You have a categorical outcome: either win or lose (1 vs. 0). You have two groups (Canberra and Sidney) and you have 300 samples from each group. The usual appropriate tests here are either Z-test for two proportions or Chi-squared-test for 2x2 contingency tables (both are equivalent).
With increasing N, the t-test will approximate the results of the above mentioned tests but I disagree with asdf that it is already equivalent for N=300 per group. Since the t-test does assume a continuous outcome I would not consider it to analyze the data here.
If we play around with following R code, we see that the p-values are quite different even for N=1000 per group.
prob1 <- 210/300
prob2 <- 200/300
sample1 <- rbinom(1e3, 1, prob = prob1)
sample2 <- rbinom(1e3, 1, prob = prob2)
t.test(sample1, sample2)$p.value # p = 0.567
chisq.test(cbind(table(sample1), table(sample2)))$p.value # p = 0.599
There is also an exact test available: Barnard's test. This is the most powerful test for equality of proportions and may be considered as the best test for this situation.
To answer your question: you are right and your wife is wrong. However, even from a statistical perspective I would not recommend to insist on it too stubbornly :-)