Mathematics and Probability as it relates to pool

Over the years I have observed that very few people have a clear concept on probability, and especially so when it relates to pool.

I want to give you a mathematical problem. You can easily solve it with a calculator, but just take a guess first before calculating.

Here is the scenario.....

The game is 9-ball. The player has made 1 ball on the break, so there are 8 balls remaining.

For each of the remaining shots, this player has a 90% chance of pocketing the ball, and a 90% chance of getting position on the next ball.

For the 9-ball, he has a 90% chance of pocketing it, and a 90% chance of avoiding a scratch.

What is the mathematical probability that he runs out?
 
Insufficient information. You would need to know the probability of making the next shot if you failed to get position.
 
Insufficient information. You would need to know the probability of making the next shot if you failed to get position.

That is the right way to think. For this problem we can just say that if you do not achieve position on the next shot, your odds of pocketing the next ball is zero (your run failed).

But we could also do a scenario in the future where you have a 50% of good shape, and a 50% of bad shape. You would then have two different probabilities for pocketing the next shot. That is a more realistic scenario, but for now this is just a basic example.

But the way you're thinking is dead on!
 
Over the years I have observed that very few people have a clear concept on probability, and especially so when it relates to pool.

I want to give you a mathematical problem. You can easily solve it with a calculator, but just take a guess first before calculating.

Here is the scenario.....

The game is 9-ball. The player has made 1 ball on the break, so there are 8 balls remaining.

For each of the remaining shots, this player has a 90% chance of pocketing the ball, and a 90% chance of getting position on the next ball.

For the 9-ball, he has a 90% chance of pocketing it, and a 90% chance of avoiding a scratch.

What is the mathematical probability that he runs out?

I don't think you can determine probability by introducing so many variables or events into your calculations.

The question should be; "With 8 balls left ,and a 90 percent chance of
making each ball, and all other things being equal, what is the probability of player running out?" The probability would be 0.39.

Why consider a .10 probability of scratching on only the 9-ball? How about miscuing? What about a spectator bumping the player? How about accidentally hitting the cue ball on a practice stroke?

Additionally , when you say that he has a 90 percent chance of pocketing the 9 aren't the other variables already calculated in that
number?

I guess your right. My concept of probability is that it is a probability and
cannot be made a certainty because of the infinite amount of variables that would be required. That's why fishing is called fishing and not catching.

Please explain your concept of probability to us.
 
I don't think you can determine probability by introducing so many variables or events into your calculations.

The question should be; "With 8 balls left ,and a 90 percent chance of
making each ball, and all other things being equal, what is the probability of player running out?" The probability would be 0.39.

Why consider a .10 probability of scratching on only the 9-ball? How about miscuing? What about a spectator bumping the player? How about accidentally hitting the cue ball on a practice stroke?

Additionally , when you say that he has a 90 percent chance of pocketing the 9 aren't the other variables already calculated in that
number?

I guess your right. My concept of probability is that it is a probability and
cannot be made a certainty because of the infinite amount of variables that would be required. That's why fishing is called fishing and not catching.

Please explain your concept of probability to us.

The 9-ball is the same as the other balls. It is just that position is not required for another ball, so we can just change the position variable to a scratch variable, so we can easily make a computation.

The point of this exercise was to make a simplified scenario. The two main variables are making the ball, and leaving the cueball in a position so that you can make the next ball. Of course there are many more variables.

You can assign a probability to having a spectator bump you, but the value assigned will be so low that it will not change the results much, therefore we can safely ignore that variable.

The point is to try and get an idea of your own chances of success on a shot, a runout, a game, a series of shots etc.

There is nothing to gain if you have to be bogged down with 500 variables. But we don't need that many. I can set up a shot and get a pretty good idea concerning my chances of pocketing the ball. I don't need to involve an endless amount of variables.
 
over the years i have observed that very few people have a clear concept on probability, and especially so when it relates to pool.

I want to give you a mathematical problem. You can easily solve it with a calculator, but just take a guess first before calculating.

Here is the scenario.....

The game is 9-ball. The player has made 1 ball on the break, so there are 8 balls remaining.

For each of the remaining shots, this player has a 90% chance of pocketing the ball, and a 90% chance of getting position on the next ball.

For the 9-ball, he has a 90% chance of pocketing it, and a 90% chance of avoiding a scratch.

What is the mathematical probability that he runs out?

0.185
.............
 
There is nothing to gain if you have to be bogged down with 500 variables. But we don't need that many. I can set up a shot and get a pretty good idea concerning my chances of pocketing the ball. I don't need to involve an endless amount of variables.

I think that is what I said in my post:
My concept of probability is that it is a probability and cannot be made a certainty because of the infinite amount of variables that would be required.
 
That's why fishing is called fishing and not catching.

I'm not sure why, but I think this is one of the best lines ever!!!


If the player is ME, then the answer is ZERO :o.

All this pool "thinking" sometimes makes my head feel like it's gonna explode :D!!!

