- #1
MathJakob
- 161
- 5
So I was thinking what would be the chances of two players on separate 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.
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.
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