# Statistics Question (RE: Shooting basketballs) **URL:** \ **Category:** Factual Questions\ **Created:** [December 14, 2005, 3:40am UTC](https://boards.straightdope.com/t/statistics-question-re-shooting-basketballs/335531 "2005-12-14T03:40:49Z")\ **Posts on this page:** 9\ **Page:** 1
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [December 14, 2005, 3:40am UTC](https://boards.straightdope.com/t/statistics-question-re-shooting-basketballs/335531/1 "2005-12-14T03:40:49Z")
I want to know how likely it is that someone that makes a given percentage of a shot will deviate from that percentage. Basically I want to know if a guy makes his shots at Z% and takes X shots what are the probabilities of him making Y shots with Y going from 0-Y. For example if we had a guy that makes 50% of his shots over the long haul takes 8 shots one game what are the respective probabilities of him making 1 shot. In this case Z=50%, X=8 and Y=1. Is there an equation that will work for any X, Y and Z? ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [December 14, 2005, 3:51am UTC](https://boards.straightdope.com/t/statistics-question-re-shooting-basketballs/335531/2 "2005-12-14T03:51:52Z")
> [@treis](#): > > Basically I want to know if a guy makes his shots at Z% and takes X shots what are the probabilities of him making Y shots with Y going from **0-Y**. That should be from 0-X ---
**Author:** ![ccwaterback](https://avatars.discourse-cdn.com/v4/letter/c/df705f/32.png) [@ccwaterback](https://boards.straightdope.com/u/ccwaterback)\ **Post date:** [December 14, 2005, 4:04am UTC](https://boards.straightdope.com/t/statistics-question-re-shooting-basketballs/335531/3 "2005-12-14T04:04:39Z")
You not only need the mean, you also need the standard deviation. So, if you can define the mean AND standard deviation AND have enough samples to make those observations statistically significant, then the values you are looking for can be calculated depending on the amount of certainty you are willing to accept. ---
**Author:** ![panamajack](https://avatars.discourse-cdn.com/v4/letter/p/47e85d/32.png) [@panamajack](https://boards.straightdope.com/u/panamajack)\ **Post date:** [December 14, 2005, 4:08am UTC](https://boards.straightdope.com/t/statistics-question-re-shooting-basketballs/335531/4 "2005-12-14T04:08:29Z")
The values are a [binomial distribution](http://mathworld.wolfram.com/BinomialDistribution.html), assuming each shot to be an independent event. Using your terminology, we have Prob (Y in X trials) = C (X, Y) \* Z/100 [sup]Y[/sup] \* ((100-Z)/100)[sup]X-Y[/sup] ---
**Author:** ![Cunctator](https://avatars.discourse-cdn.com/v4/letter/c/43a26b/32.png) [@Cunctator](https://boards.straightdope.com/u/Cunctator)\ **Post date:** [December 14, 2005, 4:09am UTC](https://boards.straightdope.com/t/statistics-question-re-shooting-basketballs/335531/5 "2005-12-14T04:09:34Z")
If you assume that each shot is random and that Z% are successful, then I think it’s just a standard binomial distribution. So the probability that, out of X attempts, Y will be successful is: Probability= X!/[Y!(X-Y)!] x Z[sup]Y[/sup] x (1-Z)[sup]X-Y[/sup] ---
**Author:** ![panamajack](https://avatars.discourse-cdn.com/v4/letter/p/47e85d/32.png) [@panamajack](https://boards.straightdope.com/u/panamajack)\ **Post date:** [December 14, 2005, 4:14am UTC](https://boards.straightdope.com/t/statistics-question-re-shooting-basketballs/335531/6 "2005-12-14T04:14:25Z")
Forget my other assumption, which **Cunctator** includes, and it goes along with what **ccwaterback** said : This assumes that each shot has Z% probability of going in. There are other ways one could get that ‘Z%’ over the long haul, but without any more information, it’s fairly reasonable. ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [December 14, 2005, 4:51am UTC](https://boards.straightdope.com/t/statistics-question-re-shooting-basketballs/335531/7 "2005-12-14T04:51:47Z")
> [@Cunctator](#): > > If you assume that each shot is random and that Z% are successful, then I think it’s just a standard binomial distribution. So the probability that, out of X attempts, Y will be successful is: > > Probability= X!/[Y!(X-Y)!] x Z[sup]Y[/sup] x (1-Z)[sup]X-Y[/sup] Thank you, thats very interesting. So there is only a .3% chance that our player makes or misses all of his shots. ---
**Author:** ![cerberus](https://avatars.discourse-cdn.com/v4/letter/c/71e660/32.png) [@cerberus](https://boards.straightdope.com/u/cerberus)\ **Post date:** [December 14, 2005, 5:12am UTC](https://boards.straightdope.com/t/statistics-question-re-shooting-basketballs/335531/8 "2005-12-14T05:12:11Z")
The Binomial/Bernoulli model may be adequate for this purpose, but: the model assumes that the probability of making a shot is constant; and that the shots are mutually independent. Realistically, the probability of making a shot depends on a number of things, and repeated shots may not be independent: a sequences of failed shots or successful shots may reduce future successful shots. The underlying concept of a Binomial experiment is that it involves a sequence of replications of the same process, with the replicated outcomes forming an independent set of trials. It may work, however, in practice, viewing the Bernoulli _p_ as some sort of “average probability of success.” ---
**Author:** ![Cunctator](https://avatars.discourse-cdn.com/v4/letter/c/43a26b/32.png) [@Cunctator](https://boards.straightdope.com/u/Cunctator)\ **Post date:** [December 14, 2005, 5:32am UTC](https://boards.straightdope.com/t/statistics-question-re-shooting-basketballs/335531/9 "2005-12-14T05:32:06Z")
> [@treis](#): > > Thank you, thats very interesting. So there is only a .3% chance that our player makes or misses all of his shots. Yes, I think that’s right. Using the example you gave: Probability of getting 0 or 8 shots = 0.00390625 Probability of getting 1 or 7 shots = 0.03125 Probability of getting 2 or 6 shots = 0.109375 Probability of getting 3 or 5 shots = 0.21875 Probability of getting 4 shots = 0.2734375 These add to 1, obviously, since they cover all of the possibilities. And as \*\*cerberus \*\* points out, there are lots of underlying assumptions implied by the use of the binomial distribution that may not be true.