# Expected Value of Election Results

• vparam
In summary: H) = p(S|...|H)p(H) = p(S)p(H) + p(H)p(S)$$Expected value:$$E(X) = 2(.4) + 1(.3) + 1(.1) + 0(.2) = 1.2$$vparam Homework Statement Suppose that in the 2050 election, polls show Party A has a 60% chance of winning the senate and a 50% chance of winning the house, and that if Party A wins the house, it has an 80% chance of winning the senate. Let X be the random variable whose value is the the number of congressional houses won by Party A. (a) Find the expected value of X. Relevant Equations ##p(S|H) = \frac{p(S \cap H)}{p(H)}## ##E(X)=\sum_{s \in S}{p(s)X(s)}## I submitted this solution, and it was marked incorrect. Could I get some feedback on where I went wrong? Let S represent the event that Party A wins the senate and H represent the event that Party A wins the house. There are 4 cases: winning the senate and house (##S \cap H##), winning just the senate (##S \cap \overline{H}##), winning just the house (##H \cap \overline{S}##), and winning neither (##\overline{S} \cap \overline{H}##). I computed the probabilities for each compound event as follows:$$p(S \cap H) = p(S|H)p(H)= (.8)(.5) = .4p(S \cap \overline{H}) = p(S)p(\overline{H}) = (.6)(.5) = .3p(H \cap \overline{S})=(p(S) - p(S \cap H))p(H) = (.6 - .4)(.5) = .1p(\overline{S} \cap \overline{H}) = p(\overline{S})p(\overline{H}) = (.4)(.5) = .2$$This means that the expected value is:$$E(X) = 2(.4) + 1(.3) + 1(.1) + 0(.2) = 1.2$$vparam said: There are 4 cases: winning the senate and house (##S \cap H##), winning just the senate (##S \cap \overline{H}##), winning just the house (##H \cap \overline{S}##), and winning neither (##\overline{S} \cap \overline{H}##). Your overbars seem to show up when I quote your post, but I can't see them in the original. vparam said: I computed the probabilities for each compound event as follows:$$p(S \cap H) = p(S|H)p(H)= (.8)(.5) = .4p(S \cap \overline{H}) = p(S)p(\overline{H}) = (.6)(.5) = .3p(H \cap \overline{S})=(p(S) - p(S \cap H))p(H) = (.6 - .4)(.5) = .1p(\overline{S} \cap \overline{H}) = p(\overline{S})p(\overline{H}) = (.4)(.5) = .2$$You have a total probability of winning the Senate as ##0.7##. And, you've assumed the probabilities are indepenent in some of those calculations. I drew a Venn diagram. vparam said: This means that the expected value is:$$E(X) = 2(.4) + 1(.3) + 1(.1) + 0(.2) = 1.2$$Two of those numbers are wrong. vparam vparam said:$$p(S \cap \overline{H}) = p(S)p(\overline{H}) = (.6)(.5) = .3$$Only if S and H are independent events. But it is easy to show that they are not independent. vparam said:$$p(\overline{S} \cap \overline{H}) = p(\overline{S})p(\overline{H}) = (.4)(.5) = .2$$Again, only if S and H are independent events. vparam PeroK said: I drew a Venn diagram. Thank you for the reminder to use a Venn diagram! The problem breaks down more easily after drawing one out. I've redone the calculations as follows:$$p(S \cap H) = p(S|H)p(H)= (.8)(.5) = .4p(S \cap \overline{H}) = p(S) - p(S \cap H) = .6 - .4 = .2p(H \cap \overline{S}) = p(H) - p(S \cap H) = .5 - .4 = .1p(\overline{S} \cap \overline{H}) = 1 - p(S \cup H) = 1 - (p(S) + p(H) - p(S \cap H)) = 1 - (.6 + .5 - .4) = .3$$This means that the expected value is:$$E(X) = 2(.4) + 1(.1) + 1(.2) + 0(.3) = 1.1$$FactChecker said: Only if S and H are independent events. But it is easy to show that they are not independent. Again, only if S and H are independent events. And thank you for pointing this out. This explains why my previous calculations don't make sense. PeroK said: Your overbars seem to show up when I quote your post, but I can't see them in the original. Is \overline the standard code for overbars, or should I use something else? Thank you both for the help! vparam said: Thank you for the reminder to use a Venn diagram! The problem breaks down more easily after drawing one out. I've redone the calculations as follows:$$p(S \cap H) = p(S|H)p(H)= (.8)(.5) = .4p(S \cap \overline{H}) = p(S) - p(S \cap H) = .6 - .4 = .2p(H \cap \overline{S}) = p(H) - p(S \cap H) = .5 - .4 = .1p(\overline{S} \cap \overline{H}) = 1 - p(S \cup H) = 1 - (p(S) + p(H) - p(S \cap H)) = 1 - (.6 + .5 - .4) = .3$$This means that the expected value is:$$E(X) = 2(.4) + 1(.1) + 1(.2) + 0(.3) = 1.1$$And thank you for pointing this out. This explains why my previous calculations don't make sense.Is \overline the standard code for overbars, or should I use something else? Thank you both for the help! I don't understand why the Latex renders only when I quote your post and not in the original. I can't see the overbars in your posts, but I can when I reply. PS it doesn't work when I take them out of the quote box!