I have a goal to raise $3,500 to pay off a gambling debt. I have a starting bankroll of $620. Here is a simple plan that I think may allow me to reach my goal in less than 47 days without taking a huge risk.
1. Each day, I will place one standard sides wager (-110) to WIN $100. If it wins, I am done for the day.
2. If the wager loses, I will try to get the loss back on the same day by playing Roulette. I will place a $1 bet on red and keep playing (using the Martingale system) until I win $110 to recover my loss for the day. This equates to about 170 spins provided I don't bust first. Once I reach $110, I am done for the day and the cycle repeats on the next day.
Knowing that I need to win 35 bets to reach my goal of $3,500 and realizing that I will probably lose half of my sides bets (but hopefully I will hit better than 50%), this means I will have to play Roulette about 17 times.
Using the Martingale system in Roulette with a $1 wager, I can lose 9 spins in a row before my bankroll is completely gone. Also, the table limit is $500 per bet. The odds of losing 9 spins in a row betting on red in a double-zero roulette wheel is .11%. Of course it will eventually happen. But I am gambling that it will not happen to me BEFORE I play roughly 17 sessions at about 170 spins per session (to account for house adavange of 5.06%). If it doesn't, then it is a mathematical certainty that I will reach my goal of $3,500 within 47 days even if lose half of my $110 sports bets.
The question: what is the probability of losing 9 times in a row on red in American roulette BEFORE completing 2,890 spins?
1. Each day, I will place one standard sides wager (-110) to WIN $100. If it wins, I am done for the day.
2. If the wager loses, I will try to get the loss back on the same day by playing Roulette. I will place a $1 bet on red and keep playing (using the Martingale system) until I win $110 to recover my loss for the day. This equates to about 170 spins provided I don't bust first. Once I reach $110, I am done for the day and the cycle repeats on the next day.
Knowing that I need to win 35 bets to reach my goal of $3,500 and realizing that I will probably lose half of my sides bets (but hopefully I will hit better than 50%), this means I will have to play Roulette about 17 times.
Using the Martingale system in Roulette with a $1 wager, I can lose 9 spins in a row before my bankroll is completely gone. Also, the table limit is $500 per bet. The odds of losing 9 spins in a row betting on red in a double-zero roulette wheel is .11%. Of course it will eventually happen. But I am gambling that it will not happen to me BEFORE I play roughly 17 sessions at about 170 spins per session (to account for house adavange of 5.06%). If it doesn't, then it is a mathematical certainty that I will reach my goal of $3,500 within 47 days even if lose half of my $110 sports bets.
The question: what is the probability of losing 9 times in a row on red in American roulette BEFORE completing 2,890 spins?