Colleagues
When rolling 2 (reg) dice there are 5 ways to get 6
(1,5),(2,4),(3,3),(4,2),(1,5)
Of these the latter 2 are but permutations of the first 2.
So there are 3 'non-equivalent' ways to get 6 from 2 dice.
When rolling 3 dice there are 10 ways to get 6
(1,1,4),(1,2,3),(1,3,2),(1,4,1)
(2,1,3),(2,2,2),(2,3,1)
(3,1,2),(3,2,1)
(4,1,1)
There are once more 3 'non-equivalent' ways to get 6 from 3 dice.
Can someone give me a formula for the number of 'non-equivalent' ways to
get a total of m from n rolls of an m sided dice?
John Haslett
(It is not obvious to me to what probabilty problem this might be (part of)
the solution; but it seems to be a legitimate question and has in fact
arisen in theoretical political science in the guise of "how many
'non-equivalent' ways can 100 seats be allocated to the 10 parties which
have nonn-zero represention in the parliament?")
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John Haslett E-mail [log in to unmask]
Professor Phone +353 1 6081114 (direct)
Department of Statistics +353 1 6081767 (sec)
Trinity College Fax +353 1 6615046
Dublin 2,Ireland
WWW: http://www2.tcd.ie/Statistics/staff/jhaslett.html
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