Playing On A Saturday

Question

So, I am trying to solve this question: James and John are playing football/soccer on a sunny Saturday. They are practicing their penalties and play 3 games.

In each game, they either score or they miss. How many possible ways can their games turn out. Now, I was stuck between 2 answers which are $2^3 $ and $4^3$. After trying to write it out, I realized that $2^3 $ is obviously wrong and the answer is $4^3$. But that is just an aside.

Now, the second part is my issue. As you know, in football/soccer, you get a point for each goal you score. Now, the question is asking that James and John have been playing this exact game for $100 $ Saturdays (weird right, I know) and within that time, they have discovered that James scores about $70\% $ of the time while John scores about $ 60\%$ of the time. Knowing that, how many points do I think they have racked up over this period.

I interpreted this as a round-about way of asking the Expected Value but with a $100$ games here, I don't know where to start. I calculated that the following probabilities:

$$P(\text{they both fail}) = 0.12$$

$$P(\text{James scores, John fails}) = 0.28$$

$$P(\text{James fails, John Scores}) = 0.18$$

$$P(\text{they both score}) = 0.42$$

Now, from here, I would simply multiply the probabilities by how many goals are in each situation but there have to be $4^{100}$ of them and I can't even count that high. Any help please?

Answer 1

Bernoulli Trials

Suppose you carry out a coin toss to settle who gets the ball at the start of a soccer game. The coin, unbeknownst to you, is biased, with $P(H) = p$, where $p$ is not necessarily $\dfrac12$. Then, the probability that heads comes up is $p$, and that of tails coming up is $1-p$. Say $p=0.3$, then $P(H) = 0.3$ and $P(T) = 0.7$. Suppose you believe your lucky side is heads, so you call heads every time. From your perspective, the event "Heads" is a success, while "Tails" is a failure.

Such kinds of random experiments, which have only two kinds of possible outcomes, one described as a success (which has a probability $p$) and the other a failure, are called Bernoulli Trials.

Binomial Distribution

Suppose due to the low probability of Heads, the first toss gives Tails. Frustrated, you beg the referee to toss the coin $100$ times, thinking that the Heads and Tails will even out due to the large number of tosses (You don't know that the coin is biased!).

Let $X$ denote the number of successes in $n$ tosses. Then, the probability that $X=k$, is the probability that exactly $k$ Heads turn up in $n$ tosses.

To find this, we can choose any $k$ of the $n$ tosses in $\displaystyle\binom nk$ ways. Then the probability that they return Heads is $p^k$ and the probability that the rest return Tails is $(1-p)^{n-k}$. Hence the probability that $X=k$ is $$P(X=k)= \binom nk p^k(1-p)^{n-k}$$ which is the binomial distribution, and here $X$ is said to be distributed binomially with parameters $n$ and $p$, or in short, $X\sim Bin(n, p)$.

The expectation of the random variable $X$ is defined as $$E(X) = \sum_{k=1}^n k\cdot P(X = k)$$ which, for $X\sim Bin(n, p)$, gives $$E(X) = \sum_{k=1}^n k\cdot \binom nk p^k(1-p)^{n-k} = np.$$ This is extremely intuitive, as if you toss a coin $n (=100)$ times with probability $p(=0.3)$ of getting Heads, then you should reasonably expect something near $np = 0.3\times 100 = 30$ Heads to turn up.

Solution to your question

To solve this, we MUST assume that the random variables $X_1$ (number of goals James scores) and $X_2$ (number of goals John scores) are independent, i.e. the number of goals scored by John doesn't affect James and vice-versa (come on, we can't possibly model psychological effects mathematically).

Then, since James gets 300 chances and scores $70\%$ of the time, $X_1 \sim Bin(300, 0.7)$. Similarly, since John gets 300 chances too but scores only $60\%$ of the time, we write $X_2\sim Bin(300, 0.6)$. Thus,

$$E(X_1) = 300\times 0.7 = 210$$ $$E(X_2) = 300\times 0.6 = 180$$ $$E(X_1 + X_2) = E(X_1) + E(X_2) = \color{red}{390}.$$

Answer 2

The expectation of the score of a player called $X$ in $n$ independent games, each with $m$ possible outcomes $\{x_1,...,x_m\}$ is $$E(X)=n\sum_{k=0}^mx_kP(x=x_k).$$

We have $n=3\times100=300$ games and $m=2$. $$E(Je)=300(0\times0.3+1\times0.7)=210$$ $$E(Jo)=300(0\times0.4+1\times0.6)=180$$ $$E(Je+Jo)=E(Je)+E(Jo)=390.$$