The other day as I was perusing the MTBoS, I found an interesting problem.

“5 distinct numbers are chosen at random from {1,2,3,4,5,6,7,8,9}.

P(k) = probability their sum = k.

What are some of the ways you can find this in general?

What sum is/are the least likely?

Which sum is/are most likely?”

http://matharguments180.blogspot.ca/2014/10/269-probability-sums.html

At first, I started trying to make different numbers with different sums. For example: choose 2,3,5,6,7 and you get 2+3+5+6+7 = 23.

Then I realized there would be many possible sums. 126 in fact.^{1}

I didn’t really feel like working with all 126 of those different sums so I wrote a quick script in Python to do it for me. Here are the results:

You can clearly see that 25 is the most likely sum with 12 ways of making 25. For fun, here are all 12 ways of making 25:

[1, 2, 5, 8, 9] |

[1, 2, 6, 7, 9] |

[1, 3, 4, 8, 9] |

[1, 3, 5, 7, 9] |

[1, 3, 6, 7, 8] |

[1, 4, 5, 6, 9] |

[1, 4, 5, 7, 8] |

[2, 3, 4, 7, 9] |

[2, 3, 5, 6, 9] |

[2, 3, 5, 7, 8] |

[2, 4, 5, 6, 8] |

[3, 4, 5, 6, 7] |

Also, 15, 16, 34, and 35 are the least likely sums. Since the table is symmetric about 25, this also means that, on average, your sum will be 25. Problem solved.

^{1}You have to select 5 numbers from the set of 9 total. This corresponds to the mathematical idea of “choose.” We have 9 choose 5 = **126 possible options**

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