# You stand at the basketball free-throw line and make 30 attempts at at making a basket. You make 3 baskets, or 10% of your shots. Is it accurate to say that three weeks later, when you stand at the free-throw line, that the probability of making a basket on your first attempt is 10%, or .10?

Jan 1, 2015

It depends. It would take multiple assumptions that are unlikely to be true to extrapolate this answer from the data given for this to be the true probability of making a shot.

One can estimate the success of a single trial based on the proportion of previous trials which succeeded if and only if the trials are independent and identically distributed. This is the assumption made in the binomial (counting) distribution as well as the geometric (waiting) distribution.

However, shooting free throws is very unlikely to be independent or identically distributed. Over time, one can improve by finding "muscle memory," for example. If one steadily improves, then the probability of the early shots were lower than 10% and the finishing shots were higher than 10%.

In this example, we still don't know how to predict the probability of making one's first shot. How much does practice help your next session? How much do you lose the muscle memory by returning three weeks later?

However, there is another concept known as personal probability. This fairly subjective concept is based on your own personal knowledge of a situation. It doesn't necessarily represent an accurate picture of reality, but rather is based on one's own interpretation of events.

To determine your personal probability, one can perform the following thought experiment. How much would someone else have to offer you for you to be willing to wager $1 on an event occurring? Whatever this value $x$is, this defines the odds of the event occurring, which is equal to $\frac{1}{x}$. One can convert this personal odds into personal probability based on the equation: "probability" = ("odds")/(1+"odds"). If you were willing to accept$9 to bet, then your personal odds would be $\frac{1}{9}$, making your personal probability:

("odds")/(1+"odds") = (1/9)/(1+(1/9)) = 1/10 = 10%