Everyone also agrees that to win one must bet more at times. When to do it, is what not all of us agree on.

Quote:ThatDonGuyI am trying to find some general solution for the probability of reaching a target amount before losing an entire bankroll based on the target and initial values, but I'm not entirely sure one exists, especially if an additional condition that the bet is capped by both the remaining bankroll and the amount needed to reach the target. It's more of a Markov Chain problem, which can be solved with simultaneous equations; the problem is, if the initial bankroll is 200 and the target is 300 (i.e. making a profit of 100 before losing the entire initial 200), there are around 6500 separate equations. Even with initial 20 and target 30, there are 163 equations.

This is a telescoping sums problem, assuming flat betting. Let's say the question is...

A player is playing double-zero roulette, making even money bets, flat betting one unit at a time. He brings 200 units to the table. His goal is to win 100 units (for a total of 300). He plays until he achieves that goal or goes bust trying. What is his probability of success?

((10/9)^220 - 1) / ((10/9)^300 - 1) = 0.0002184745

Another way to denigrate D'Alembert is to start at one unit. I suggest 5 units to negate the effects of early success.

Hi TDGQuote:ThatDonGuyI am trying to find some general solution for the probability of reaching a target amount before losing an entire bankroll based on the target and initial values, but I'm not entirely sure one exists, especially if an additional condition that the bet is capped by both the remaining bankroll and the amount needed to reach the target. It's more of a Markov Chain problem, which can be solved with simultaneous equations; the problem is, if the initial bankroll is 200 and the target is 300 (i.e. making a profit of 100 before losing the entire initial 200), there are around 6500 separate equations. Even with initial 20 and target 30, there are 163 equations.

This is almost exactly described in my blog post.

I do assume negligible house edge. It's derived quite simplistically and has the exact conditions you ask for.

Quote:OnceDearHi TDG

This is almost exactly described in my blog post.

I do assume negligible house edge. It's derived quite simplistically and has the exact conditions you ask for.

I don't see any formula, or even a way that the probability is calculated - just a single run.

I also noticed this:

"I assigned a starting Bankroll of $1,000,000 ( You can vary that in the model. )

I decided an initial wager of $10 ( You can vary that in the model. )

I decided a progression increment of $1. ( You guessed it )

Whenever a bet won, the next bet was $1 less than the previous bet, or $1, whichever was the greater.

Whenever a bet lost, the next bet was $1 more than the previous bet, or total remaining bankroll, whichever was the smaller.

The model simulated potentially 50,000 spins."

You then noted that you never doubled the bankroll. This is not surprising.

A run of N wins and N losses that starts with a win does not change the bankroll unless you happen to reach the target within the run

A run of N wins and N losses that starts with a loss adds N to the bankroll.

However, you need to add 1,000,000 to the bankroll in 50,000 spins.

10 + 11 + 12 + ... + (10 + (N-1)) = 1,000,000

10 N + (1 + 2 + ... + (N-1)) = 1,000,000

10 N + (N-1) N / 2 = 1,000,000

(20 + (N-1)) N = 2,000,000

N^2 - N - 1,999,980 = 0

N = about 1414

This means the wins need to exceed the losses by 1414 at some point in 50,000 spins.

Are you seeing the same post as me? Oncedear's rule of thumb. For some reason, I get scrolled forward to the appropriate post. So scroll down till you see it, while I wander off to amend the link.Quote:ThatDonGuyQuote:OnceDearHi TDG

This is almost exactly described in my blog post.

I do assume negligible house edge. It's derived quite simplistically and has the exact conditions you ask for.

I don't see any formula, or even a way that the probability is calculated - just a single run.

https://wizardofvegas.com/member/oncedear/blog/5/#post1370

Quote:OnceDearAre you seeing the same post as me? Oncedear's rule of thumb. For some reason, I get scrolled forward to the appropriate post. So scroll down till you see it, while I wander off to amend the link.

https://wizardofvegas.com/member/oncedear/blog/5/#post1370

You mean, "Oncedear's Rule - What's the Chances?"

All that says is, the probability of success can never exceed the starting bankroll divided by (the starting bankroll + the target profit).

I was referring to calculating a specific probability of success.

Quote:ThatDonGuyYou mean, "Oncedear's Rule - What's the Chances?"

All that says is, the probability of success can never exceed the starting bankroll divided by (the starting bankroll + the target profit).

I was referring to calculating a specific probability of success.

For a zero edge game...

Probability of Success = (Starting Bankroll)/(Starting Bankroll + Target Profit)

For a low house edge game, it gives a decent probability estimation if you wager a modest proportion of bankroll on each wager, especially so with a progressive such as Marty or d'Alembert

I acknowledge that it's far more complex, and beyond me to derive an equation where there is a house edge.