In the American Roulette wheel, the winning odds of betting colours(black/red) is 47.36%. Consider this method, if the minimum single bet is \$10 and if I were to bet \$10 and double my next bet of what my previous bet is $10 if I happen to lose my previous bet for a total of 3 times going forward making the total bet count of 4, will this be a profitable play in the long run.
Like this way:
Initial bet: \$10, if I win I put \$10 again for the next bet. If I lost I put double the amount of my previous bet for my next bet, which is \$10 making it \$20. If I win this bet of \$20, I go back to betting \$10, but if I lost I will double this amount for my next bet making it \$40. I'm going to do this for a total of 3 consecutive times in the event of a lost, so it's 10 + 20 + 40 + 80 = \$150. So if each single bet gives me 47.36%, 0.4736 * 4 = 1.8944 * 100 =189.4% meaning if I were to do this doubling for 3 consecutive times for a total of 4 bet counts, each single bet technically gives me 189% chance of winning. Am I right, is this the correct way of calculating it?
I was also thinking about the application of this same method to the dozen bets. For a dozen bet, the winning odds is 31.57%, but the payout is 2 to 1 vs. only 1 to 1 for colour bets. If the same method is applied to betting, which of the two types of bets will be more profitable in the long run?