Doesn't the above assume(incorrectly) that after you remove an A-vote then the probability remains the same(5/10) for getting another A-vote?
I mean it's not a 10-sides-dice that is thrown 3 times for the probability to stay the same. After you remove one A-vote(the one 5/10 of the (5/10)^2) the probability to get another A-vote should lower a bit from 5/10.
My way of calculating the desired probability results in that the result depends on the total numbers of votes, let's say
total number of votes it is 10N(we don't choose N, for the sake of simplicity to not have to deal with fractions like 5N/10, but it's equivalent whatever we choose). And it is:
(We have 5N A-votes, 3N B-votes and 2N C-votes)
Which is equal to:
$\displaystyle \frac{{6{N^2}\left( {25(5N - 1) + 21(3N - 1) + 16(2N - 1)} \right)}}{{20N(10N - 1)(10N - 2)}} =$
$\displaystyle \frac{{3N\left( {25(5N - 1) + 21(3N - 1) + 16(2N - 1)} \right)}}{{10(10N - 1)(10N - 2)}} = $
$\displaystyle \frac{{3N\left( {125N - 25 + 63N - 21 + 32N - 16} \right)}}{{10(10N - 1)(10N - 2)}} = $
$\displaystyle \frac{{3N\left( {220N - 62} \right)}}{{10(10N - 1)(10N - 2)}} = $
$\displaystyle \frac{{3N\left( {110N - 31} \right)}}{{5(10N - 1)(10N - 2)}}$
That can't be simplified any more and it depends on the value of N. The value of the votes. And it is different if we talk about 10 voters or 100.
Am i mistaken?