Calculation of Odds: Probability and Deviation

Calculating odds is a science, and statisticians/analysts, who can do the job competently, receive good annual salaries (GBP 50k-80k; €60k-95k: Quantitative Analyst).

These jobs are paid so well because the results form the backbone of each bookmaker’s business. The better an analyst understands his job, the bigger the potential profit margins for the bookmaker. In order to make long-term profits, a good understanding of odds calculation is therefore also essential for any bettor.

Man with 4 arms and juggling with calculator, abacus, note pad and penImage: alphaspirit (Shutterstock)

Probability of Football Match Results

Odds are based on the probability that a certain event occurs: for example a home win, a draw, or an away win in football games based on historical data. But what is the current probability of each of these outcomes and how are percentages computed? Also, from where does one get the data from?

The following diagram shows the distribution of results for the English Premier League for home win, draw, and away win during the last five complete seasons:

Results: English Premier League - 2005 to 2010 – Diagram

Looking at the graph, one can see that the distribution of results is rather similar for each year.

Without considering factors such as matches between strong or weak teams, teams under new management, rain affected games or anything else one can think of, fixtures in the English Premier League, according to the statistics, show on average 24.46% of all games were drawn, 27.35% ended in an away win and 48.16% were won by the home team:

Results: English Premier League - 2005 bis 2010 – Data from: www.sportpress.com

Only twice (of 15) did results fall outside of ± 2% residual from the average figures, being drawn games in 2005/2006 (-4.2%) and away wins in 2009/2010 (-3.4%):

Absolute deviation: English Premier League 2005 to 2010

Comparison of the expected results with observations

Now, it is time to compare the expected results (based on the average results of 5 years) with the observed results of the current 2010/2011 season up to 19 February 2011:

Expected vs. Observed: English Premier League 2005 to 2010

There are quite large differences from the expected values (averages or means) to the observed results (actual results for 2010/2011) in both the drawn matches and away wins categories. However, I am pretty sure that by the end of the season, the figures will adjust themselves more in line with the 5-year average figures and that the differences showing now can be explained by having compared only two-thirds of the current season with the average results of a full-year. However, this indicates that there is an uneven distribution of home wins, draws and away wins at different times over the season, which should balance out by the close.

So, if the odds of a single match are calculated, this seasonal effect must be considered, but for this article and for your general understanding, I shall not complicate matters by touching on it further.

Mean (average value), Errors and Residuals (relative and absolute deviation)

For those of you with difficulties understanding what arithmetic mean and errors/ residuals are, herewith a few additional explanations with examples:

  1. Arithmetic Mean
    The arithmetic mean, often referred to as simply the mean or average when the context is clear, is a method to derive the central tendency of a sample. In probability theory, the mean is also called expected value (or expectation, or mathematical expectation, or the first moment).

    The mean (average value) is a known value and can therefore be used as an expectancy value.

    Simply speaking, one can use the mean to predict future events (e.g. the distribution of football results) quite accurately.

    Example using the 5-Year English Premier League table above for home games:
    (50.53% + 47.89% + 46.32% + 45.29% +50.79%) divided by 5 = 48.16% (= mean/ average)

    Therefore, it “is expected” that the 2010/2011 season will produce 48.16% home wins.
  2. Errors and Residuals (relative and absolute deviation)

    Statistical errors and residuals are two closely related and easily confused measures of the deviation.

    The error of a sample (e.g. observed football results for a certain period of time) is the ‘relative’ deviation from the function value (mean), while the residual is the difference (absolute deviation) between the sample and the function value (mean).

    For example, the RESIDUAL (absolute deviation) for home wins during 2005/2006:
    50.53% (2005/2006 home wins) minus 48.16% (average value for 5-years) = 2.37% (= residual)

    For example, the ERROR (relative deviation) for home wins during 2005/2006:
    2.37% (residual) divided by 48.16% (mean) = 4.92% (= error)

    The error (relative deviation) is the proportional deviation between the observed value (the actual results) and the expectancy value (the mean/ average of all years). The error (relative deviation) puts residuals into relation to each other.

Relative deviation: English Premier League 2005 to 2010

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Last Update: 11 March 2011

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23 Responses to “Calculation of Odds: Probability and Deviation”

  1. scott
    29 May 2016 at 10:58 am #

    The minimum odds are computed as follows:
    Home win: 2.08 calculated odds multiplied by (1 minus ‘error’ 4.14%) = 1.99

    The maximum odds are computed as follows:
    Home win: 2.08 calculated odds multiplied by (1 plus ‘error’ 4.14%) = 2.17

    am i just being really stupid beacuse i cant seem to get these same answers when i do this sum??

    many thank!

    • Soccerwidow
      30 May 2016 at 7:29 am #

      Hi Scott, here’s the calculation a little bit more in detail. Hope it helps.

      2.08 calculated odds multiplied by (1 minus ‘error’ 4.14%) = 1.99
      1 minus ‘error’ 4.14% >> 1 – 0.0414 = 0.9586
      2.08 × 0.9586 = 1.9939 (rounded: 1.99)

      2.08 calculated odds multiplied by (1 plus ‘error’ 4.14%) = 2.17
      1 plus ‘error’ 4.14% >> 1 + 0.0414 = 1.0414
      2.08 × 1.0414 = 2.1661 (rounded: 2.17)

      Best wishes,

  2. Jimbo
    20 October 2016 at 5:03 pm #


    I just came across an app that offers the chance to win 50,000 to correctly predict the odds of two entire leagues.

    As an example it says if I can correctly predict 100% premier league results this weekend (10 games) + (10 games of the Spanish League)

    What is the average chance per game or the chance winning the 50,000?

    • Soccerwidow
      28 October 2016 at 6:32 am #

      Do you have to predict the expected odds (antepost), or predict the actual results which will be played? These are two completely different things.

  3. erez
    7 December 2016 at 10:59 am #

    if the home team have advantage in the odds ?

    for example odds are 2 but this home team get 1.75 ?

    • Soccerwidow
      8 December 2016 at 7:58 am #

      Hi erez, I don’t understand your question. Sorry!

      Generally speaking, home favourites are mainly overpriced, meaning that bookmakers price them at a higher chance to win than their true chances are. For example, odds of 1.75 stand for a 57.1% chance to win, whilst odds of 2.0 for a 50% chance. If bookmakers offer 1.75 odds for a team which should be actually priced at 2.0 then they are ensuring that they have the mathematical advantage on their side.

      You may be interested into our HDA Simulationtables. Just have a look.

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