Ballot Theorem – 3 (26/365)

In the last post I showed how to count the number of valid vote sequences given that candidate A ends up with only one extra vote over candidate B. In general, candidate A may end up with n+k votes. How do the counts change?

In terms of balanced parantheses this means we have k extra open parantheses to make use of. Let’s say n=3 and k=3 then

  1. There are C_3 ways to arrange 3 votes for each candidate

  2. For each of the C_3 valid ways we can insert an extra open paranthesis in 7 possible places. After this, we can insert the next extra open paranthesis in 8 possible places. So, we now have 8 \times 7 \times C_3 arrangements.

  3. The last extra parantheses we know we have to place it at the beginning. But there are k choices for the first paranthesis. So now have 3 \times 8 \times 7 \times C_3 possible arrangements.

  4. Finally, since the extra parantheses are also indistinguishable we have to divide by (8!)(7!).

Thus the number of ways to arrange n+k votes for candidate A and n votes for candidate B where candidate A always has the higher number of votes is

\displaystyle  \frac{k \times (n+k-1)\times \dots \times (n+1)}{(n+k)\times \dots \times (n+2)} \times \frac{(2n)!}{(n+1)!n!} = \frac{k(n+k-1)!}{(n+k)!n!}

The probability that candidate A always has the higher number of votes than candidate B and ends up with a and b votes respectively is

\displaystyle  \frac{(a-b)(a+b-1)!}{(a+b)!n!} \div \frac{(a+b)!}{(a+b)!n!} = \frac{a-b}{a+b}

We have in fact arrived at the solution without the use of martingales. Next time, let’s see why this example is used in the chapter on martingales.

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