Whose your daddy?

[Bob Jewett]

Here's another problem with attempts at mathematical determination of real world phenomena. How was the 30% for any individual shot determined? It has to be based on the real world experience of shooting shots. What was the pool that the average for each individual shot was determined from? What was the distribution of the pool? Was the distribution really averaging every third shot or were there clusters of shots made versus missed. All of that would change the actual probability of making a certain number of shots in a row.

Maybe the player is coming down with flu. Maybe the table was moved last night and has a new roll. Maybe ....

The point is to guess as well as you can how likely something is, based on limited knowledge. If you know how much data went into assumed percentages, you can say how accurate those percentages are. For example a FargoRate with a robustness of only 200 is expected to have something like 25 points of error compared to the player's true average performance in competition. With 10,000 games recorded, the rating is expected to be much closer to the player's true ability.

But, the particular point being discussed is a theoretical one: given that the player is exactly x% to run a rack, and there is no real-world, extraneous interference, and everything goes according to theory, what is his chance to string N racks together in T tries?

Such theoretical understanding can help with real-world predictions.

 

[maha]

yes and that can be figured out without a computer program. just a pocket calculator. and pencil and paper for those who did that sort of things before either.

and once i figured out one time, how many tries it would take for a smaller chance happening to be 50% i went and beat one of the highest rated gambling bettors out of more than five figures. it wasn't intuitive

 

[Chris]

Here's another problem with attempts at mathematical determination of real world phenomena. How was the 30% for any individual shot determined? It has to be based on the real world experience of shooting shots. What was the pool that the average for each individual shot was determined from? What was the distribution of the pool? Was the distribution really averaging every third shot or were there clusters of shots made versus missed. All of that would change the actual probability of making a certain number of shots in a row.

It's a probability theory problem. Bob (the one in the problem, the esteemed Bob Jewett) is not a real person, nor was anyone's shooting monitored over thousands or millions of shots to determine their success probability. Stringing together racks in pool or making basketball free throws are just aids to better visualize the problem.

 

[Chris]

It can be done with recursion, but you have to look farther back than just the previous rack. You have to go back eleven racks for the chance the run has already occurred, and then there is necessarily a miss followed by ten runs, so

S(n+1) = S( n) + (1-S(n-10)) * p^10 * q

I'm not doubting your solution, but I would have to start at the other end, i.e., S(1), S(2), etc., to wrap my head around it. I don't do very well working backwards with recursion.

 

[Chris]

yes and that can be figured out without a computer program. just a pocket calculator. and pencil and paper for those who did that sort of things before either.

Many of the advances in statistics and data science over the last few decades haven't been discovery of new algorithms, but rather the new existence of sufficiently cheap processing power and computer memory that allowed algorithms other than least squares regression to become practical.

Restricting oneself to only what can be done with pencil, paper, and a pocket calculator means most algorithms are completely off limits. On many occasions, I completed more calculations before lunch, with the aid of computer programming, than Sir Isaac Newton performed across his entire lifetime.

 

[maha]

of course chris you are absolutely right about that.

my original thing was you dont need to write a program to figure that one out.