Nt1330 Unit 1 Case Study

Why do bookmakers have different odds? If the odds are different, this may suggest one bookmaker is better than the other at predicting results. Consider the phrase `you can never beat the bookie', one would think intuitively that they can beat the bookmaker as many bookmakers have different odds for the same result of a certain match. Therefore, if there are many different predictions then why can't my prediction be better? Bookmakers have different margins, some are better to bet with than others. A bookmaker's margin is the difference between the offered odds and the true odds. Bookmakers adjust the odds from the true odds they have calculated so they can have an edge over the people betting on the match. Bookmakers also adjust the odds …show more content…
\cite{jolliffe07} Suppose that a set of probability forecasts is made of $n$ binary events of interest such as ``precipitation tomorrow'' or ``damaging frost next month.'' Denote the $n$ forecasts by $\{p_1, p_2,..., p_n\}$, where each $p_i$ is a probability between $0$ and $1$. The corresponding observations $\{x_1, x_2,..., x_n\}$ are coded as $1$ if the event occurs and $0$ if it does not. To assess the quality of the forecasts, various measures or scores can be constructed that quantify the difference between the set of forecasts and the corresponding set of observations. …show more content…
The BS for a sample of $n$ binary forecasts, where the event either occurs or does not occur, is given by
\begin{equation}
\ BS=\frac{1}{n}{\sum_{t=1}^{n}}(p_{i}-x_{i})^2
\end{equation} \vspace{12pt}

\cite{brier50} The Brier score was first introduced by Glenn W. Brier in 1950 to measure the accuracy of weather predictions. It can be applied to all forms of probability forecasting not just weather. The Brier score is the difference between the sum of the squares between forecasts and observations. To demonstrate how to Brier score works, the following example explains it: \vspace{12pt}

Consider a football match, you forecast Team A to score at least once in the match at $80\%$ and they do go on to score in that match. With there only being one forecast, $n=1$ therefore the Brier score is then simply as follows

\begin{equation}
\ BS=(0.8-1)^2=0.04

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