$$p(S \cap \overline{H}) = p(S) - p(S \cap H) = .6 - .4 = .2$$FactChecker PeroK said: I don't understand why the Latex renders only when I quote your post and not in the original. I can't see the overbars in your posts, but I can when I reply. PS it doesn't work when I take them out of the quote box!$$p(S \cap \overline{H}) = p(S) - p(S \cap H) = .6 - .4 = .2$$I'm not sure why that's the case since I'm very new to Latex and it seems to show up properly on my end. I wrote up everything again using an alternate notation instead for visibility. Probability calculations:$$p(S \cap H) = p(S | H) p(H)= (.8)(.5) = .4p(S \cap H^{\mathrm{C}}) = p(S) - p(S \cap H) = .6 - .4 = .2p(H \cap S^{\mathrm{C}}) = p(H) - p(S \cap H) = .5 - .4 = .1p(S^{\mathrm{C}} \cap H^{\mathrm{C}}) = 1 - p(S \cup H) = 1 - (p(S) + p(H) - p(S \cap H)) = 1 - (.6 + .5 - .4) = .3$$Expected value computation:$$E(X) = 2(.4) + 1(.1) + 1(.2) + 0(.3) = 1.1$$1.1 is right. Good job working out the probabilities. Also, his is a trick question! Expectancies add linearly. Let ##S## be the number of senate seats won, and ##H## the number of house seats won (each is either 0 or 1. This is called an indicator variable and is a very common idea to use) Then you want to compute ##E(S+H)=E(S)+E(H)##. The correlations are irrelevant. vparam and PeroK PeroK said: I don't understand why the Latex renders only when I quote your post and not in the original. I can't see the overbars in your posts, Same here. PeroK said: but I can when I reply. PS it doesn't work when I take them out of the quote box!$$p(S \cap \overline{H}) = p(S) - p(S \cap H) = .6 - .4 = .2$$When I reply to you, the overline of the first H shows.$$p(S \cap \overline{H}) = p(S) - p(S \cap H) = .6 - .4 = .2
And now it doesn't. (There is an overline of the first H. If it matters, I am using the Firefox browser.)

## 1. What is the expected value of an election result?

The expected value of an election result is a statistical concept that represents the average outcome of a particular election. It takes into account the probability of each possible outcome and calculates the weighted average of those outcomes.

## 2. How is the expected value of an election result calculated?

The expected value of an election result is calculated by multiplying the probability of each possible outcome by the value associated with that outcome, and then summing these values together. This calculation can be represented mathematically as: E(x) = ∑(P(x) * x), where E(x) is the expected value, P(x) is the probability of outcome x, and x is the value associated with outcome x.

## 3. Why is the expected value of an election result important?

The expected value of an election result is important because it can help predict the most likely outcome of an election. It also allows for a comparison of different election scenarios and can inform decision-making processes.

## 4. What factors can influence the expected value of an election result?

The expected value of an election result can be influenced by a variety of factors, such as polling data, historical voting patterns, economic conditions, and campaign strategies. It can also be affected by unforeseen events or changes in public opinion.

## 5. Can the expected value of an election result accurately predict the outcome?

While the expected value of an election result can provide a good estimate of the most likely outcome, it is not a guarantee. Other factors, such as voter turnout and last-minute changes in public opinion, can impact the actual outcome of an election. However, the expected value can still be a useful tool for understanding potential outcomes and making informed decisions.

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