Covariance of a Roulette Spin

Question

$N$ = number of spins

$N_R$ = number of spins that land on red

$N_B$ = number of spins that land on black

$r=$ chance of landing on red

$b=$ chance of landing on black

$Cov(N_R, N_B | N=n) = -n\cdot r \cdot b$

Why is that?

Answer 1

For $n = 5$ spins of an American roulette wheel $E(R) = E(B) = 5(18/38) = 2.368421,\,$ $SD(R) = SD(B) = \sqrt{5(18/38)(20/38)} = 1.116484,$ and $$Cov(R,B) = -5(18/38)^2 = -1.121884.$$

A simulation with a million 5-spin games approximates these values to about three significant digits, confirming what is easy to prove using binomial distributions and also in good agreement with the formula you hope to prove for the covariance. (See below.)

While indicator variables are useful for doing a general proof of the covariance structure of a multinomial distribution, it has occurred to me that a more elementary approach will work for this simpler tri-variate roulette problem. Consider the equality

$$Var(R + B) = V(R) + V(B) - 2Cov(R, B).$$

Using binomial arguments it is possible to express $Var(R + B),\,Var(R),\,$ and $Var(B)$ in terms of $n, r,$ and $b.$ Then solve for $Cov(R, B).$ The algebra is moderately tedious, but I got it to work.

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If you want to use indicator random variables for the formal proof, it is possible that thinking about results for individual spins as in the simulation will help you set set them up.

set.seed(4418)
wheel=c(0,1,2); cp = c(2,18,18)/38 # red=1, black=2, green=0
n = 5;  m = 10^6;  x = y = numeric(m)
for(i in 1:m) {
  col = sample(wheel, n, repl=T, pr=cp)
  x[i] = sum(col==1);  y[i] = sum(col==2) }
mean(x);  mean(y);  sd(x);  sd(y);  cov(x,y)
## 2.367798
## 2.36911
## 1.115676
## 1.11556
## -1.11976
cor(x,y)
## -0.8996922

Particular results for the first half dozen 5-spin games were as shown below. The third and fifth games happened to have green outcomes. Of course, it's the possibility of green outcomes that keeps the correlation from being $-1.$

 head(cbind(x,y))
      x y
 [1,] 4 1
 [2,] 0 5
 [3,] 1 3
 [4,] 2 3
 [5,] 4 0
 [6,] 5 0