Probability of Hitting SafelyDate: 11/06/2001 at 09:44:38 From: Jacob Garvin Subject: Probability Calculate the probability that a baseball player batting .357 would hit safely in at least 56 straight games, during the course of a 162- 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 thinking. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ 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 2^162. 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 ponder (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 (162/3.7)*.000028. ------------ 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 http://www.mathforum.org/dr.math/problems/becker.7.07.99.html - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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