If there was a way to systematically double your buy in within 29 spins would this give you an advantage if your odds are 37/1 in your favor every spin? If so A.)What is the percentage of advantage you have. B.) Should I factor in standard deviation considering your entire buy in is at risk until you double your money at 29 spins.
Let alone standard deviation for now. First you must understand that you odds can never be 37/1.
Maybe I should re-word. 37 possible outcomes and only 1 of the outcomes would result in a loss of my buy in every single roll, but I am able systematically double my buy in at 29 spins by manipulating my bets. What is the % of the advantage?
Obviously I don't know the way you bet, but I can assure you two things.
First, there is no advantage. The dis-advantage is always 2,7% against the player.
Secondly, necessarily not all your 36 numbers would pay the same. Because if you bet the same amount on each number then in case of a hit you will win nothing (the payout on the winning number is equal to the units you lost on the other numbers). Technically at least one of your 36 numbers should produce a loss. Not a total loss, but a loss.
But let's say that you have devised a way to bet on 36 numbers and produce a profit in case they hit. According to the mathematics, the average profit would be less than 1/36 (or 2,7% of your bet).
Let me tell you this: Theoretically (if you played with very low denominations like one can online) you can double your bankroll much sooner than in 29 spin by betting 35 numbers, if you reinvested your profit. However this doesn't prove anything. There is no "advantage". No matter your betselection or betting amount, your risk of ruin when you try to double your bankroll, is 48,7%, like with any Simple Chance.
I'm not a zealot of "the house edge rules roulette" approach, but in your case, since you asked for a calculation of the "advantage", I have to tell you you cannot beat the house edge.