Consider an even money game that pays 4:3 if event A occurs. What is the probability of event A?
I am thinking that here,
probability = denominator of odds / (denominator of odds + numerator of odds)
but I find other sources say
odds = probability / (1-p)
Not sure which to use
How do I know which one to use
They are the same thing.
e.g.
suppose there are 2 possible outcomes to some experiment, A and B and
prob of event A is 5/19
then the prob of B is 1 - 5/19 = 14/19
odds in favour of event A = (5/19) ÷ (14/19)
= (5/19)(19/14) = 5/14 or 5 : 14
or
odds in favour of event A = (5/19) ÷ (1 - 5/19)
= (5/19) ÷ (14/19) , which is the same as above
So for yours, the odds in favour of event A = 4 : 3
working backwards:
the prob of A = 4/7
the prob of NOT A = 3/7
Notice I added the two terms of the given odds ratio to form my denominator
which what your first method suggests.
To determine the probability of event A in an even money game that pays 4:3 if event A occurs, you can use either approach depending on the information available to you.
Approach 1: Using the odds
If you are given the odds of 4:3, you can use the formula:
probability = denominator of odds / (denominator of odds + numerator of odds)
In this case, the odds are 4:3, where the denominator is 4 and the numerator is 3. Plugging these values into the formula:
probability = 4 / (4 + 3) = 4 / 7 ≈ 0.5714
So, the probability of event A is approximately 0.5714.
Approach 2: Using the probability
If you are given the probability of event A, you can use the formula:
odds = probability / (1 - probability)
Let's denote the probability of event A as p. Plugging this into the formula:
4:3 = p / (1 - p)
Now, we can solve this equation for p:
4(1 - p) = 3p
4 - 4p = 3p
7p = 4
p = 4 / 7 ≈ 0.5714
So, the probability of event A is approximately 0.5714.
Both approaches yield the same result. Therefore, you can use either of them to find the probability.