I suppose you think if you hit four blacks in a row on roulette you should always go red because it's red's turn to come up?
Although statistically it's totally irrelevant what the prior spins are, betting like this is how I pay for my expenses whenever I visit Vegas. Sure, I don't make a fortune doing it (usually just 20 here, 40 there), but I've never walked away from a roulette table a loser.
Then you haven't played enough roulette, or actually kept accurate records.
I would remember losing. Now, we're only talking about around $3,200 over like 4 different trips but still, it payed for my meals, taxis and, well, some of the alcohol. I never stay at the table, I swoop in, red or black, collect, and leave. I've stood around and watched people lose startling amounts of money on those tables.
Let me see if I understand you. Is this the gist of what your practices are? You walk down to the roulette wheel and wait for the ball to hit four blacks in a row, then place $800 on red and walk away a winner with an extra $800 in your pocket. I could certainly believe that has happened to you four times.
If you are making "20 here, 40 there" on a red/black bets, then taking in $3200 would require around 800 to 1600 winning bets on red/black. To simplify the math, I'll say each bet was $40, so we need at least 800 more winning bets than losing bets. If you have actually only made those 800 winning bets, then wow - that's freaky! The odds of that happening are something like 1 in 6.7x10^240 (two to the power of eight hundred). I doubt very much you have NEVER lost a roulette bet, because the odds against that are so incredibly incredibly incredibly incredibly high (I should really add a whole lot more "incredibly" terms).
So what might you actually be doing? Based on your reported winnings and bet size, we need 800 more wins than losses (assuming the numbers reported are accurate). A quick google search turns up a nice discussion about coin flips (link below) that I can use to get some calculations from. If we ignore the house advantage due to the green "0"/"00" slots we can just use a 50/50 probability. The variance of a Bernoulli distribution is: Var(X) = np(1-p) where n is the number of trials/bets, p is the expected probability of outcome 1 and (1 - p) is the expected probability of outcome 2. With p=1/2=0.5 we have Var(X) = n(1/2)(1-1/2) = n/4. The square root of the variance is the standard deviation and about 2/3 of all trials of a given size should fall within one standard deviation of the expected result (Wikipedia link below). If someone got 800 more wins than losses, that would not be too surprising as long as the number of bets they made was large enough so that they were still within a few standard deviations of the expected result of equal numbers of wins and losses. Only 1 out of 400 billion trails fall outside of 7 standard deviations, so 8 standard deviations should be very very very conservative (and makes the division of 800 simple!) So with a standard deviation of 800/8 = 100, this gives a variance of 100^2 =10,000 and the number of bets (n) is around 40,000. A more reasonable estimate of 4 standard deviations (1 out of 16,000 trials fall outside of 4 standard deviations), gives a standard deviation of 200, variance 40,000 and a number of bets of 160,000.
flipping coins calcs: https://ca.answers.yahoo.com/q...
standard deviation odds: https://en.wikipedia.org/wiki/...
Basically, in order to have 800 more wins than losses, you need to make a whole lot of bets, unless the odds are significantly different from 50-50.
For me at least, it seems likely that your memory or record keeping is at fault and that in actuality you have not come out $3,200 ahead on bets of $40. Or even more likely, I am misunderstanding what you are claiming. I really am interested in what you are doing, and recognize that any "theoretical" or "statistical" calculation I might make would not invalidate your actual experiences, but I would love to reconcile the math with the actuality.