Let's pretend (we'll return to whether this is reasonable in a minute) that this team has 5 trials or attempts a game and has a 40% chance of success in each trial.

\(n = 5, p = 0.4\) \(X =\) # of goals scored (a random variable) \(E[X] = np = 2\) goals

\(P(X = k) = {n \choose k} p^k (1-p)^{n-k}\)

So,

\(P(X = 0) = {5 \choose 0} .4^0 (1-.4)^{5-0} = 0.6^5 =\) 0.07776 \(P(X = 1) = {5 \choose 1} .4^1 (1-.4)^{5-1} =\) 0.2592

With 10 attempts

\(n = 10\) \(p = 0.2\) \(E[X] = np = 2\) goals (just as before)

\(P(X = 1) = {10 \choose 1} .2^1 (1-.2)^{1-1} =\) 0.2684355

Blue: \(n = 50, p = .04, E[X] = np = 2\)

Red: \(n = 500, p = .004, E[X] = np = 2\)

Instead of of a number of attempts, \(n,\) and rate of success, \(p,\) we'll think about an expected number of successes \(\lambda.\) and let n go to infinity.

\(P(X = k) = {n \choose k} p^k (1-p)^{n-k}\)

\(p = \frac{\lambda}{n}\)

\(P(X = k) = \lim_{n \to \infty} {n \choose k} (\frac{\lambda}{n})^k (1-\frac{\lambda}{n})^{n-k}\)

We'll break this into four parts:

\(P(X = k) = \frac{\lambda^k}{k!} \cdot A \cdot B \cdot C\)

where \(A =\lim_{n \to \infty} k! \cdot {n \choose k} (\frac{1}{n})^k\) and \(B = \lim_{n \to \infty} (1-\frac{\lambda}{n})^{n}\) and \(C = \lim_{n \to \infty} (1-\frac{\lambda}{n})^{-k}\)

\(A = \lim_{n \to \infty} k! \cdot {n \choose k} (\frac{1}{n})^k\)

\(\lim_{n \to \infty} k! \cdot \frac{n!}{k! (n-k)!} (\frac{1}{n})^k\)

\(\lim_{n \to \infty} \frac{n!}{(n-k)!} (\frac{1}{n})^k\)

\(\lim_{n \to \infty} n \cdot (n-1) \cdot (n-2) \cdot ... \cdot (n-k+1) (\frac{1}{n})^k\)

\(\lim_{n \to \infty} \frac{n \cdot (n-1) \cdot (n-2) \cdot ... \cdot (n-k+1)}{n^k}\)

\(\lim_{n \to \infty} \frac{n}{n} \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdot ... \cdot \frac{n-k+1}{n} = 1 \cdot 1 \cdot 1 \cdot ... \cdot 1 = 1\)

So A just equals 1.

\(B = \lim_{n \to \infty} (1-\frac{\lambda}{n})^n\)

subtitute \(j = -1 \cdot \frac{n}{\lambda}\) and we have

\(B = \lim_{j \to \infty} (1+\frac{1}{j})^{-1 \cdot \lambda j} = \lim_{j \to \infty} [(1+\frac{1}{j})^{j}]^{-1 \cdot \lambda}\)

But what is \(\lim_{j \to \infty} (1+\frac{1}{j})^{j}\) ?

So \(B = e^{-\lambda}\)

What about C? Recall, \(C = \lim_{n \to \infty} (1-\frac{\lambda}{n})^{-k}\).

and as \(n \to \infty\) this becomes \(1^{-k}\) or \(\frac{1}{1^k}\) or just 1.

Putting it all together we have:

\(P(X = k) = \frac{\lambda^k}{k!} \cdot 1 \cdot e^{-\lambda} \cdot 1\)

\(P(X = k)= \frac{\lambda^k }{k!} e^{-\lambda}\)

\(P(X = k)= \frac{\lambda^k }{k!} e^{-\lambda}\)

\(\lambda = 2\)

\(P(X = 0)= \frac{2^0 }{0!} e^{-2} = e^{-2}\) = 0.135

\(P(X = 1)= \frac{2^1 }{1!} e^{-2} = 2 e^{-2}\) = 0.271

\(P(X = 2)= \frac{2^2 }{2!} e^{-2} =\frac{4}{2} e^{-2}\) = 0.271

\(P(X = 3)= \frac{2^3 }{3!} e^{-2} = \frac{8}{6} e^{-2}\) = 0.18

\(P(X = 4)= \frac{2^4 }{4!} e^{-2} = \frac{16}{24} e^{-2}\) = 0.09

\(P(X = 5)= \frac{2^5 }{5!} e^{-2} = \frac{32}{120} e^{-2}\) = 0.036

\(P(X = k)= \frac{\lambda^k }{k!} e^{-\lambda}\)

For Rare Events when \(p\) is very small or \(n >> k\)

and

rate of successes is constant

Soccer goals and Hockey goals are “rare” relative to the number of plays in the game… but is a teams chance of scoring (per minute) the same across games?

The Rangers scored 3.0 goals per game and allowed 2.3. Pittsburg scored 2.6 goals per game and allowed 2.5. League Average is about 2.55 goals per game.

We expect the Rangers to score: 3.0 + 2.5 - 2.55 = 2.95 goals per game We expect the Penguins to score: 2.6 + 2.3 - 2.55 = 2.35 goals per game

But what do the distributions look like?

Blue (Rangers): \(\lambda = 2.95\) goals per game

Red (Penguins): \(\lambda = 2.35\) goals per game

It gives the home team ~0.25 more goals and the road team ~0.25 fewer.

The Rangers scored 3.0 goals per game and allowed 2.3. Pittsburg scored 2.6 goals per game and allowed 2.5. League Average is about 2.55 goals per game.

At home we expect the Rangers to score: 3.0 + 2.5 - 2.55 + 0.25 = 3.2 goals per game

On the road we expect the Penguins to score: 2.6 + 2.3 - 2.55 - 0.25 = 2.1 goals per game

On the road we expect the Rangers to score: 3.0 + 2.5 - 2.55 - 0.25 = 2.7 goals per game

At home we expect the Penguins to score: 2.6 + 2.3 - 2.55 + 0.25 = 2.6 goals per game

Blue: \(\lambda = 3.2\) goals per game

Red: \(\lambda = 2.1\) goals per game

Blue: \(\lambda = 2.7\) goals per game

Red: \(\lambda = 2.6\) goals per game