Winning an election does not mean having led throughout the count. This puzzle asks for something finer: not just who wins in the end, but how often the lead was never even threatened.
The Count That Never Turns
Riddle statement
In an election there are two candidates, A and B.
A receives 7 votes.
B receives 5 votes.
The 12 votes are counted in a random order.
What is the probability that A is always ahead of B throughout the entire count?
That is: after the first vote, after the second, after the third… A always has more accumulated votes than B. B never ties or overtakes A.
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Solution
The probability is $1/6$.
All orders of the 7 votes for A and the 5 votes for B are equally likely. There are:
possible orders.
Bertrand's ballot theorem says that if A receives $a$ votes and B receives $b$ votes, with $a>b$, then the fraction of counts in which A remains strictly ahead throughout is:
The formula can be obtained by pairing, through a reflection at the first tie, the counts that lose the lead with paths ending on the opposite side. The only paths left unpaired are those that never touch a tie.
Here $a=7$ and $b=5$:
This corresponds to 132 favorable orders out of the 792 possible orders.
Answer: the probability is $1/6$.