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SphinxnihpS
Banned
I_Need_D_8 said:
Easy? this problem is far more difficult than the above. It requires the above just as a function of a much larger simulation.
Throw the 7 stripes plus the 8-ball, plus the cueball on the table and you get a random arrangement that can be expressed but not predicted by the above calculation. Since it can not be predicted, you must now run a simulation of many itterations. Each itteration will produce a different pattern of balls on the table. Now how do we shoot them? How hard do we hit them? Which first, which last? What English, etc, ad nauseum? With only very few variables you could easily construct a simulation that requires eons to run, and far more time to interpret the output. In other words, it's better to just go do it in reality.
SphinxnihpS said:Easy? this problem is far more difficult than the above. It requires the above just as a function of a much larger simulation.
Throw the 7 stripes plus the 8-ball, plus the cueball on the table and you get a random arrangement that can be expressed but not predicted by the above calculation. Since it can not be predicted, you must now run a simulation of many itterations. Each itteration will produce a different pattern of balls on the table. Now how do we shoot them? How hard do we hit them? Which first, which last? What English, etc, ad nauseum? With only very few variables you could easily construct a simulation that requires eons to run, and far more time to interpret the output. In other words, it's better to just go do it in reality.
Sorry I wasn't clear. I was referring to the 'order' in which the balls could be run. I'm sure it's a factor of 7, but how does the 6 different pockets figure in?
SphinxnihpS
Banned
The number of pockets does not matter. The answer is simply 7! (because the 8 ball is always last), or 5040 different ways to run a rack of eightball, which is more than the amount of ways to leave a lover.
Last edited:
SphinxnihpS said:
That is exactly what my calculation states. I said @ the end... "Take the answer and power it to the number of balls on the table." Or am I wrong?
9 Ball (9 balls + 1 cueball): 4.35E+60 is like saying 4.35 x 1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000.
JV<-- still laughing over that damn bunny...
Verysmallrocks
m0ney $hot
I saw this problem long ago when I was somewhat of a lurker here and always wondered what the answer was. I recently remembered it and spent some time trying to dig up this thread. the answer may surprise you!
I used a table resolution of 1/64 inch rather than 1/16 inch to make the answer a little more real world. I suspect that 1/64 of an inch is about the lower limit you can move a pool ball on a clothed table.
But for this problem you must account for the other balls, since no two balls can occupy the same space. So the answer for a table with a resolution of 1/64 inch must subtract the answer for a table with the resolution of one ball.
My table fits about 24 balls end to end on the short rail and 48 balls end to end on the long rail (roughly). This gives 1152 discrete ball positions for any ball to occupy. The 1/64 table is 2.25 (ball diameter) times 64. So the 1/64th resolution has 23,887,872 discrete ball positions.
On the break, any number of balls may go in pockets or off the table. So you may have anywhere from 0 to 10 (nine plus the cue ball) balls remaining on the table after any given break shot.
To figure out 10 ball combinations is relatively easy. 10! x the number of discrete positions for both resolutions, then subtract the lower resolution from the higher resolution.
To figure out anything less is a little bit trickier, but similar.
To figure out all possible combinations requires figuring out all combinations of 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and 0 and adding them all together.
The answer for my table is 87,935,657,238,036, or approximately 8.79 x 10^13. Far less than the OP suggests.
I used a table resolution of 1/64 inch rather than 1/16 inch to make the answer a little more real world. I suspect that 1/64 of an inch is about the lower limit you can move a pool ball on a clothed table.
But for this problem you must account for the other balls, since no two balls can occupy the same space. So the answer for a table with a resolution of 1/64 inch must subtract the answer for a table with the resolution of one ball.
My table fits about 24 balls end to end on the short rail and 48 balls end to end on the long rail (roughly). This gives 1152 discrete ball positions for any ball to occupy. The 1/64 table is 2.25 (ball diameter) times 64. So the 1/64th resolution has 23,887,872 discrete ball positions.
On the break, any number of balls may go in pockets or off the table. So you may have anywhere from 0 to 10 (nine plus the cue ball) balls remaining on the table after any given break shot.
To figure out 10 ball combinations is relatively easy. 10! x the number of discrete positions for both resolutions, then subtract the lower resolution from the higher resolution.
To figure out anything less is a little bit trickier, but similar.
To figure out all possible combinations requires figuring out all combinations of 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and 0 and adding them all together.
The answer for my table is 87,935,657,238,036, or approximately 8.79 x 10^13. Far less than the OP suggests.
Your math is wrong. With only two balls there are nearly (23e6)^2 positions and with 10 balls there are nearly (23e6)^10. You are not counting all the possible combinations correctly.
Obviously your books only have to do with robots and not pool, duh.
That's awesome. Never heard of that one. I might try it on a good player.
I am just here looking for the answer, if someone could help me please.
I am just trying to find out how many different shots I need to practice, if anyone knows please let me know, thank you.:grin:
I am just trying to find out how many different shots I need to practice, if anyone knows please let me know, thank you.:grin:
You only need to practice the last one you missed.:thumbup:
There is no particular number -- the finer you feel like dividing the table up, the more shots you have. To the best of the knowledge of the interwebs, Einstein made no comment about pool.
Not included in any of the analysis above is the shot you intend to play and the position you intend to get from it. If you intend to draw straight back one diamond, there are a bunch of ways just to play that exact position.