= Solution
Four groups have $\binom42=6$ pairwise comparisons. Testing each at <test size> $\alpha=0.05$ without adjustment generally increases the <familywise error rate>, the chance of at least one <type I error>. If the six rejection events were independent, that chance would be $1-(1-0.05)^6\simeq0.265$; shared samples make the actual comparisons dependent, so this expression is illustrative rather than exact.
A single omnibus <ANOVA> tests equality of all four means with one stated <test size>, avoiding that six-test inflation for the global question. If the aim is to identify differing pairs, follow with suitably adjusted comparisons, such as <Bonferroni correction> or a simultaneous pairwise method. A significant omnibus test does not automatically license six unadjusted claims. The benefit concerns control of <type I error>, rather than a guarantee that the model's independence and common-variance assumptions hold.
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