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Probability of Hitting Safely

Date: 11/06/2001 at 09:44:38
From: Jacob Garvin
Subject: Probability

Calculate the probability that a baseball player batting .357 would 
game regular season, assuming 4 at bats per game. 

I don't even know where to start. There must be a lot of different 
cases involved. I know he would have 648 at bats in the year, and 
between 231 and 232 of those at bats equals a .357 batting average.

Thanks a lot!

Date: 11/06/2001 at 11:31:50
From: Doctor Mitteldorf
Subject: Re: Probability

Jacob -

A good place to start is to calculate the probability that he'll get 
a hit in any given game. There are 4 at-bats, each with a .357 
probability of success. What is the probability that at least one of 
them will result in a hit?

A next step is to take the answer above and raise it to the 56th 
power, for 56 straight games.   

Write back after you've gone this far - the next part requires more 

- Doctor Mitteldorf, The Math Forum

Date: 11/06/2001 at 13:02:00
From: Jacob Garvin
Subject: Batting average

I did those steps and ended up with .82906^56 = .000027594.  

Any suggestions as to what is next?

Date: 11/07/2001 at 21:26:23
From: Doctor Mitteldorf
Subject: Re: Batting average

Jacob -

So far, so good.  I got the same answers you did. I also spent a few 
hours this morning trying to think of how to continue to an exact 
solution, and couldn't come up with one. Then I consulted a reference, 
and found out there's an advanced mathematical literature on the 
subject, leading me to believe that it's an "interesting problem."

I had some ideas for ways to make progress with it, and to get an 
approximate answer. Here are two of my thoughts:

1. A way you might look for an approximate answer is this: Start with 
"coin flips" instead of ball games, so that each trial has an equal 
chance of success or failure.  Then you just need to count the number 
of strings of H-T contain the string of 56, out of a universe of 

You might think of it this way: there are 2^162 total possibilities. 
If you fix the first 56, then ther are 2^106 possibilities for the 
remaining games. So (2^106)/(2^162) is the probability of getting the 
first 56 heads. But you could equally well start with #2 or #3...all 
the way up to #107.  This leads to 107*(2^106) possibilities. 

But what about the string in which the first 57 are all heads?  We've 
counted that as part of our #1, and we've counted it again as part ouf 
our #2. We've double-counted all such strings, so we ought to correct 
by subtracting the number of strings with 57 heads in a row, which by 
the same method we just used is 106*(2*105). This leads us to a 
probability of [107*(2^106)-106*(2*105)] / (2^162).

This is not exactly the right answer; but strangely enough, the same 
approach DOES give exactly the right answer. So I'll leave you to 

  (a) when does this approach give exactly the right answer?

  (b) when is it appropriate to continue by "correcting the 
      correction" so you get 
      [107*(2^106)-106*(2*105)+105*(2*104)-104*(2*103)+...] / (2^162)?

  (c) how would you modify this approach to take into account a 
      probability for "heads" that is not exactly 0.5?

2. Here's how I thought you might get a reasonable approximate answer:  
You've calculated that the probability of hitting in 56 consecutive 
games is .82906^5 6= .000027594. How many chances to do this does a 
batter get in a 162-game season? In other words, how many INDEPENDENT 
trials does the season contain? 

Well, it certainly isn't more than 107, because there are only 107 
games in which he could begin his hitting streak. And it isn't less 
than 162/56 = 2.89, because he could fit 2.89 complete trials of 56 
games in the season. So the probability we seek is .000027594 times 
some number between 2.89 and 107. We can make a good guess by asking, 
on average, how many games go by before the batter has to start over?  
If he has a probability of 0.83 of hitting in each game, then, "on 
average" his streak of games with hits will end after about 3.7 games, 
since 0.83^(3.7) is approximately 0.5.  So our estimate is 


There are many other approaches you might take to getting a good 
approximate answer, and many ways to have fun with this question and 
learn from it.  

When you're done with that, you might look at this page from our 
archives, where our resident statistics expert answered a similar 
question a few years ago.

   Consecutive Failures in Bernoulli Trials

- Doctor Mitteldorf, The Math Forum
Associated Topics:
College Probability
High School Probability

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