Haloform reaction with limited bromine
Let's take a look at the first product you propose: 1-bromopropan2-one.

The mechanism includes deprotonation by hydroxide to form an enolate. The resulting enolate is better stabilized with more electronegative atoms (Br in this case) and will form more readily than another molecule of acetone turning into an enolate ion. Thus, as the reaction proceeds, the product distribution will be determined by the most stable enolate anion. This will be the tribromoacetyl intermediate enolate ion and will drive the reaction forward. So a whole mol of bromine will be converted completely to the tribromoacetyl intermediate with a 1/3 mol of acetone. Notice how only a 1/3 will react to give the monosubtituted intermediate, than 1/3 mol of bromine will react to give disubstituted product, and lastly the last 1/3 mol of bromine will react to form the trisubstitued product. The last step is displacement by hydroxide via the nucleophilic acyl substitution mechanism (not SN2!) to form acetate and release bromoform. This reaction isn't really great if bromine isn't in excess.