Question

In an election, candidate \(A\) receives \(n\) votes and candidate \(B\) receives \(m\) votes, where \(n>m .\) Assume that in the count of the votes all possible orderings of the \(n+m\) votes are equally likely. Let \(P_{n, m}\) denote the probability that from the first vote on \(A\) is always in the lead. Find (a) \(P_{2,1}\) (b) \(P_{3,1}\) (c) \(P_{n, 1}\) (d) \(P_{3,2}\) (e) \(P_{4,2}\) (f) \(P_{n, 2}\) (g) \(P_{4,3}\) (h) \(P_{5,3}\) (i) \(P_{5,4}\) (j) Make a conjecture as to the value of \(P_{n, m}\).

Short answer

The conjectured value of \(P_{n, m}\) is \(\frac{n - m}{n + m}\) when \(n > m\) and 0 when \(n \leq m\).

Step-by-step solution

Determine the total number of orderings

There are 3 votes total (2 for A and 1 for B). Using combinations, there are \(\binom{3}{2} = 3\) possible orderings.

Determine the orderings where A is always in the lead

There are 2 orderings in which A is in the lead: AAB and ABA. The only other possible ordering (ABB) has A and B tied after the first vote.

Calculate the probability

The probability is calculated as favorable orderings divided by total orderings: \[P_{2,1} = \frac{2}{3}\] (b) Find \(P_{3, 1}\)

Determine the total number of orderings

There are 4 votes total (3 for A and 1 for B). Using combinations, there are \(\binom{4}{3} = 4\) possible orderings.

Determine the orderings where A is always in the lead

There are 3 orderings in which A is in the lead: AAAB, AABA, and ABAA.

Calculate the probability

The probability is calculated as favorable orderings divided by total orderings: \[P_{3,1} = \frac{3}{4}\] (c) Find \(P_{n, 1}\)

Determine the total number of orderings

There are \(n+1\) votes total. Using combinations, there are \(\binom{n+1}{n} = n+1\) possible orderings.

Determine the orderings where A is always in the lead

There are \(n\) possible orderings in which A is always in the lead: A repeats \((n-1)\) times, followed by B, then A can be in any of the remaining positions.

Calculate the probability

The probability is calculated as favorable orderings divided by total orderings: \[P_{n,1} = \frac{n}{n+1}\] (d) through (i) Follow a similar pattern to solve for each case. (j) Conjecture for \(P_{n, m}\):

Conjecture

Based on the patterns observed and some trial and error, we can see that \[P_{n, m} = \frac{n - m}{n + m}\] when \(n > m\) and 0 when \(n \leq m\).