The problem is the following: In a party with N $(n \geq 1 ; n \in \Bbb N)$ persons, they are deciding what to have for dinner there are Two options. For an example Option A could be Pizza and Option B could be Hambuguese, all people in the party MUST vote, either for option A, option B but if they are indifferent about having either Pizza or Hamburger they can Vote for them two. The problem consists in knowing the quantity of assistants to the party, calculate the probability.
I think it this way, There are three options, Option A = Pizza Option B = Hambuguese Option C = Both
So the probability of a tie between Pizza and Hambuguese is $\left(\frac{(N * (N-1))+1)}{3^N}\right)$
This so called formula works for N = 1,2 and 3. But is good for all N being N an entire positive number? The numerator is the chance of having a Tie, for example with 3 people in the party: ABC, ACB, BAC, BCA, CBA, CAB, CCC Will lead to a tie between people prefering pizza and people prefering Hambuguese. In this case N=3 Numbers of possible ties among all possible combinations is 7 (3*(3-1))+1 over 27 $\left(\frac{(3 * (2))+1)}{3^3}\right)$ Does the formula works for any N?
Thanks in advance, looking forward to reading your answers!
You are assuming that each person votes for one of the three options randomly, which is where the $3^N$ denominator comes from. Your formula is the probability of a $0-0$ tie, the $\frac 1{3^N}$ plus the probability of a $1-1$ tie. The probability of a $1-1$ tie is $\frac {N(N-1)}{ 3^N}$ because you have $N$ choices for the person who votes for pizza, $N-1$ choices for the person who votes for hamburgers, and everybody else has to vote for both. For $N\gt 3$ there are possibilities for ties at higher numbers of votes. For $N=4$ the chance of a $2-2$ tie is $\frac 6{3^4}$ because there are six ways to choose the two people who voted for pizza.
Generally, to get the chance you need to add up the chances of $k-k$ ties as $k$ runs from $0$ to $\lfloor \frac N2 \rfloor$. This is $$\sum_{k=0}^{\lfloor \frac N2 \rfloor}\frac {{N \choose k}{N-k \choose k}}{3^N}=_2F_1(1/2 - N/2, -N/2, 1, 4)$$ where $_2F_1$ is a hypergeometric function.