Skip to main content

Probability and Cumulative Dice Sums

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, \( \frac{p}{q} \). Additional factors can be easily chained using this form to provide more complex estimates. For example, let \( p_e \) be an environmental factor, then:

\begin{align}
\frac{p}{q} &= \frac{p_1}{q_1} \cdot \frac{q_2}{p_2} \cdot \frac{q_e}{p_e}
\end{align}

Chess

The most common rating system in chess is the Elo rating system. This has also been adopted for various other uses, e.g. ``hot or not'' websites. This system assigns ratings \( R_1, R_2 \) to players 1 and 2 such that the probability of player 1 beating player 2 is approximately:

\begin{align}
p &= \frac{e^{R_1/C}}{e^{R_1/C}+e^{R_2/C}}
\end{align}

Here \( C \) is just a scaling factor (typically \( 400/\ln{10} \) ). The Elo rating is connected to log5 via setting \( e^{R/C} = p/q \). We then recover:

\begin{align}
\frac{p}{q} &= e^{R/C}\\
p &= \frac{e^{R/C}}{1+e^{R/C}}\\
R &= C\cdot \ln(p/q)
\end{align}

Note that \( p \) is also the probability of this player beating another player with Elo rating 0. The Elo system generally includes enhancements accounting for ties, first-move advantage and also an online algorithm for updating ratings. We'll revisit these features later.

Psychology

The Bradley-Terry-Luce (BTL) model is commonly used in psychology. Given two items, the probability \( p \) that item 1 is ranked over item 2 is approximately:

\begin{align}
p &= \frac{Q_1}{Q_1+Q_2}
\end{align}

In this context \( Q_* \) typically reflects the amount of a certain quality. That this model is equivalent to the previous models is immediate:

\begin{align}
Q &= e^{R/C} = p/q\\
R &= C\cdot \ln(Q) = C\cdot \ln(p/q)\\
p &= \frac{Q}{1+Q}
\end{align}

Psychometrics

The dichotomous (two-response) Rasch and item response models are commonly used in psychometrics. For the Rasch model, let \( r_1 \) represent a measurement of ability and \( r_2 \) the difficulty of the test item. The Rasch model estimates the probability of correct response \( p \) as:

\begin{align}
p &= \frac{e^{r_1-r_2}}{1+e^{r_1-r_2}}
\end{align}

The one-parameter item response model estimates:

\begin{align}
p &= \frac{1}{1+e^{r_2-r_1}}
\end{align}

These are clearly equivalent to each other and to the previous models.

Comments

Popular posts from this blog

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...

Probability and Cumulative Dice Sums

Let a die be labeled with increasing positive integers \(a_1,\ldots , a_n\), and let the probability of getting \(a_i\) be \(p_i>0\). We start at 0 and roll the die, adding whatever number we get to the current total. If \({\rm Pr}(N)\) is the probability that at some point we achieve the sum \(N\), then \(\lim_{N \to \infty} {\rm Pr}(N)\) exists and equals \(1/\rm{E}(X)\) iff \((a_1, \ldots, a_n) = 1\). The direction \(\implies\) is obvious. Now, if the limit exists it must equal \(1/{\rm E}(X)\) by Chebyshev's inequality, so we only need to show that the limit exists assuming that \((a_1, \ldots, a_n) = 1\). We have the recursive relationship \[{\rm Pr}(N) = p_1 {\rm Pr}(N-a_1) + \ldots + p_n {\rm Pr}(N-a_n);\] the characteristic polynomial is therefore \[f(x) = x^{a_n} - \left(p_1 x^{(a_n-a_1)} + \ldots + p_n\right).\] This clearly has the root \(x=1\). Next note \[ f'(1) = a_n - \sum_{i=1}^{n} p_i a_n + \sum_{i=1}^{n} p_i a_i = \rm{E}(X) > 0 ,\] hence this root is als...

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...