November 21, 2024

Fractional odds

Fractional odds explained

First off the fractional odds,f.o=a/b are simple to calculate and also f.o figures determine profit as ( a linear) function of stake

These figures can be understudied in many ways
One is next: The denominator(b) is the amount bet and the numerator(a) is the amount that your wager will yield

  • a/b     ‘a’ to ‘b’   against   if a>b     or worse than evens
  • 1-1      evens odds                if b=a
  • a/b     ‘b’ to ‘a’       on (or in favur)          if b>a   better than evens

Example:
7/2 ….. seven to two against( Whom? punter of cause)
if we stake 2€ then the profit will be 7€ in the winning case

2/7 …. two to seven in favor
stake 7€ will produce 2€ of profit

Tiny deeper insight
a)against case
4/2
meaning:     for investing every 2€ we got 4€ of profit
revenue: 4+2=6</em> <!-- /wp:paragraph -->  <!-- wp:paragraph --> b) <strong>on</strong> case    2/4meaning: for investing 4€ we got 2€ of profit,  revenue: 2+4=6€ <!-- /wp:paragraph -->  <!-- wp:paragraph --> <strong>Note: same revenue has different profits</strong> <strong>and of course different risks</strong> <!-- /wp:paragraph -->  <!-- wp:heading {"style":{"spacing":{"margin":{"bottom":"25px"}}}} --> <h2 style="margin-bottom:25px"><strong>Relation fractional odds with <em><mark style="background-color:rgba(0, 0, 0, 0);color:#f12525" class="has-inline-color">implied</mark></em>  probability</strong></h2> <!-- /wp:heading -->  <!-- wp:paragraph --> Instructive example <!-- /wp:paragraph -->  <!-- wp:paragraph {"style":{"typography":{"lineHeight":"1.3"}}} --> Rolling 1 dice has 6 possible outcomesSuppose we bet on 5 dots land upThis event has only 1 outcome in our favor, and 5 against us, so f.o({5})=5/1Suppose we bet on 5 or 3 dots landing up in one rollNow 2 outcome goes in our favour and 4 against, f.o({3},{5})=4/2=2/1Suppose we bet on 5 or even number dots land up in one rollbet now has 4 outcomes go in our favor and 2 against, f.o=2/4=1/2 <!-- /wp:paragraph -->  <!-- wp:paragraph --> Let's calculate the corresponding probabilities in all 3 cases <!-- /wp:paragraph -->  <!-- wp:list {"ordered":true} --> <ol><li>f.o=5/1prob=1/6 => d.o=1/prob….. d.o=6</li><li>f.o=2/1prob=2/6 => d.o=6/2=3</li><li>f.o=1/2prob=4/6 d.o=3/2 = 1.5</li></ol> <!-- /wp:list -->  <!-- wp:paragraph --> In the betting world, we are dealing with so-called bookie's probability, p<sub>bookie ,- '</sub>probability' under the influence of Dutch bookRelation between d.o and p<sub>bookie</sub>  :                                                     <strong>d.o= 1/ p<sub>bookie</sub></strong> <!-- /wp:paragraph -->  <!-- wp:paragraph --> From the other side connection p<sub>bookie</sub> (implied probability) and decimal odds is:                                          <strong> p<sub>bookie</sub>=1/d.o</strong> <!-- /wp:paragraph -->  <!-- wp:paragraph {"style":{"typography":{"fontSize":"19px"}}} --> e.g.  if d.o=2 for the home win then the bookmaker gives <em>0.5=50%</em> chances for the same event <!-- /wp:paragraph -->  <!-- wp:heading {"level":3,"style":{"spacing":{"margin":{"bottom":"25px"}}}} --> <h3 style="margin-bottom:25px"> The connection between <em>implied probability</em> and fractional odds</h3> <!-- /wp:heading -->  <!-- wp:paragraph {"style":{"typography":{"lineHeight":"1.3"}}} --> Let's find a general rule for the relationship between (game) odds in the form a/b or a-b and probability in generalIt is natural to start from the general form for f.o=a/baleatory event (in the game) with fractional odds a/b shows us that game has a+b outcomes <!-- /wp:paragraph -->  <!