### Mitchell reported expected probability

Today at 2:00, the "Mitchell Report" will be released, dealing with Major League Baseball's "performance-enhancing drug" issues.

This provides an excellent opportunity for playing with binomial probability distributions. There are 30 Major League teams, and reportedly as many as 70-80 players named. The binomial probability formula says

P(k out of n) = (n! / k!(n-k)!) * (p^k)*(q^(n-k))

where n is the number of trials, k is the number of successes, p is the probability of success and q is the probability of failure. In the simplistic case, assuming random distribution of named players, and 80 players named, we can construct a probability table using n = 75, p = (1/30), and q = (1-p) as follows:

Players named | Probability of that many players | Expected number of teams |
---|---|---|

0 | 7.87% | 2 |

1 | 20.34% | 6 |

2 | 25.96% | 8 |

3 | 21.78% | 7 |

4 | 13.52% | 4 |

5 | 6.62% | 2 |

6 | 2.66% | 1 |

7 | .91% | 0 |

8 | .27% | 0 |

9 | .07% | 0 |

10 | .02% | 0 |

We'd expect to see 2 teams with no named players and 1 team with 6, just as a matter of simple probability.

Now, it isn't, of course, that simple. Most players have player for more than one team. If Roger Clemens (to take one name that has been alleged) is in the list, he played for the Red Sox, Blue Jays, Yankees and Astros. If we assume that the average named player has played for two teams, than p changes from 1/30 to 2/30. And the table changes to:

Players named | Probability of that many players | Expected number of teams |
---|---|---|

1 | 3.03% | 1 |

2 | 8.01% | 2 |

3 | 13.93% | 4 |

4 | 17.91% | 5 |

5 | 18.16% | 5 |

6 | 15.13% | 5 |

7 | 10.66% | 3 |

8 | 6.47% | 2 |

9 | 3.44% | 1 |

10 | 1.62% | 0 |

11 | .68% | 0 |

12 | .26% | 0 |

Now we don't expect any teams to have no players named. The odds are that every team will have a player named who either is, or has been, affiliated with the team.

Labels: Mitchell Report, MLB, steroids

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