• There is an event that can occur 0 or more times during a given discrete process.
  • An observation that the average number of events per process is p.


  • Using only p above is it possible to state the probability that the event will not occur during a given process? If so, what is it? If not, what more do I need to know?
  • What is the common nomenclature for describing problems of this type?


Clayton Kershaw of the LA Dodgers has a WHIP of 0.67. That is the number of walks plus hits per inning pitched. (w+h/i) Statistically, what is the probability that no walks or hits occur during an inning Kershaw pitches?

Note: I think I know how to work this problem if an event can occur exactly once or not at all, but the fact that some times it may occur many times is throwing me off.

  • 1
    $\begingroup$ This seems like a natural place to use a Poisson Process. To be clear, though: this is a model only. There is no guarantee that any physical process will follow a Poisson process, or any other. Still, it often fits observed data quite nicely. $\endgroup$
    – lulu
    Commented Jun 22, 2016 at 21:51
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    $\begingroup$ For example: applying this to Clayton, using your numbers, we get $e^{-.67}\sim .5117$ for the probability that neither walk nor hit occurs in a given inning (a scary number indeed, for non-Dodger fans). $\endgroup$
    – lulu
    Commented Jun 22, 2016 at 21:56
  • $\begingroup$ Hard to know about the right model. Many baseball watchers are likely to believe that pitchers suffer from episodes of wildness, that is, that walks are not independent. But that may be inaccurate, just like the "hot bat" theory, now largely debunked. $\endgroup$ Commented Jun 22, 2016 at 22:05
  • $\begingroup$ The WHIP stat is relevant only if one believe that innings (or 1/3 innings in fact) are independent. It is likely not true completely, but it does give saber metric folks numbers to chew on. Interestingly, if we take an inning without a walk or a hit as an approximation of a perfect inning that still means the chance that Kershaw will throw a perfect game as ~1/500. $\endgroup$ Commented Jun 23, 2016 at 23:36

1 Answer 1


If the discrete process is a Markov process with no state (that is, if the number of success events is simply the result of $N$ independent trials, then the number of successes is described by a Bernoulli distribution with $p = \mu/N$, and indeed this is well approximated by a Poisson process if $N$ is large and $\mu N$ is not too small.

The exact answer in that case, however, would be simpler: $(1-p)^N$.

However, nobody has told you that the process is the result of $N$ independent trials. For example, if the process chooses success or failure for trial 1, each with probability $\mu/N$, and then replicates that result on each of the remaining $N-1$ trials, then the probability of zero successes is as large as it can be, namely $1-\mu/N$.


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