Probability of exactly 2 goals in soccer

Question

Suppose we know that the probability of x goals in a given soccer game is:

• $10$% when $x = 0$

• $20$% when $x=1$

• $25$% when $x = 2$
• $45$% when $x >2$

Every week $45$ games are played, contestants will choose $8$ matches from these $45$ and are given the following:

A) if the selected $8$ matches all have exactly $2$ goals scored each then the contestant will win \$$50$k

B) If there is EXACTLY $8$ matches with $2$ goals in each game then the contestant will win \$$100$k (i.e. if there is $>8$ matches with $2$ goals each then this scenario will not be possible, but A might be if the contestant chose the correct $8$ games)

I am trying to find the probability of scenario A and B.

Perhaps I am missing something, but is this not a binomial with parameters $Bin(45, 0.25)$ i.e. in any given game we have a probability of $0.25$ of getting exactly two goals.

Then scenario B will be $P(X=8)$ and scenario A will be $P(X>8)$?

It seems a little too simple - any advice

Answer 1

If $X$ is the number of matches with exactly two goals, then you are correct that $X$ is a binomially distributed random variable with parameter $p:=0.25$. However, there is a catch. A contestant has pre-chosen the matches, and cannot change his mind. Even if $X\geq 8$, it is no guarantee that this contestant will win anything.

To give an answer to this problem, we just look at the eight chosen matches. The probability to win the prize $B$ equals the probability that all the chosen matches turn up with exactly two goals, but no other matches (outside the chosen matches) result in two goals. That is, $$\text{Prob}(\text{winning }B)=p^8(1-p)^{45-8}=p^8(1-p)^{37}\,.$$

On the other hand, to win $A$, all eight chosen matches must turn up with exactly two goals, but at least one of the other matches result in exactly two goals. This means $$\text{Prob}(\text{winning }A)=p^8-\text{Prob}(\text{winning }B)=p^8-p^8(1-p)^{37}=p^8\big(1-(1-p)^{37}\big)\,.$$