Will this system work to get the advantage at roulette?

Question: A friend told me a roulette system based on the third 12-number column of the betting layout having eight red and four black positions. You bet $5 on this column and $5 on black, and should win an average of $25 in every 38 spins. Is this true?

Answer: I've run into this error before. Here's how the system is usually explained for 38 "statistically-correct" spins. Note that columns pay 2-to-1; blacks pay 1-to-1 and either lose half or everything on 0 or 00 depending on the rules at particular casinos.

(a) Zero and double zero are expected twice; in casinos where outside numbers lose only half on these results, you’d be out $5x2=$10 on the columns and $2.50x2=$5 on the blacks. A $15 net loss.

(b) Black is expected 18 times. Of these, 14 should be in the first two columns, so blacks win and columns lose, a break-even. Four should be in the third column, winning $5x4=$20 on blacks and $10x4=$40 on columns. A $60 net win.

(c) Red is expected 18 times. Of these, 10 should be in columns one and two, losing both bets – $10x10=$100. Eight should be in the last column, winning $10x8=$80. A $20 net loss.
According to this logic, you should lose $15 on 0 and 00, win $60 on blacks, and lose $20 on reds – a $25 overall profit.

The error is in (c). With red in the first two columns, both bets lose – $10x10=$100 – as indicated. What about red in the third column? The column bet earns $10x8=$80 but the black bet loses $5x8=$40, netting $80-$40=$40. So expected loss is $100-$40=$60, not $20. In all, you should therefore lose $15 on (a), win $60 on (b), and lose $60 on (c) – a $15 net loss. Separately, columns win $10x12=$120 and lose $5x26=$130; blacks win $5x18=$90 while losing $5x18=$90 plus $2.50x2=$5 – the same $15 net loss.

The results are worse if black loses everything instead of half on 0 or 00. You can plug in the numbers to see how it affects the results.