0

I’ve learned that to calculate implied probability based on a gambling odds, you use the following formula:

Implied probability of win %= stake/ total payout

For example, if my fractional odds for team a to win are 5/1 (bet 1 dollar to win 5 dollars if team a wins), then the implied probability = 1/6.

Is there any intuition to why this is the case considering probability is the number of successful outcomes/ total number of outcomes?

user60519
  • 49
  • Assume the payout is $~\dfrac{1}{k}~$ where $k$ is any positive real number. This implies that you wager $1$, and if you win, you get back $(1 + k)$. This implies that in order for the proposition to be a break even proposition, your probability of winning must be $~\dfrac{1}{k+1}.$ This is because you want the probability of winning times the payout to equal $1$, which is your original investment. You have that $$\frac{1}{k+1} \times (1+k) = 1.$$ – user2661923 Dec 17 '22 at 16:51
  • @user2661923 Thanks a lot for the response. Could you please break that down a little bit for me?
    • Why would the payout be assumed to be 1/k? & can you please explain that last formula you made and the intuition behind it? I appreciate your help.
     – user60519 Dec 17 '22 at 17:04
  • I am assuming that if you wager $1$, that you will get back $(k+1)$, where $k$ is any positive number. So, I am making $k$ into a variable. The idea is that any payout ratio can be expressed in the form of $1$ to $k$, where $k$ is some positive real number. Looking at it from that perspective, my comment indicates that for the betting proposition to be break even, the probability of winning must be $\frac{1}{1+k}.$ As an example, if you are required to invest $2$ to win $3$, that would be expressed as $1$ to win $1.5$, so in that instance, $k$ would equal $1.5.$ – user2661923 Dec 17 '22 at 17:13
  • Similarly, if you are required to invest $4$ to win $1$, that would be expressed as $1$ to $(1/4)$, so in that instance, $k$ would equal $(1/4).$ – user2661923 Dec 17 '22 at 17:15
  • So, in general, if you are required to invest $A$ to win $B$, this is equivalent to investing $1$ to win $\dfrac{B}{A}.$ So, you have that then, $k = \dfrac{B}{A}.$ – user2661923 Dec 17 '22 at 17:18
  • @user2661923 Thanks so much. And the way you related your equation to probability was the idea of expected value (where expected value= probability x outcome)? I am just unsure of how you made the jump of your payout ratio x payout equation to a probability. – user60519 Dec 17 '22 at 19:19
  • If you invest $1$, and the total payout is $(1 + k)$, and the break even probability is $P$, then $P$ must satisfy the equation $$P \times (1 + k) = 1.$$ This is because you want your expected return to exactly match your original investment of $1$. – user2661923 Dec 17 '22 at 19:23
  • Beautiful, thanks for your help. – user60519 Dec 17 '22 at 21:21
  • Take a look at this – Rodrigo de Azevedo Dec 18 '22 at 14:05

0 Answers0