I'm wondering if I have cheated here?

A sporting team, 41 games, 45% chance of winning a home game, 30% chance of winning away game. 18 home games, 23 away games. Assume that each game is independent

What is the probability that they would win at most 11/41 games?

I thought this sounds like a binomial problem, expected to win 18 x 0.45 + 23 x 0.3 = 15. 15/41 = 0.366

Then the solution is to look at the CDF of B(41,0.366) where X <= 11?

  • $\begingroup$ Can you really expect game wins to be independent? There's a lot of evidence to support they are not -- hot teams tend to recruit better and win more games, thus increasing $p$ in later trials. As an analogy from sport, Alabama football demonstrates this well. $\endgroup$ Mar 14, 2019 at 1:15
  • 1
    $\begingroup$ sorry should have said (assume they are independent), I fully agree but let's assume for this example $\endgroup$
    – James
    Mar 14, 2019 at 2:17

2 Answers 2


The approach you describe is misleading. It assumes the probability of winning a particular game, independent of whether home or away, is 0.366. While the expected number of wins you derived (15) is correct (since the expectation operator is linear and therefore $E(X+Y)=E(X)+E(Y)$), it does not help address questions requiring the distribution function to be answered, such as determining $P(W\leq 11)$.

Try instead to derive the probability mass function (pmf) of a new variable $W$, denoting total wins, that is the sum of two independent binomials, $X$ and $Y$, which denote home and away wins, respectively. Then we can write the pmf for $W$ as $$P(W= w) = \sum_{k=0}^{w} P(X =k)P(Y=w-k),$$ which is found by simply adding the probabilities of each possible way to observe $w$ wins, noting that $(X =k)$ and $(Y=w-k)$ are independent events. $P(X=k)$ and $P(Y=w-k)$ are the (18,.45) and (23,.3) binomial pmfs evaluated at $k$ and $w-k$, respectively. To obtain the cdf for $W$, we add the values of of the pmf of $W$ from 0 up to the value of interest. This is given by

$$P(W \leq w) = \sum_{z=0}^wP(W= z) =\sum_{z=0}^w \sum_{k=0}^{z} P(X =k)P(Y=z-k).$$

Evaluating this at $w=11$ gives $$P(W \leq 11) =\sum_{z=0}^{11} \sum_{k=0}^{z} P(X =k)P(Y=z-k).$$

To clarify any possible confusion about notation, $$P(X=k)= {18 \choose k} (.45)^k (.55)^{18-k}$$ if $k\in \{0,1,2,...,18\}$ and zero otherwise. Similarly, $$P(Y=z-k)= {23 \choose z-k} (.3)^{z-k} (.7)^{23-z+k}$$ if $(z-k) \in \{0,1,2,...,23\}$ and zero otherwise.


You could see the final number of won games as a Poisson binomial distribution. This is a generalization of the binomial distribution.

  • The binomial distribution is the sum of Bernouilli processes with equal probabilities
  • The Poisson binomial distriution is the sum of Bernouilli processes with different probabilities

You can approximate this reasonably well with a normal distribution that has the same mean, $\mu = \sum p_i$, and variance $\sigma^2 =\sum p_i(1-p_i)$.

In your case it looks like this:


You can see that the difference is not that big. When you integrate those probabilities you get values close to 0.124

> # probabilities x <= 11:
> # True sum of two binomials
> sum(dbibinomial(0:11, n=18, n2=23, p=0.45, p2=0.3))
[1] 0.1241174
> # approximation with normal distribution (summing discrete values)
> sum(dnorm(0:11, 41*p, sqrt( 18*0.45*(1-0.45)+23*0.3*(1-0.3) ) ))
[1] 0.124287
> # approximation with normal distribution (integrating)
> pnorm(11.5, 41*p, sqrt( 18*0.45*(1-0.45)+23*0.3*(1-0.3) ) )
[1] 0.125356
> # approximation with single binomial
> sum(dbinom(0:11, size=41, prob = p))
[1] 0.1271134

where the probability distribution for the sum of two binomials is computed as:

# function to compute distribution of sum of two binomials
dbibinomial <- function(x,n,n2,p,p2) {
  xs <- max(0,x-n2):min(x,n)
  ps <- dbinom(xs,n,p)*dbinom(x-xs, n2, p2)
dbibinomial <- Vectorize(dbibinomial)

The effect of using the Poisson binomial distribution is that the variance will be lower in comparison to using the binomial distribution. $$\sum_{i=1}^n p_i(1-p_i) = n\bar{p}(1-\bar{p}) - \sum_{i=1}^n (\bar{p}-p_i)^2$$ or $$\text{Var}(X_{\text{Poisson binomial}}) = \text{Var}(X_{\text{binomial}}) - \sum_{i=1}^n (\bar{p}-p_i)^2 $$

Smaller variance

So, the variance of the nummer of winning games $X$ will be smaller than expected when you use a binomial distribution.

Seperating the games into sets of home and away games is already a good step. But, you should actually also vary the winning probability based on the strengths of the opponent teams. The effect is that the variance will be even less.

An extreme case is when you have only probabilities $p_i=1$ or $p_i=0$ to win making the value $X = \sum(p_i)$ zero variance (ie certainty). Of course it is not realistic but it does illustrate the effect more intuitively; how the final sum of wins will be less variable (to even no variation at all) when the win probability varies among different games.

Larger variance

However, there is also an effect that increases the variance. The number of 0.366 wins for a particular team is probably not a very certain value or constant in time.

So, on the one hand, when you look at a team that wins 36.6% of the games, then you could compute the mean and standard deviation for 41 games $41 \cdot p \pm \sqrt{41 \cdot p \cdot (1-p)} \approx 15 \pm 3$ and then expect the deviation to be less than this $\sigma = 3$ based on the idea of a Poisson binomial distribution having smaller variance.

However, on the other hand, the performance of a team is likely not to be fixed. Thus you will get random variations in the probabilities from year to year (or also within a year). This would be much comparable to a beta binomial distribution which has a larger variation.

So, if you do this calculation in order to predict results, e.g. for betting on games, then you should design a more complex model that incorporates both the systematic variations from game to game, and as well the random variations due to a performance not being constant and fully predictable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.