Why is the formula for expectation as is in this question?

Question

The problem I am looking at comes from a STEP paper and is as follows

I managed to do the first part correctly, however, I was unsuccessful in doing the second part. The answers say that we should have

however, shouldn't the expectation be $\sum x P(X=x)$? I don't understand how they got their formula. I see that they summed the winnings for that ammount of people times the probability of that many people wining but I don't know why this gives the expectation (as it seems to differ from the aforementioned formula). Could somebody explain why this is the case and what is the basic principle for future reference?

Answer 1

It seems to me the fundamental problem is how you read the formula $\sum x P(X=x)$.

If you are taking the expected value of a binomial distribution (for example), let's say the binomial distribution for $n$ trials with probability $p,$ you would take the sum like this:

$$ E(X) = \sum_{x = 0, 1, 2, 3, \ldots, n} x P(X=x). $$

This is because the possible outcomes of the binomial distribution are $0, 1, 2, 3, \ldots, n.$ These are the possible values of "number of successes" in a sequence of $n$ trials. It is not possible to have the outcome $\frac12$ or the outcome $\frac{17}3.$

In short, to get the expected number of successes, you take each possible number of successes and multiply by the probability of getting that number of successes, and add up all those products.

But this problem is not asking for an expected number of successes. It is asking for expected net loss. Since net loss is the same as the cost ($N$) minus the net winnings, the answer is also $N$ minus the expected winnings, which is how other people are answering this question.

But the possible values of the expected winnings are not $0, 1, 2, 3, \ldots, n.$ The possible winnings are $W$ (if no other ticket wins), $W/2$ (if one other ticket wins), $W/3$ (if two other tickets win), and so forth.

So to get the expected winnings, you take each possible amount of winnings and multiply by the probability of getting that amount of winnings, then add up the products like this:

$$ E(X) = \sum_{x = W, \frac W2, \frac W3, \ldots} x P(X=x). $$

And that is exactly how people are solving this problem.

Answer 2

The probability that there are $k$ winners is approximately $p_k=e^{-2.8}(2.8)^{k-1}/(k-1)!$ (using Poisson model). The win for the syndicate is then $W/k$. The expected win for the syndicate is $$\sum_{k=1}^\infty p_k\frac{W}k=W\frac{e^{-2.8}}{2.8}\sum_{k=1}^\infty\frac{(2.8)^k}{k!} =\frac{5W}{14}e^{-2.8}(e^{2.8}-1)=\frac{5W}{14}(1-e^{-2.8}).$$ The syndicate's expenditure is of course $N$.

Answer 3

Let's do a slightly more general answer. Suppose that the syndicate buys $N$ tickets and that the rest of the ticket buyers buy $\lambda N$ tickets, such that the total tickets purchased is $(\lambda +1)N$. We can get the easy part of the question out of the way instantly:

Find the most probable number of winning tickets.

Well, since we assume that the non-syndicate buyers buy randomly, we can say that on average, $1$ in $N$ of them will be winners. And, since the syndicate buys each number exactly once, they will have exactly $1$ winning ticket. So the most probable number of winning tickets is $$\frac{\lambda N}{N}+1=\lambda+1.$$

Find the expected loss of the syndicate.

The easiest way to do this is to find the expected winnings of the syndicate, then subtract their cost, which is £$N$. Let $X$ be the winnings of the syndicate, a random variable. You are correct when you state that the formula for expected value of a discrete random variable is $$\sum_{x\in\mathcal{X}}x\cdot \mathrm{Pr}(x)$$ Where $\mathcal{X}$ is the set of all possible values $X$ can take. A good place to start is to figure out what this is. Remember that the amount of money that each winning ticket is awarded depends on how many people selected that winning number. We know that the syndicate will have exactly one winning ticket. Let the number of non-syndicate winners be $M$ (another random variable). Then, $$X=\frac{W}{M+1}\implies \mathrm{E}[X]=W\cdot\mathrm{E}\left[\frac{1}{M+1}\right]$$ Using LOTUS, we can write this as $$\mathrm{E}[X]=W\cdot\sum_{m\in\mathcal{M}}\frac{1}{m+1}\mathrm{Pr}(m)$$ So what is $\mathcal{M}$? Well, since there are $\lambda N$ non-syndicate players in total, $m$ can be integer in the range $[0,\lambda N]$. Now the more important question: Given $\lambda N$ players, what is the probability that exactly $m$ of them win? Well, we assume the tickets are independent, and it is clear that the probability that any individual ticket is a winner is $1/N$, so we can use the binomial distribution: $$\mathrm{Pr}(m)={}_{\lambda N}\mathrm{C}_m \left(\frac{1}{N}\right)^m\left(1-\frac{1}{N}\right)^{\lambda N-m}$$ Which, assuming $N$ is large, can be approximated by a Poisson distribution: $$\mathrm{Pr}(m)\approx \frac{\lambda^m e^{-\lambda}}{m!}$$ Therefore $$\mathrm{E}[X]=W\cdot \sum_{m=0}^{\lambda N}\frac{1}{m+1}\frac{\lambda^m e^{-\lambda}}{m!}$$ Rewriting, $$\mathrm{E}[X]=\frac{We^{-\lambda}}{\lambda}\cdot\sum_{m=0}^{\lambda N}\frac{\lambda^{m+1}}{(m+1)!}$$ This is a classic sum (albeit missing the first term), and again assuming $N$ is fairly large $$\mathrm{E}[X]\approx \frac{We^{-\lambda}}{\lambda}(e^{\lambda}-1)=\frac{W}{\lambda}(1-e^{-\lambda})$$ Since the syndicate had to spend £$N$ to make these winnings the expected loss of the syndicate is $$\text{Expected loss}=N-\frac{W(1-e^{-\lambda})}{\lambda}$$ In our example, we are told $3.8N$ tickets are purchased in total. Since the syndicate bought exactly $N$ of those, in this example, $\lambda=3.8-1=2.8=\frac{14}{5}$. Thus in this example $$\text{Expected loss}=N-\frac{W(1-e^{-2.8})}{14/5}=N-\frac{5W(1-e^{-2.8})}{14}.$$ Introducing random numbers into a problem can often lead to confusion. I would recommend trying problems totally symbolically first, then plugging in the numbers once the hard work is done.