Editing Bertrand's ballot theorem (section)

==Example==
Suppose there are 5 voters, of whom 3 vote for candidate ''A'' and 2 vote for candidate ''B'' (so ''p'' = 3 and ''q'' = 2).  There are ten equally likely orders in which the votes could be counted:
*''AAABB''
*''AABAB''
*''ABAAB''
*''BAAAB''
*''AABBA''
*''ABABA''
*''BAABA''
*''ABBAA''
*''BABAA''
*''BBAAA''
For the order ''AABAB'', the tally of the votes as the election progresses is:
{| class="wikitable" border="1"
|-
!  Candidate
|  ''A''
|  ''A''
|  ''B''
|  ''A''
|  ''B''
|-
!  ''A''
|  1
|  2
|  2
|  3
|  3
|-
!  ''B''
|  0
|  0
|  1
|  1
|  2
|}
For each column the tally for ''A'' is always larger than the tally for ''B'', so ''A'' is always strictly ahead of ''B''. For the order ''AABBA'' the tally of the votes as the election progresses is:
{| class="wikitable" border="1"
|-
!  Candidate
|  ''A''
|  ''A''
|  ''B''
|  ''B''
|  ''A''
|-
!  ''A''
|  1
|  2
|  2
|  2
|  3
|-
!  ''B''
|  0
|  0
|  1
|  2
|  2
|}
For this order, ''B'' is tied with ''A'' after the fourth vote, so ''A'' is not always strictly ahead of ''B''.
Of the 10 possible orders, ''A'' is always ahead of ''B'' only for ''AAABB'' and ''AABAB''. So the probability that ''A'' will always be strictly ahead is
:<math>\frac{2}{10}=\frac{1}{5},</math>
and this is indeed equal to <math>\frac{3-2}{3+2}</math> as the theorem predicts.