% Chance / Probability of multiple events occuring 1.5 times or more. Baseball Question!

Question

I have the data, and the answer, but can't figure out the math driving it. Baseball Hitter has a x% chance of meeting or exceeding 1.5 hits,rbis,and runs in a game (think player proposition bets like gamblers bet on). Here is the projected data given for the hitter in the game:

Hits Doubles Triples HomeRuns Runs RBIs

1.09 0.17 0 0.16 0.5 0.64

Additional information that I don't know if is needed or not: Plate Appearances in game = 4.31. At Bats in game = 3.92

Here is the answer to my question: Probability of a player getting at least 1.5 Hits,RBIs,Runs = 0.652683 Probability of a player NOT getting 1.5 Hits,RBIs,Runs =0.34731683

How do I arrive at this answer? I've tried everything as far as solving for probability that I have learned, read about, etc. I cannot arrive at these numbers in the answer. I also do understand that I may be using the term probability incorrectly as that is the % chance that this will or will not occur... Thank you in advance!

Also, I am assuming that you may only need Hits 1.09 and Runs 0.5 and RBIs .64 to solve this? This is because Hits is the sum of all of the other kinds of hits (singles, doubles, triples, home runs).... If you needed to solve for Singles, you would just need to subtract Doubles, Triples, and HomeRuns from Hits = .76 singles

I've spent hours and hours trying to solve this (find the correct equation). I've had no luck even through articles, youtube, etc. Thanks again!

Answer 1

Hello Greedy,

Define a random variable X to be the total number of hits, runs, and rbi's the player gets in any randomly selected game. Hits, runs, and rbi's could each be considered individual random variables, and they are certainly not independent, but we are still justified in using the data given to conclude that the expected value (or mean) of X is

μ = E[X] = 1.09 + .5 + .64 = 2.23

It is important to note that although the average value of X (per game) is a decimal number, the actual value of X in any one given game must be a whole number. That is, X is a discrete random variable.

The simplest method to approach the problem is to assume that X follows a Poisson distribution. A discrete random variable X has the Poisson distribution if its probability mass function is given by

(Eq.1) p(x) = (e^-μμ^x)/x!, for x = 0,1,2,3,4,......

In the above formula, the parameter μ is the mean value of X, and, as noted previously, in our problem, X is the number of hits + runs + rbi's in any given game. We want the probability that X meets or exceeds 1.5. That is, we want P(X≥1.5). But since X takes on only whole number values, this is the same P(X≥2) (the probability that X is greater than or equal to 2.) That is,

(Eq.2) P(X≥1.5) = P(X≥2)

= P(2) + P(3) + P(4) + P(5) + ....

Since X≥2 corresponds, at least in theory, to infinitely many possible values of X, we would have to apply the Poisson formula infinitely many times. Instead of calculating the probability of X≥2 directly, calculate the probability of the complement of the event X≥2 and subtract this from 1. The complement of "X≥2" is "X<2", which is equivalent to X = 0 or 1. Therefore,

(Eq.3) P(X≥2) = 1 - P(X<2)

= 1 - [p(0) + p(1)]

From the Poisson distribution given above (Eq.1), we have

p(0) = (e^-2.23·2.23⁰)/0!

p(0) ≅ .1075

and

p(1) = (e^-2.23·2.23¹)/1!

p(1) ≅ .2398

Plug these results in to Eq.3 to obtain

P(X≥2) = 1 - [.1075 + .2398]

P(X≥2) = 1 - .3473

P(X≥2) ≅ .6527 (rounded to the fourth decimal place.)

Therefore, the probability that X is at least 1.5 [the same as P(X≥2)] is .6527. Formally, any probability p is a number between 0 and 1, inclusive (0≤p≤1). However, it is often convenient to informally report a probability in percent terms, which in our case, would be 65.27%.

Hope that helps! Please let me know if you need any additional explanation.

William