martingale

So this question has been driving me crazy for some time now, the odds of rolling 10 consecutive N-F #'s is once in 357 ten roll sequences..i understand it may happen back to back or even 3 times in a row etc but the same way that the 12 hits once in every 36 rolls average i look at the probability of 10 nf #s once in 3,570 rolls. Over the course of 3,560 rolls i say 3560 because 10 consecutive rolls would result in a little over 5k loss. That leaves us with 3560 rolls to make money. The field will hit 44 percent of the time. Thats about 1566 rolls will generate a unit of profit the rest will not affect. If the unit is 5 dollars thats over 7,500 win every 3560 rolls but you can expect to lose 5k + so thats a net profit of about 2500 average. Now that looks good on paper but thats a low profit for time invested and money risked which is why martingalw is always debunked..but we have not factored in 2s paying double and 12s paying triple..the odds of hitting those numbers are twice in every 36 rolls..how much is being wagered at the moment they hit is hard to calculate but every roll will have a wager regardless. And those 2s and 12s do roll and they will bring that average profit of 2500 to a higher sum regardless of when they hit. How much on average is the question. Im assuming 12k to 20k from personal experience..now risking 5k to win close to 20k on avegare is not that crazy. The casino risk 30 dollars to win one dollar on snake eyes, why cant a player risk 5k to win between 50 to 2,000 + dollars a session/day until busting out on 10 nf rolls and starting over with another 5 k from profit. Ofcourse they may lose that very same day they begin but eventually having enough money to rebuy several times and take some big hits, over time the odds of coming out ahead are strong. Please share your thoughts on this and what if the field pay triple on both 2 and 12 like some casinos have occasionaly and what if the martingale was extended to 11 rolls but with a neutral progression. So every roll after the first loss will result in a break even. First progression 5,10,20,40,80,160,320,640,1280,2560

2nd progression 5,5,10,20,40............2560

Those 2 and 12s would make up for the times winning one unit was sacrificed for giving yourself 11 rolls instead of 10. I seem to prefer breaking even after 11 than having only 10 rolls to recuperate and make one unit more often, and the probability of rolling 11 nfns must be alot lower than 1 in 357 sequences which i dont know how to calculate. Please share your thoughts on everything above, am i mistaken on anything?

Your initial assumption has a problem. You go from "the odds of rolling 10 consecutive non-field numbers (i.e. 5, 6, 7, 8) is once in 357 ten-roll sequences" to "the probability of 10 non-field numbers is once in 3570 rolls," which is incorrect - you forgot to take into account the fact that you can roll 10 consecutive non-field numbers starting in the middle of one of the 357 ten-roll sequences.