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.
1You 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
