The Bdalt System
Basic Gist: Complicated, requires large bankroll, but probably can be adjusted for small bankroll.
Max Loss: $150
2. I usually buy in for $2,000. My loss limit is 1/2 of my buy in (usually $1,000). If I'm only willing to lose a maximum of $1,000, why buy in for $2,000? There are 2 reasons: First, for me there is the psychological reason. If I'm down $700 or so and had only bought in for $1,000, the few black and green chips I had left in my rack would appear naked and puny to me.
And the overall losing feelings I had would become even worse. Even though I'm getting close to the point of quitting, I don't want to have the losing feeling that one gets when playing with very few chips.
That may not make sense to some, but I like to always have a decent amount of chips in my rack. The second reason is the effect $2,000 buy ins have on comp ratings.
What are the odds that I'll make one of the initial 2 numbers? If the 2 numbers are 4 and 10, I have 6 chances out of 36 of catching one of them. If the numbers are 5 or 9, I have 8 chances out of 36 of catching one of them.
And if they are 6 or 8, I have 10 chances out of 36 of catching one of them. The chances of a 7 are 6 out of 36. So probability has it that most of the time I'll catch one of the 2 numbers before a 7 comes up.
If initially the number that he hit (repeated) for me was a 4 or 10, I would have been paid a total of $105. If the number was a 5 or 9, my pay back would be $80. I would receive $65 if the number was 6 or 8. After winning on that first number, my money at risk on the table (if the roll has lasted long enough to establish the 3 bets) would amount to $90 for one $5 line bet with $25 odds and two $5 come bets with $25 odds on each (unless a number was a 5 or 9 where we would have $30 odds).
So, my win on the very 1st number hit goes a long way to funding those 3 follow-up bets. If I hit any of these 3 numbers I'm ahead. And of course, if more numbers are repeated, it's even better.