Skip to main content

Probability and Cumulative Dice Sums

What's the Value of a Win?

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation. Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc.

Let our Pythagorean win expectation formula be \[ w=\frac{P^e}{P^e+1},\] where \(w\) is the win fraction expectation, \(P\) is runs/allowed (or similar) and \(e\) is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with \(g\) games is \[W = g\cdot w = g\cdot \frac{P^e}{P^e+1},\] so for one estimate we only need to compute the value of the partial derivative \(\frac{\partial W}{\partial P}\) at \(P=1\). Note that \[ W = g\left( 1-\frac{1}{P^e+1}\right), \] and so \[ \frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}\] and it follows \[ \frac{\partial W}{\partial P}(P=1) = \frac{ge}{4}.\] Our estimate for the run value of a win now follows by setting \[\frac{\Delta W}{\Delta P} = \frac{ge}{4} \] giving \[ \Delta W = 1 = \frac{ge}{4} \Delta P.\] What is \(\Delta P\)? Well \(P = R/A\), where \(R\) is runs scored over the season and \(A\) is runs allowed over the season. We're assuming this is a league average team and asking how many more runs they'd need to score to win an additional game, so \(A\) is actually fixed at \(L\), the league average number of runs scored (or allowed). This gives us \[1 = \frac{ge}{4} \Delta P = \frac{ge\Delta R}{4L}.\] Now \(L/g = l\), the league average runs per game, so we arrive at the estimate \[\Delta R = \frac{4l}{e}.\] In the specific case of MLB, we have \(e = 1.8\) and \(l = 4.3\), giving that a win is approximately \(\Delta R = 9.56\) runs.

Bill James originally used the exponent \(e=2\); in this case the formula simplifies to \(\Delta R = 2l\), i.e. we get the particularly simple result that a win is equal to approximately twice the average number of runs scored per game.

Applying this estimate to the NBA, a win is approximately \( \Delta R = \frac{4\cdot 101}{16.4} = 24.6\) points. Similarly, we get the estimates for a win of 4.5 goals for the NHL and 5.1 goals for the Premier League.

Comments

  1. I think you've assigned the incorrect goals/win to the wrong league. NHL I think is 5.1 and Premier League is 4.5. Thanks for sharing!!

    ReplyDelete

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...

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...

Sergey Brin, Please Pick up your Paychecks

The state of California is currently holding over $6 billion  in unclaimed property belonging to millions of people. What type of property and who are the rightful owners? According to California's official unclaimed property website, these assets fall into the following categories: Bank accounts and safe deposit box contents Stocks, mutual funds, bonds, and dividends Uncashed cashier's checks or money orders Certificates of deposit Matured or terminated insurance policies Estates Mineral interests and royalty payments, trust funds, and escrow accounts People forget, people die, people move around. But $6 billion is a staggering amount of money; some of these amounts have to be really large. Let's try to find some interesting examples. This is official California UCP search form . Programmer and database types will notice one problem immediately - no fuzzy string matching . If your name or address was misspelled on the assets, or munged in the recording proce...