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I'm really struggling to find examples of converting implied probability back into European decimal odds.

Say I have the implied probability of a horse to win at 66% of the book (no overrounds for now) , and there are 4 horses racing in total. How would I go about converting this back into decimal (european) odds?

Glen_b
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Mr Scalp
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    I took the liberty of replacing your tag with one relevant to the question. Some of the terminology in your last paragraph is not clear to me (not being a gambler on horse races), so it might help to clarify the intent there. – Glen_b Nov 27 '18 at 00:06
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    What is "implied probability"? Is it any different from probability? – Peter Flom Nov 27 '18 at 11:22

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Leaving aside bookmaker margins (overrounds) as you suggest, throughout; we may take that the total probabilities implied by the odds for a collection of events$^\dagger$ are 1.

First let me introduce odds in the sense statisticians use the term:

If $p$ is the probability of an event, then

$$\text{Odds of the event} = \frac{p}{1-p}\,.$$

Statisticians use this definition quite widely, and often use the natural logarithm of such odds in statistical models. (I introduce this because if you go searching statistical pages on the relationship between odds and probability, this is usually going to be the thing you'll encounter.)

This corresponds to "odds for" the event; typically gambling odds (at least as far as I am used to the term) are "odds against" $(1-p)/p = \frac{1}{p} - 1$; this would correspond to what Wikipedia calls "Fractional odds".

However, decimal odds are different again (representing total payout on a 1 unit bet, rather than what is returned above the original unit stake), essentially that drops off the "-1" from the fractional odds.

Let $d$ be this ratio (d for decimal). We therefore have:

$$d = \frac{1}{p}\,.$$

This equation converts $p$, the implied probability to decimal odds.

And to go back the other way, therefore, we have

$$p = \frac{1}{d}\,.$$

Example 1: if the decimal odds are $5$, that implies a probability of $\frac{1}{5}$ (the mathematical odds would be $\frac14$ and the fractional odds against would be $4$ i.e. 4 to 1).

Example 2: if the implied probability is $0.4$ the decimal odds will be $1/0.4 = 2.5$, the mathematical odds would be $\frac23$ and the fractional odds against are $0.6/0.4 = 1.5$ i.e. 6 to 4.

$\dagger$ (a set of mutually exclusive, exhaustive events)

Henry
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Glen_b
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  • I think you may find that decimal odds or European odds are the reciprocal of the probability, i.e. $\frac1p$, and are therefore always at least $1$: see Wikipedia – Henry Nov 27 '18 at 00:49
  • I slightly adjusted your examples - in particular I think you should have $4$ rather than $6$ in Example 1. – Henry Nov 27 '18 at 08:01
  • In example two, where did you get the 0.6 from? Edit: Ok I get it now, it really is that simple huh, Thank you very much for the explanation – Mr Scalp Nov 27 '18 at 11:36
  • $0.6 = 1-0.4$... – Glen_b Nov 27 '18 at 11:40