls width apart from each other and/or the rail.
 
vapoolplayer said:
Considering you can fit 968 balls on a 50”x100” playing surface….

That’s if they all only had close to an exact 2.25” spot to sit.

Now you have 10 balls that have the room not being taken by about 958 balls.

And, just moving them .25” or less would constitute a “different” position.

“A lot” would be a huge understatement.
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Not fixed balls, won't it be in the hundreds of millions ?
 
SBC said:
Easy
Surface area of pool table is 50x100= 5,000 square inches

A 2.25 inch diameter ball takes up 3.976 inches of surface area at it's circumference...

Take 5,000 inches divided by the surface area of a ball....that's 1,258. That's how many balls that could fit on a 9 foot table.

To figure all the possible combinations you need to do a mathematical calculation credit a Permutation Without Repetition.

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
Or 9.577 million trillions

It's a shit ton, but there's the math.
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Okay and this is only to get the location of the balls one time, Now what about the different configuration if each ball is moved in any direction one pin head sitting its spot ? Now break the 9 balls and how many patterns could there be ?
 
SBC said:
Wouldn't yield a significant difference

Bigger consideration is the additional space balls take up when they are less than a balls width apart from each other and/or the rail.
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Again, just for conversation sake…..

How would it not make a significant difference?

Let’s say I could fit 44 balls from one corner to the other corner on a head rail. Sitting next to one another.

But I remove 43 and then start barely moving the ball as mentioned above.

That would end up with possibly ~ 400 locations for that ball instead of 44.
 
rjb1168 said:
Who really cares?
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Anybody that sets up a shot after game over to practice a miss, wouldn't it take close to forever to find that same spot to place the ball that will shoot the same. I keep hearing about patterns after break and how some people can break so much same each time. Don't we put this in our subconscious as simple ? Also how easy the pool game can be ? Guy
 
SBC said:
A 2.25 inch diameter ball takes up 3.976 inches of surface area at it's circumference...

Take 5,000 inches divided by the surface area of a ball....that's 1,258. That's how many balls that could fit on a 9 foot table.
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If they could fit together without leaving any gaps.

But with round objects there are gaps.

pj
chgo
 
vapoolplayer said:
Again, just for conversation sake…..

How would it not make a significant difference?

Let’s say I could fit 44 balls from one corner to the other corner on a head rail. Sitting next to one another.

But I remove 43 and then start barely moving the ball as mentioned above.

That would end up with possibly ~ 400 locations for that ball instead of 44.
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I think the point is that to a human there is no difference between 1e25 and 1e50, which is probably the range of permutations depending on how we fix size and position. Both are infinite numbers to our ape brains.
 
Help us.
 

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Hey you need to redo your calculations. Since the balls are 2.25” you have to calculate on a 47.75”x97.75” platform. Not a 50”x100” one. And shelf size also will play a role in the calculation.
 
vapoolplayer said:
That makes sense.

Big number is big number.
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There are more possible pool ball permutations than grains of sand on Earth, but fewer ball permutations than the number of atoms in the observable universe.

I think an interesting twist is how many pool ball locations matter to a decent player? Rail or not rail matter, but an inch here or there for most shots does not. So you could do an area analysis to define theses areas and wind up with something much less than 1000 possible ball positions. I bet it's closer to 100 ball positions that matter. But that still results in 100!/90! Permutations, which is like 6e19 or
62815650955529472000. At one minute per rack, that's 119840603929200 years to go through all of them,.which is longer than the current age of the universe.
 
kling&allen said:
It's one of my favorite things about pool. I love chess also, but chess openings are always the same and largely played from memory, so many games can be repetitive at my low skill level. Ever rack of 9 ball seems different to me.
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You need to watch Corey Deuel more.

As for the OP.... Those who think moving a ball a tenth of an inch doesn't make any difference to its position should try spotting a second ball at one pocket. Leave it a tenth of a millimeter from the first ball and your opponent will be down upon you like a rabid weasel.

I'll go for a grid that is in 100ths of an inch. That's a little too coarse to make sure two balls are frozen, but let's see what it gets us.

The first ball can be in any of 4800 locations across the table and 9800 along the length. Call it 5000 and 10000 for convenience. That first ball can go in any one of 50,000,000 locations. There are about as many locations -- within a little -- for each additional ball That gives us
50,000,000 to the 16th power for all the possible locations as a very conservative estimate. 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
unless I slipped a decimal point. It would take a long time to practice each one.