-- wp:heading {"level":4,"style":{"typography":{"fontSize":"25px"}}} --> <h4 style="font-size:25px"><strong>Calculating the implied  probability for a bet placed at odds</strong></h4> <!-- /wp:heading -->  <!-- wp:paragraph {"style":{"typography":{"lineHeight":"1.3"}}} --> from the mass of a+b outcomes <mark style="background-color:#f5ebc5" class="has-inline-color">b</mark> goes in our favor so corresponded probability is:b/(a+b)So from the given f.o, the probability we read like this:(f.o=a/b)prob = <mark style="background-color:#f5ebc5" class="has-inline-color">b</mark>/(a+b)shorter:                                                          <strong><em> (f.o)<sub>prob</sub> = b/(a+b)</em></strong> <!-- /wp:paragraph -->  <!-- wp:paragraph {"style":{"typography":{"lineHeight":"1.3"}}} --> Decimal odds of an outcome are equivalent to the decimal value of the fractional odds plus one <!-- /wp:paragraph -->  <!-- wp:paragraph --> Here's why:                         1/(f.o)<sub>prob </sub>=(a+b)/b= a/b +1  = f.o +1  = d.o <!-- /wp:paragraph -->  <!-- wp:paragraph -->  <!-- /wp:paragraph -->  <!-- wp:paragraph {"style":{"typography":{"fontStyle":"normal","fontWeight":"600"}}} --> <em>So, for a given event and appropriate odds in form d.o- to find out 'probability' is enough make:1/d.o</em> <!-- /wp:paragraph -->  <!-- wp:paragraph --> <strong>The decimal odds of an outcome are equivalent to the decimal value of the fractional odds plus one</strong> <!-- /wp:paragraph -->  <!-- wp:paragraph {"fontSize":"large"} --> F.O= D.O - 1D.O= F.O +1 <!-- /wp:paragraph -->  <!-- wp:paragraph --> In accordance with that:<strong><em>Revenue</em></strong><em> = Stake x Decimal Odds figure</em><strong><em>Revenue</em></strong><em> = Stake x (f.o +1)=f.o*stake + stake=Profit + stake</em> <!-- /wp:paragraph -->  <!-- wp:paragraph --> So we have: <!-- /wp:paragraph -->  <!-- wp:paragraph {"style":{"typography":{"fontSize":"25px"}}} -->                                                    <strong> Profit=Stake*f.o</strong> <!-- /wp:paragraph -->  <!-- wp:heading {"level":3,"style":{"spacing":{"margin":{"bottom":"25px"}}}} --> <h3 style="margin-bottom:25px">One useful exploration of decimal odds</h3> <!-- /wp:heading -->  <!-- wp:paragraph {"style":{"typography":{"lineHeight":"1.3"}}} --> Let's prove, as practice, that <span style="text-decoration: underline;">even odds</span> <strong>1-1</strong> are quoted in decimal odds as <strong>2.00</strong>1-1 (one to one) 1€ (or) of wage follow 1 € ($) profit!
So we have got:(stake=)1 +(profit=)1 = (revenue=)2

2=stake(=1)*d.o => d.o=2

So when we notice in the betting offer that the victory of, for example, the home team is odds=2, then the bookmaker gives a 50% chance of winning that very team

How does it d.o=2 work in the field?

To answer that, let’s dive into one of my databases with 116 739 played games in the period 2005-2020 taken from the source Football-Data.co.uk
Decimal odds 2.00 of 1/1 in fractional form is provided by Bet365

As presented in 116 739 records has 2651+1587+1365=3 603 those with even odds
Well, in 15 years even odds took place 3603/116739 =0.0309= 3.09%
in that amount, home winning is 47% for the same period, almost even-50%

Top four order:

1. N1- Eredivisie 56% home winning with even odds, d,o(H)=2
2. I1 –Serie A 52%
3. SP1-LaLiga 52%
4. I2 –Serie B 51%

In the graph below, as an example is singled out Seria A

The below curve is basically the same as the curve above
The only difference is that natural numbers replace the averages
The aleatory nature of the game is clearly reflected

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