I am wondering how to convert sport odds to the percentage format.

There's this example:

Odds for the home team victory: 2.50
Odds for the draw:`3.25
Odds for the away team victory: 3.00

To convert these odds into percentages, I can do this:

1/2.50 = 0.40 => 40% that the home team wins
1/3.25 = 0.31 => 31% for draw
1/3.00 = 0.33 => 33% that the away team wins

But if the percentages are added, then:

40% + 31% + 33% = 104%

What am I doing wrong that the sum is not equal to 100%? Is the procedure I follow wrong?

  • $\begingroup$ It is the bookie's edge. If it is under 100% it is a "sure bet" for the gambler, and there is a web site for this where they combine the websites of all bookies to check if there is a surebet possibility by betting on WinHome on betting site A, Draw on betting site B, WinAway on betting site C. If it is more than 100% then it means bookie is making money on overall population of gamblers. They are also mitigating their risk (on average they win but there is volatility) of loss on a single event by modifying the odds based on the bets they have received so far as time gets nearer to the event. $\endgroup$ Mar 4 '15 at 11:27

The odds you have are in decimal format, which the bookmaker calculates as:

$$ d_E = \frac{1}{p_E + o_E} $$

where $d_E$ is the decimal odds for event $E$, $p_E$ is the bookmakers estimated probability of event $E$, and $o_E$ is the over-round which the bookmaker adds to the decimal odds for event $E$. The over-round effectively reduces the odds, making them unfair for the bettor.

In your example, with events $H$ (home win), $D$ (draw), and $A$ (away win), you have:

$$ \frac{1}{d_H} + \frac{1}{d_D} + \frac{1}{d_A} = (p_H + p_D + p_A) + (o_H + o_D + o_A) = 1.04 $$

of course, $p_H + p_D + p_A = 1$. One option is assuming $o_H = o_D = o_A = o$ so you have $1 + 3o = 1.04$ and so $o = 0.04/3$, which you can use to uncover the bookmaker's underlying probabilities, $p_H$, $p_D$, and $p_A$. In practice however, there is no way of knowing how the bookmaker adds over-round onto betting events, it is likely they add more over-round to events which are more likely to be bet, for example a UK bookmaker's odds on England to win the World Cup.


If the odds describe real events, you most likely got them from some betting site. You are searching for overround. If your implied probabilities do not add up to 1 but are greater 1, then the bookmaker makes a sure profit of all above 1.

A simple way to work is to re-normalize, i.e. divide every inverse odd again by the sum, 1.04 in your example.


If you just want to estimate the probabilities, it seems indeed reasonable to just devide each percentage by 1.04.

However, if you want to find out more about the true percentages that the bookkeeper uses, you can do something more.

Let's assume that the bookkeeper does not want to accept a negative expected value for any of his bets

Now we can conclude the following:

40, 31 and 33 are upper bounds for the underlying true percentages.

40-4, 31-4 and 33-4 are lower bounds for the underlying true percentages (otherwise one of the other two odds would give him a negative expected value)

40+31-4, 31+33-4 and 40+33-4 are lower bounds for the underlying true percentages of a pair of possible outcomes (otherwise the other odd would give him a negative expected value)

Sidenote: I expect a bookkeeper to especially pad bets with a high risk factor, so if it would be 60% vs 40% vs 4%, it would be safe to say that the 4% contains relatively much padding.


I think they are odds against. i.e.

2.5 = (Prob(away team wins) + Prob(draw)) / Prob (Home team wins)

You can compute P(Away team wins) + Prob(draw) by taking 1/(2.5+1).

If you do that for all 3, you can P(W) = 0.285, P(D) = 0.235, P(L) = 0.25.

They do not sum up to 1, probably because the odds are rigged. But you can easily normalise them i.e. P*(W) = P(W) / P(W) + P(D) + P(L) = 0.37

  • $\begingroup$ is this method reversible? If I know the probabilities and the target margin, how to I calculate the odds (2.5, 3.25, 3.00) $\endgroup$
    – Fierce82
    Sep 6 '18 at 11:33

There is a major issue with the odds that are offered in this scenario. Where did they come from?

Suppose I bet \$1 on the home team. With odds of 2.5, I would win \$2.50 and get my \$1 back, for a collection of \$3.50. So, let's look at the chart below:

Winner - Odds - Payback on $1
Home   - 2.5  - $3.5
Draw   - 3.25 - $4.25
Away   - 3.00 - $3.00

Therefore, I can bet \$3 (\$1 on each), and have a positive return. Therefore, there is no way this is odds from a bookmaker, which is why I wonder where the odds come from?

  • $\begingroup$ The odds are decimal, so if you bet £1 at odds of 2.5, you get £1.50 profit and your £1 stake back. $\endgroup$
    – Jeff
    Aug 31 '15 at 16:48

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