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Leading candidate never trailing in a 100 to 80 count

Question

In an election candidate AA receives nn votes and candidate BB receives m<nm<n. The ballots are shuffled and counted one at a time, with a running total for each. Let Q(n,m)Q(n,m) be the probability that AA's running total is never below BB's at any point after counting begins (ties are allowed). Compute Q(100,80)Q(100,80).

Short answer

21/10121/101. All orderings are equally likely, so count lattice paths. A path is bad iff it touches the line where BB leads by one; reflecting the part up to the first touch gives a bijection with all paths to a shifted endpoint, so Q(n,m)=1−mn+1=n+1−mn+1Q(n,m)=1-\tfrac{m}{n+1}=\tfrac{n+1-m}{n+1}. Here that is 21101\tfrac{21}{101}.

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

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