I think you are using 'correlation' in a colloquial, rather than
a technical sense. Here is a test of hypothesis based on your Comment, with a bit of
rationalization for it.
By random chance, the probability of agreement would be 1/3. So the
null distribution is $X \sim Binom(40, 1/3),$ where $X$ counts
agreements. If you have $X = 30$ agreements, then that's more than
the 'expected' $E(X) = np = 40(1/3) = 13.33.$
The P-value of a test
would be $P(X \ge 30) \approx 0.$ If colors choices were at random
without regard to team favoritism, it would be almost impossible
to get 30 'agreements'. Probability computations below are from
R statistical software:
1 - pbinom(29, 40, 1/3) # 'pbinom' is binomial CDF
## 8.474878e-08
x = 30:40; sum(dbinom(x, 40, 1/3)) # 'dbinom' is binomial PDF
## 8.474878e-08
And from Minitab 17:
Test and CI for One Proportion
Test of p = 0.3333 vs p > 0.3333
Exact
Sample X N Sample p 95% Lower Bound P-Value
1 30 40 0.750000 0.612940 0.000
The interpretation of the one-sided 95% CI is that you have 95%
'confidence' that the true proportion $p$ of 'agreements' exceeds $0.61.$
Note: One might use a normal approximation to get the P-value. Since it turns
out to be so small, an approximation would be OK. But in general, for
$n$ as small as 40, I think exact computation is better.
Addendum: (response to Comment) Here are computations in R for your
proposed chi-squared test. You have 2 cells, hence df = 2-1 = 1.
The chi-squared goodness-of-fit statistic is $Q = \sum_{i=1}^2 (X-E)^2/E,$
where X's are observed counts and E's are expected counts. $Q \sim Chisq(1),$
provided $E_i > 5.$ The computed value is $Q = 31.25;$ the 5% critical value
is $q^* = 3.8415$. Reject 'fit' because $Q > q^*;$ P-value of this
approximate test is essentially $0.$
Obs = c(30, 10); prob = c(1/3, 2/3)
n = 40; Exp = n*prob
Q = sum((Obs-Exp)^2/Exp); Q
## 31.25
qchisq(.95, 1) # critical value
## 3.841459
1 - pchisq(Q, 1) # aprx P-value
## 2.268475e-08