Probabilities and Pool

I wonder if any sums could be done to suggest whether having the break is actually good or bad?

Looking at the GB9 Tour stats (The official British 9-ball tour) the top percentage for Break and Runs is Daryl Peach with 32%. I know there are other variables such as being able to play safe and still win the rack, but how many racks lose to dry breaks or a bad shot for the opponent.

Is it even possible to call the break a disadvantage? Surprised the Break and run percentage isnt higher than 32% when the UK players are as good as anyone else in the world...
 
So, my conclusion is that all the balls on the table are ONE problem to solve. Not
nine individual problems. Or, as the case may be,if your turn at the table comes and
there are 5 balls left, the 5 balls are ONE problem.The layout of balls on the table
becomes a parlay.

Thus if you come to the table after a dry break and the 4 ball is an awkward shot a
plan should be made before you start to run 3 balls then play safe.

This may not be a revolutionary thought to most of you but it may change
my strategy a bit. As soon as you say to yourself , "I think I can make this shot,"
you gotta duck. :duck: DeLeon is right.
 
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No question there. Didn't need to ask. You shoot the shot.

True. But it does slightly contradict what the old timer told you. In other words, if the shot probability is less than 90%, one shouldn't always play safe. That was my point.
 
True. But it does slightly contradict what the old timer told you. In other words, if the shot probability is less than 90%, one shouldn't always play safe. That was my point.

My point was how close the fellows instincts were to being right on. My thought at the time was that he was full of crap. I mean why would I pass on an 80 percent shot? I couldn't see the forest for the trees. As you know, pool is not statistically hard and fast.

I used to play a computer pool game that had an instructional program on how to shoot nine- ball. It spread the balls at random on the table and all were clear except that the 8B and 9B were frozen .
One by one the computer pocketed the balls. But it never attempted to break apart the 8B and 9B. Instead it pocketed the 7B and then played position so that it could hit the 8B free of the 9B and left it virtually impossible for the opponent to hit the 8B. The computer had planned that safety from the start.
I was stunned by the beauty of it but at the time didn't understand the significance of it. Instead of thinking 2,3,or 4 balls ahead we should be thinking 9 balls ahead.
 
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Unnecessary nitpicking

He is making an order of magnitude estimate. Your point is mere nitpicking...it adds no value to his analysis.

Not so. His post includes the following,



This statement shows that the artimetic mean, generally referred to as the average is what is being exponentiated, and this is in error. Here's an example of why.


Shot Player 1 Player 2
Number Prob Prob
Shot 1 0.91 0.95
Shot 2 0.92 0.95
Shot 3 0.93 0.95
Shot 4 0.94 0.95
Shot 5 0.95 0.95
Shot 6 0.96 0.95
Shot 7 0.97 0.95
Shot 8 0.98 0.95
Shot 9 0.99 0.95

Runout 62.8% 63.0%

Each player has an average success rate of .95 on their shots, but their runout percentages are different. That’s because you can't merely exponentiate the average success rate to get to the runout probability.
 
Actually that 95% "average" was just what I used for each individual shot to approximate some of the percentages. I think in practice for me to determine whether I'm 93%, or 95%, or 97% to make a certain shot and get position is not that relevant.

I just tried to say in general I think a top level player would be about 95% to execute most routine shots successfully, so for a series of 9 shots for a top player I could estimate something like 95%, 95%, 90%, 85%, 95%, 90%, 90%, 95%, 95%. I think some shots would be closer to 100% (straight on, 1 foot away, stop shot type shots), where others would still be seemingly pretty routine (making a ball at a normal angle and following out 1 rail for position to center of table) - those were the types of shots I would estimate at 95% for a top player, 90% for a shortstop/A player, etc.

Is it exact? No. What I didn't do was just assume 95% across the board, I tried to simulate getting out of position and having to shoot a 66% shot for instance and saw how that impacted the overall percentage of running out.

As others have commented (and thanks for those), it just gives you something to think about. Obviously your shot selection will alter that shot's percentage (another differentiator of top players), and your goal won't be to run out on every rack depending on the position of the balls etc. I know for me it just hammered home that I needed to get back to basics, work on my consistency, and pay more attention to my routes and shot selection and tweaking with english in order to keep things as routine as possible throughout the run.

Scott
 
I know for me it just hammered home that I needed to get back to basics, work on my consistency...

That's the important message here, I think. The real value of the math you correctly elucidated in your first post, is that the difference between 90% and 95% is actually huge, since it exponentiates over the course of the run, and there's no value (in fact there's negative value) in running halfway out and missing. There is no substitute for consistency in this game. There are important complements to consistency, but absolutely no substitute.

-Andrew
 
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