Maniac

Not to "pick on" Maniac, but I truly believe this is apropos for most of the pool playing population...I believe the higher echelon in the game have learned over time what shots/moves/positioning gives them a better "chance" at running out or winning - thus they're taking into account the probabilities...however I HIGHLY doubt they have the capabilities of doing the actual mathematical calculations in their heads as they're playing...I think FEW pool players really can!!!

So with that being said, I DO look for a pattern that APPEARS to have a higher probability of winning me the game...but to do the math on it...nope...it's all guesswork...kind of like my pool game...it's all "poke-&-hope"! :o

Jason
 
Given the original sanerio, if it's Shane Van Boneing shooting, he's got probably a 90% chance of running the table. On the other hand, if it's me shooting I'd give it more like 45%. Get what I'm saying ?
 
I don't think you can determine probability by introducing so many variables or events into your calculations.

The question should be; "With 8 balls left ,and a 90 percent chance of
making each ball, and all other things being equal, what is the probability of player running out?" The probability would be 0.39.

Why consider a .10 probability of scratching on only the 9-ball? How about miscuing? What about a spectator bumping the player? How about accidentally hitting the cue ball on a practice stroke?

Additionally , when you say that he has a 90 percent chance of pocketing the 9 aren't the other variables already calculated in that
number?

I guess your right. My concept of probability is that it is a probability and
cannot be made a certainty because of the infinite amount of variables that would be required. That's why fishing is called fishing and not catching.

Please explain your concept of probability to us.

.39 is the correct answer, 90% chance of making the next shot already takes into account the position. so .9 to the 8th power. the way the question is, you should say it is an 81% chance of making all the shots, taking into account the position.

it's still a good question. so basically, if the balls are all open. a good player should get out 39% of the time. that sounds right.
 
Over the years I have observed that very few people have a clear concept on probability, and especially so when it relates to pool.

I want to give you a mathematical problem. You can easily solve it with a calculator, but just take a guess first before calculating.

Here is the scenario.....

The game is 9-ball. The player has made 1 ball on the break, so there are 8 balls remaining.

For each of the remaining shots, this player has a 90% chance of pocketing the ball, and a 90% chance of getting position on the next ball.

For the 9-ball, he has a 90% chance of pocketing it, and a 90% chance of avoiding a scratch.

What is the mathematical probability that he runs out?

You forgot to mention if an aiming system is to be used.:D
 
All of this discussion of probabilities based on averages is interesting, but all it does for me is put in perspective how incredible it is that running racks is even possible.

Once you get to a level where you're thinking about running racks, a typical run-out is composed of several very high probability shots with equally high probability position routes and one or two difficult shots or position routes. That means that the "math" involved is heavily weighted on a couple of shots and has little to do with the other, nearly automatic, ones. If the probability of dealing effectively with the tough shots and getting out is higher than the probability of playing effective defense, you go for it. If not, you D up your opponent.
 
.39 is the correct answer, 90% chance of making the next shot already takes into account the position. so .9 to the 8th power. the way the question is, you should say it is an 81% chance of making all the shots, taking into account the position.

it's still a good question. so basically, if the balls are all open. a good player should get out 39% of the time. that sounds right.

The way I read it, pocketing the ball doesn't imply automatic position, which is why the position probability was explicitly stated. I'm sticking with .9^16 (or .81^8 if you prefer). :smile:

I do agree that a good player will get out more often than .185 if there is a 90% probability of obtaining position on each ball, which implies that the balls are open, but I attribute that to the fact that a good player will make more than 90% of shots that he/she has ideal position on. If we give them a 95% chance of making each shot they have ideal position on, which I think is more realistic for a good player, their odds of getting out jump up to about 30%, which I also think is more realistic.

Aaron
 
Over the years I have observed that very few people have a clear concept on probability, and especially so when it relates to pool.

I want to give you a mathematical problem. You can easily solve it with a calculator, but just take a guess first before calculating.

Here is the scenario.....

The game is 9-ball. The player has made 1 ball on the break, so there are 8 balls remaining.

For each of the remaining shots, this player has a 90% chance of pocketing the ball, and a 90% chance of getting position on the next ball.

For the 9-ball, he has a 90% chance of pocketing it, and a 90% chance of avoiding a scratch.

What is the mathematical probability that he runs out?

90%?

I don't know how to solve it easily with a calculator. But if the player has a 90% chance of getting shape and the resulting shot is also 90% then I say getting out is 90% certain.
 
think of it like this: 1 time out of ten he will screw up the shot. 1 time out of ten he will screw up the leave. If he shoots the shot thousands of times, some of those he will screw up the leave, some he will screw up the shot, one in twenty times (or so) that he screws up one, he will screw up the other.

His odds are 0.9x0.9 for doing both right (81%).

But this is all just a middle-school exercise, regardless. the system has been so overly simplified that it has no realistic.

dld

.......... :d

Incomplete.jpg
 
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