Skip to main content

Probability and Cumulative Dice Sums

Post a Comment

Who Controls the Pace in Basketball, Offense or Defense?

During a recent chat with basketball analyst Seth Partnow, he mentioned a topic that came up during a discussion at the recent MIT Sloan Sports Analytics Conference. Who  controls the pace of a game more, the offense or defense? And what is the percentage of pace responsibility for each side? The analysts came up with a rough consensus opinion, but is there a way to answer this question analytically? I came up with one approach that examines the variations in possession times, but it suddenly occurred to me that this question could also be answered immediately by looking at the offense-defense asymmetry of the home court advantage.

As you can see in the R output of my NCAA team model code in one of my public basketball repositories, the offense at home scores points at a rate about \( e^{0.0302} = 1.031 \) times the rate on a neutral court, everything else the same. Likewise, the defense at home allows points at a rate about \( e^{-0.0165} = 0.984\) times the rate on a neutral court; in both cases the neutral court rate is the reference level. Notice the geometric asymmetry; \( 1.031\cdot 0.984 = 1.015 > 1\). The implication is that the offense is responsible for about the fraction \[ \frac{(1.031-1)}{(1.031-1)+(1-0.984)} = 0.66 \] of the scoring pace. That is, offensive controls 2/3 of the pace, defense 1/3 of the pace. The consensus opinion the analysts came up with at Sloan? It was 2/3 offense, 1/3 defense! It's nice when things work out, isn't it?

I've used NCAA basketball because there are plenty of neutral court games; to examine the NBA directly we'll have to use a more sophisticated (but perhaps less beautiful) approach involving the variation in possession times. I'll do that next, and I'll also show you how to apply this new information to make better game predictions. Finally, there's a nice connection to some recent work on inferring causality that I'll outline.

Comments

Post a Comment

Popular posts from this blog

A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem . Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way. Standard notation: probability of A given B is written \( \Pr(A \mid B) \) probability of B is written \( \Pr(B) \) Bayes' Theorem: Using the notation above, Bayes' Theorem can be written:  \[ \Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)} \] Let's apply Bayes' Theorem to the Monty Hall problem . If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random. Assume we pick Door 1 an...
20 comments
Read more

Baseball, Chess, Psychology and Pychometrics: Everyone Uses the Same Damn Rating System

Here's a short summary of the relationship between common models used in baseball, chess, psychology and education. The starting point for examining the connections between various extended models in these areas. The next steps include multiple attempts, guessing, ordinal and multinomial outcomes, uncertainty and volatility, multiple categories and interactions. There are also connections to standard optimization algorithms (neural  nets, simulated annealing). Baseball Common in baseball and other sports, the log5 method provides an estimate for the probability \( p \) of participant 1 beating participant 2 given respective success probabilities \( p_1, p_2 \). Also let \( q_* = 1 -p_* \) in the following. The log5 estimate of the outcome is then: \begin{align} p &= \frac{p_1 q_2}{p_1 q_2+q_1 p_2}\\   &= \frac{p_1/q_1}{p_1/q_1+p_2/q_2}\\ \frac{p}{q} &= \frac{p_1}{q_1} \cdot \frac{q_2}{p_2} \end{align} The final form uses the odds ratio , \( \frac{p}{q}...
Post a Comment
Read more

Simplified Multinomial Kelly

Here's a simplified version for optimal Kelly bets when you have multiple outcomes (e.g. horse races). The Smoczynski & Tomkins algorithm, which is explained here (or in the original paper): https://en.wikipedia.org/wiki/Kelly_criterion#Multiple_horses Let's say there's a wager that, for every $1 you bet, will return a profit of $b if you win. Let the probability of winning be \(p\), and losing be \(q=1-p\). The original Kelly criterion says to wager only if \(b\cdot p-q > 0\) (the expected value is positive), and in this case to wager a fraction \( \frac{b\cdot p-q}{b} \) of your bankroll. But in a horse race, how do you decide which set of outcomes are favorable to bet on? It's tricky, because these wagers are mutually exclusive i.e. you can win at most one. It turns out there's a simple and intuitive method to find which bets are favorable: 1) Look at \( b\cdot p-q\) for every horse. 2) Pick any horse for which \( b\cdot p-q > 0\) and mar...
Post a Comment
Read more