# Probability of Winning Election if Outcomes not Equally Likely

SebastianLinde 09/13/2018. 3 answers, 212 views

I just started learning probability, so my level is not very high. I am doing a homework problem, and my answer is different than the book's. I can't understand why. I see how the answer in the book makes sense, but I also see how my procedure makes sense. Could someone please point at what I am doing wrong?

Problem:

Four candidates, A, B, C, and D, are running in an election. A is twice as likely to be elected as B. B and C are equally likely. C is twice as likely as D. What is the probability that C will win?

The sample space is $\{A,B,C,D\}$. Since all the events are mutually exclusive, $S = A\cup B\cup C\cup D$. So, $P(S) = P(A)+P(B)+P(C)+P(D)$. Since $P(S) = 1$, $P(A)+P(B)+P(C)+P(D) = 1$. Since A is twice as likely as B, $P(A) =2\times P(B)$. Since B and C are equally likely, $P(B)=P(C)$. Thus, $P(A) = 2\times P(C)$. Since C is twice as likely as D, $P(C) = 2\times P(D)$. Therefore, the probability of the sample space in terms of C is: $1 = 6P(C)$. So, $P(C) = \frac{1}{6}$.

Let $p = P(D)$. Since C is twice as likely as D, $P(C) = 2p$. Since B and C are equally likely, $P(B) = 2x$. Since A is twice as likely as B, $P(A)= 4p$. Since $P(S) = 1$, $p + 2p + 2p + 4p = 1$. So, $p = \frac{1}{9}$ and $P(C) = \frac{2}{9}$.

Thanks!

N. F. Taussig 09/14/2018.

You correctly found that $\Pr(A) = 2\Pr(C)$, $\Pr(B) = \Pr(C)$, and that $\Pr(C) = 2\Pr(D)$. Since these events are mutually exclusive and exhaustive, $$\Pr(A) + \Pr(B) + \Pr(C) + \Pr(D) = 1$$ Substituting $2\Pr(C)$ for $\Pr(A)$, $\Pr(C)$ for $\Pr(B)$, and $\frac{1}{2}\Pr(C)$ for $\Pr(D)$ gives \begin{align*} 2\Pr(C) + \Pr(C) + \Pr(C) + \frac{1}{2}\Pr(C) & = 1\\ 4\Pr(C) + 2\Pr(C) + 2\Pr(C) + \Pr(C) & = 2\\ 9\Pr(C) & = 2\\ \Pr(C) & = \frac{2}{9} \end{align*} I suspect that you made an incorrect substitution for $\Pr(D)$. If you wrote $2\Pr(C)$ for $\Pr(D)$, you would have obtained the incorrect answer $\Pr(C) = 1/6$.

Henry 09/13/2018.

You correctly have $P(B)=P(C)$ and $P(A)=2P(C)$ but you have accidentally misused $P(C)=2P(D)$, which is equivalent to $P(D)=\frac12P(C)$

Done correctly you would then have $$2P(C)+P(C)+P(C)+\frac12P(C)=1$$ which, since $2+1+1+\frac12 = \frac92$, would lead to $$\frac92P(C)=1$$ and thus $$P(C)=\frac29$$

Phil H 09/13/2018.

The probabilities are in the ratio of $2,1,1\ \text{and}\ 0.5$ and must sum to $1$. $$P(A) = \frac{2}{2+1+1+0.5} = \frac{4}{9}$$

$$P(C) = \frac{1}{4.5} = \frac{2}{9}$$