OK...you buy in at a roulette table, and, like Oscar, the table "owes" you a unit. You put 1 unit on an even chance. It loses. Repeat the bet. It loses. Repeat...it loses. You lose five straight EC bets, so now you are down 6 units, counting the unit that the table "owes" you. You continue, and on the next bet, you win one unit on the even chance. You are now down 5 units.

Oscar calls for a two unit bet on the EC. Under this method, you bet one unit on a dozen. It loses. Down 6 units. You repeat the bet, and you win. Down 4 units.

Oscar calls for a 3 unit bet on EC. Under this method, you put one unit on an EC, and one unit on a dozen. It loses. Down 6 units. Same bet, and the EC wins, but the dozen loses. No change, still down 6 units. On the next bet, the dozen wins, but the EC loses. You won 1 unit. Down 5 units.

Oscar calls for a four unit bet on the EC. Under this method, bet one unit on a dozen, and one unit on a column. Both bets win, and now you are down 1 unit (basically what the table "owes" you). You bet one unit on an EC, it wins, and now you are up 1 unit.

I don't know how feasible this method is. The jury is still out on this question. I can see a situation where you're down 30 or 40 units, and you placing one unit on a single street. It will take forever to make your one unit profit. It probably needs a computer simulation to see how it works out.

Kynge