I have a game where I can bet on two teams:

  • One team has odds $1.7$
  • The other has $1.5$

I understood that the one that pays off $1.5$ has an higher probability of victory, so it pays less. But how are these odds calculated? How can I calculate the probability of victory for each team from these odds?


We assume your bet is fair, and we denote $A$ as the event of team One winning, $B$ as the event of team two winning. The Fair Bet Odds Rule says:

In a fair bet, the payoff odds equal the chance odds.

  • Decimal odds

If $A$ happens your profit is $1.7-1=0.7$, if not your profit is $-1$. Assuming a fair bet we get: $$0.7P(A)-(1-P(A))=0$$ and we get $1.7P(A)=1$ and $P(A)=\frac{10}{17}$. Similarly if $B$ happens you win $0.5$ and lose $1$ if it doesn't. The equation is thus $$0.5P(B)-(1-P(B))=0$$ and we get $1.5P(B)=1$ and $P(B)=\frac{2}{3}$. We note that probabilities add up to more than 1 in this case since the bookmaker has to take its share.

  • Pay off odds

In some settings you deal with payoff odds. "1.7 to 1" in this case is equivalent to decimal odd of $2.7$. In this scenario:

If $A$ happens you win $1.7$ if it does not happen you lose $1$. The equation is thus: $$1.7P(A)-1(1-P(A))=0 $$ and $P(A)=\frac{10}{27}$, Similarly if $B$ happens you win 1.5 and lose 1 if it does not. The equation is thus $$1.5P(B)-1(1-P(B))=0$$ and $P(B)=\frac{2}{5}$. In this scenario your probabilities add up to less than one and the $difference\times stake$ is "the house percentage".

Special thanks to @Rodrigo de Azevedo for clarifying the difference between decimal and payoff odds.

  • 1
    $\begingroup$ Those probabilities do not add up to $1$. $\endgroup$ – Rodrigo de Azevedo Aug 25 '18 at 21:35
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    $\begingroup$ This is still wrong. If $A$ happens, you win $1.7 - 1 = 0.7$, not $1.7$. $\endgroup$ – Rodrigo de Azevedo Aug 26 '18 at 3:44
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    $\begingroup$ If you don't consider profit, why $-1$ in the expected value? Note that the "probabilities" are the reciprocals of the odds. $1.7$ becomes $\frac{10}{17}$, for example. The "probabilities" add to more than $1$ because the bookmaker must make a profit. $\endgroup$ – Rodrigo de Azevedo Aug 27 '18 at 5:16
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    $\begingroup$ No. Suppose you bet 1 USD. Cost is 1 USD. If you win, you get 1.7 USD, which is revenue. Profit is 0.7 USD. If you lose, you get no revenue. Profit is -1 USD. Consider a football match, say, Real Madrid x Barcelona in some final. No chance of a draw. Equal chances of winning. Thus, odds are 2 and 2 if the bookmaker takes no cut. However, bookies must earn a living, so they reduce the odds to, say, 1.9 and 1.9. The "probabilities" become 52.6% for each team, which adds up to more than 100%, of course. $\endgroup$ – Rodrigo de Azevedo Aug 27 '18 at 5:41
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    $\begingroup$ Pitman is not using decimal odds! "10-to-1" is equivalent to decimal odds of 11. Take a look at this. $\endgroup$ – Rodrigo de Azevedo Aug 27 '18 at 6:54

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