Gambling odds on sports betting are designated with a number in the set $(-\infty , -100) \cup [+100 , +\infty)$. If one places a wager of $w$ dollars at $p \in (-\infty , -100) \cup [+100 , +\infty)$ odds and your bet is successful, then your winnings $E(w , p)$ are calculated as follows: $$ E(w , p) : = \begin{cases} \left( 1 + \dfrac{100}{p} \right)w & p < -100\\ \left(1 + \dfrac{p}{100} \right)w & p \geq +100. \end{cases} $$ So large positive odds correspond to events that are less likely to happen and thus have greater payouts while large negative odds correspond to events that are more likely to happen and thus have smaller payouts.

+100 odds on an event correspond to a 50% chance of that event occurring, but how does one convert odds to the probability that event will occur in general? Basically, is there a function $L : (-\infty , -100) \cup [+100 , +\infty) \rightarrow [0,1]$ that takes the odds $p$ of an event and maps it to the probability it will occur?

I am particularly interested in applying this to betting on who will score the first basket of a basketball game. Betting on a single player scoring the first basket has relatively high (positive) odds and therefore a good payout should the bet be successful. While betting on a single player is risky and not the most likely to happen, betting on multiple players increases the likelihood of winning while good individual odds can offset the losses on the other bets.

Take for instance the following odds (which are real, but writing out the names will take too long) on who will score the first basket in tomorrow's Bulls v. Cavaliers game:

  • Player 1: +490
  • Player 2: +500
  • Player 3: +550
  • Player 4: +600
  • Player 5: +850
  • Player 6: +850
  • Player 7: +1000
  • Player 8: +1100
  • Player 9: +1300
  • Player 10: +1500

Is there a function $L$ with the properties I have described above and satisfies $$ 1 = L(490) + L(500) + L(550) + L(600) + L(850) + L(850) + L(1000) + L(1100) + L(1300) + L(1500)? $$ Will I need to adapt the function $L$ if I was to look at a different game with different odds?

  • $\begingroup$ Odds are not quite probabilities in disguise. Are you sure you want to do it? $\endgroup$ Dec 12, 2021 at 19:37

1 Answer 1


Assuming the bet has zero expected gain, and let $q$ be the probability of winning, then we must have:

$$ q \times E(1,p) = 1$$

or equivalently:

$$ q = \frac{1}{E(1,p)}$$

Now of course, gambling houses set the odds in such a way that $q \times E(1,p) < 1$, at least in their estimation of the true probability $q$.

As to whether your final equality $1 = \sum L(blah)$ is possible, I am too lazy to go through the calculation. But if these are actual odds posted by a gambling house, you can be sure $1 < \sum L(blah)$ because otherwise they would lose money guaranteed, vs a sharp bettor.


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