Quote:
Originally Posted by mike alexander
(I preface this with the warning that I can never keep the formulas for permutations, subluxations and combinations straight. Although I approve of using factorials in formulae as a way of making math more exciting.)
I think using a pair of dice as illustration is a different problem, since there are six ways to get a total of 7. But that is not the question here.
My example: a and b are running and six people vote. The final tally has to be one of the following:
a b
6 0
5 1
4 2
3 3
2 4
1 5
0 6
There are 7 possible vote totals, of which 1 is a tie. Probability of a tie is 1/7.
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With 6 votes cast for two different candidates, you have 2
6 possible elections.
Of these, there are 6!/((6-n)!n!) ways to get n votes.
a b #of ways to get this
6 0 1 aaaaaa
5 1 6 aaaaab aaaaba aaabaa aabaaaa abaaaa baaaaa
4 2 15 aaaabb aaabab aaabba aabaab aababa aabbaa abaaab abaaba ababaa abbaaa baaaab baaaba baabaa babaaa bbaaaa
3 3 20 aaabbb aababb aabbab aabbba abaabb ababab ababba abbaab abbaba abbbaa repeat with a's and b's swapped to get the rest
2 4 15 bbbbaa bbbaba bbbaab bbabba bbabab bbaabb babbba babbab bababb baabbb abbbba abbbab abbabb ababbb aabbbb
1 5 6 bbbbba bbbbab bbbabb bbabbb babbbb abbbbb
0 6 1 bbbbbb
so given no voter bias, the chances are ~1.5% for one candidate to get all votes, ~9.5% for one to get all but one vote, 23.4% for one to get all but 2, and a whopping 31.25% or almost 1/3 change to get a tie.