I've seen someone quote Albert Einstein saying there are 6 million shots on the pool table. I don't think he said that.
 
vapoolplayer said:
Again, just for conversation sake…..

How would it not make a significant difference?

Let’s say I could fit 44 balls from one corner to the other corner on a head rail. Sitting next to one another.

But I remove 43 and then start barely moving the ball as mentioned above.

That would end up with possibly ~ 400 locations for that ball instead of 44.
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Actually you are on to something.

My surface area calculation doesn't take into account the space between balls. So it's less.

22 balls wide, 44 balls long = 968

So the answer is
968x967x966x965x964x963x962x961x960x959=
6.8944225e+29
Or
689,442,250,000,000,000,000,000,000,000
 
Bob Jewett said:
You need to watch Corey Deuel more.

As for the OP.... Those who think moving a ball a tenth of an inch doesn't make any difference to its position should try spotting a second ball at one pocket. Leave it a tenth of a millimeter from the first ball and your opponent will be down upon you like a rabid weasel.

I'll go for a grid that is in 100ths of an inch. That's a little too coarse to make sure two balls are frozen, but let's see what it gets us.

The first ball can be in any of 4800 locations across the table and 9800 along the length. Call it 5000 and 10000 for convenience. That first ball can go in any one of 50,000,000 locations. There are about as many locations -- within a little -- for each additional ball That gives us
50,000,000 to the 16th power for all the possible locations as a very conservative estimate. 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
unless I slipped a decimal point. It would take a long time to practice each one.

I've seen someone quote Albert Einstein saying there are 6 million shots on the pool table. I don't think he said that.
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LOL, at the Corey Deuel statement.

And you hit the nail on the head with the grid. Just how much movement is enough to warrant a "different" shot?

A ball is 2.25 inches wide, table is 50" (W) x 100" (L).

50" - 2.25" = 47.75". Divide that by however far it's decided you need to move the ball.

If we go with your .01 that would be 4775 + 1 (original spot) for a total of 4776 possible locations in the first row.

Going the long way it's 97.75/.01 = 9775 so there are 9776 spots going the long ways down the rail.

Multiply those and you get 46,690,176 possible spots given a 100ths of inch allowance for movement between shots.

That does not include the space in the pockets of course.

If you change the .01 to .001 it becomes

W - 47751 options
L - 97751 options
Total would be 4,667,708,001 - a little over 4.6 billion possible shots - not including pockets ;)

And if you want to keep going - if it's .0001 it becomes

W - 477,501
L - 977,501
Total - 4.66758E+ 11

Excel for the win ;)

EDIT: LOL, that doesn't include diagonal movement. Headed out the door, will have to think about that later.
 
Bob Jewett said:
You need to watch Corey Deuel more.

As for the OP.... Those who think moving a ball a tenth of an inch doesn't make any difference to its position should try spotting a second ball at one pocket. Leave it a tenth of a millimeter from the first ball and your opponent will be down upon you like a rabid weasel.

I'll go for a grid that is in 100ths of an inch. That's a little too coarse to make sure two balls are frozen, but let's see what it gets us.

The first ball can be in any of 4800 locations across the table and 9800 along the length. Call it 5000 and 10000 for convenience. That first ball can go in any one of 50,000,000 locations. There are about as many locations -- within a little -- for each additional ball That gives us
50,000,000 to the 16th power for all the possible locations as a very conservative estimate. 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
unless I slipped a decimal point. It would take a long time to practice each one.

I've seen someone quote Albert Einstein saying there are 6 million shots on the pool table. I don't think he said that.
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This^^^

Consider that there is an INFINITE number of positions for just ONE ball on the table, we’d have to establish what constitutes a ball being in a different position. 0.0001”? 0.1”.

I like Bob’s solution but we’d need to consider the positions where two balls would occupy the same space, which would eliminate a bunch of configurations in Bob’s calculation. But agreed it’s mind-numbingly “a lot”.

Why is this even relevant?
 
dquarasr said:
... but we’d need to consider the positions where two balls would occupy the same space, which would eliminate a bunch of configurations in Bob’s calculation. ...
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Not really. All the balls together occupy less than 2% of the area of the table. That's down in the noise for the kind of approximation I did.
 
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