Roullette math

What would be the Odds on roulette if say all I wanted to do was make and stop at 200 dollars by placing 5 dollars on 1-12 and 5 dollars on 2-12 with a progression of 5 $5-$25-$125 and starting back a 5 dollars if the 125 didn't hit. Reason being starting over after 125 I figure I wouldn't have that much of a hole to dig myself out of especially If it hits 1-12 at 125 dollars

1-12 and 13-24, I hope you mean?

What do you mean odds? Probability of success depends also on how much you're willing to lose.

Probability to lose one bet specifically is 14/38, since you'd have 24 numbers covered. Losing 3 in a row is (14/38)^3. So slightly more likely than 5% to lose 3 straight.

Question 1: what's your initial bankroll? You're more likely to make 200 if you're willing to risk 5000 than if you're only willing to risk 500.

Question 2: if one of your two bets wins, do you reset just that bet, or both of them?
For example, if your first bet is 5 on 1-12 and 5 on 13-24, and the number is 10, is your next bet 5 on 1-12 and 25 on 13-24, or 5 each on both of them?

Quote: rivbogambler

What would be the Odds on roulette if say all I wanted to do was make and stop at 200 dollars by placing 5 dollars on 1-12 and 5 dollars on 2-12 with a progression of 5 $5-$25-$125 and starting back a 5 dollars if the 125 didn't hit. Reason being starting over after 125 I figure I wouldn't have that much of a hole to dig myself out of especially If it hits 1-12 at 125 dollars

https://wizardofvegas.com/member/oncedear/blog/#post1370 gives you an approx answer

Lets say with a bankroll of 5000 with a progression of 5 x on 1-12 and 13-24 with a starting bet of 10 so 5 dollars on each being 5 25 and I would start back over at 125 even if it still lost reason being I figure it would not take as long climbing out of a 310 dollar hole from the previous bets. I'm just trying to figure the math of it hitting the 0s or 3rd 12 3 times in a row

It didn't really give me the answer I was looking for but it did help thanks

Quote: rivbogambler

It didn't really give me the answer I was looking for but it did help thanks

Hi,
I'm afraid you do need to state the bankroll that you are prepared to put 'at risk' before it is in any way possible to answer your question.

Quote: rivbogambler

Lets say with a bankroll of 5000 with a progression of 5 x on 1-12 and 13-24 with a starting bet of 10 so 5 dollars on each being 5 25 and I would start back over at 125 even if it still lost reason being I figure it would not take as long climbing out of a 310 dollar hole from the previous bets. I'm just trying to figure the math of it hitting the 0s or 3rd 12 3 times in a row

RS answered the question of losing 3 in a row: (14/38)³ = 343/6859 = 0.05001

Now the question of winning $200 with such a system, assuming the following:
-Bankroll of $5000
-Both bets resets to $5 after any of them wins
-If the available bankroll is not enough to cover the bet, the largest bet is made with the remaining bankroll
-Likewise, if the difference between $200 and the current winnings is less than the bet, the smallest bet that will reach the goal on a win is made

After 100,000 simulations with those assumptions (hopefully done correctly):
12408 lost $5000
13087 lost $4995
41614 won $200
32891 won $205

For an expected loss of $1123

Basically a 25.5% chance to lose $5000 and a 74.5% chance to win $200.

Thank you