This is a problem that kills bankrolls, over and over - yet it keeps being done. Incorporate this information into your methods and you will see an improvement in results. What I mean by the "last" of something is exactly what it looks like. Let's take the game where Player 1 bets on Dozen #1, Player 2 bets on Dozen #2 and Player 3 bets on Dozen #3 Before the game even begins, I can tell you the outcome. Why ? Because it's how the game and random work. One player will win on the first spin (excluding the 0/00 of course, they aren't needed to explain this) One player will win right about on average (around 3 spins) and the last player will take a loss. This happens every time and you can calculate it in any way that you'd like to. Even in a perfect world where all 3 dozens show in 3 straight spins. The player who wins first will win 2 units The player who wins next would win 1 unit. The last player to win would just break even. And this is in a perfect world, where all 3 appear every 3 spins. Sadly, that isn't how it works. So without the game even beginning, I can tell you that 1 player will make a profit, 1 player will break even, and the last player will lose. How the dozens appear isn't relevant, the "last" to appear will never appear on average at a rate that will return a profit. That is why betting on "sleepers" is dangerous. (Betting on them once they appear is not). Now with this in mind, ("law of 3rd" things that some people refuse to believe in) - we can assume that 2/3 of the numbers on the table will appear in a cycle of spins. Of what's left - a few of them or most certainly 1 of them is possibly 200+ spins away from appearing. So why bet on it ? Why would anyone have ever bet on it ? This logically reduces to - never bet on a number that hasn't appeared yet. Which also supports the "repeaters" methods and why they work over cold/sleeper ones. Here is a practical example, your results may vary. In example #1, I will play all 3 dozens and remove the winner until all 3 have appeared. In example #2, I will play all 3 dozens and stop when 2 of the 3 have appeared. I will then restart and bet the 2 dozens that DID appear, and avoid the last dozen which is the potential sleeper. (reference "Avoid the sleeper" system from the past) Now of course, when 1 of the 2 wins - I have 1 location left being bet on, and that would go against what I just pointed out. So any win results in a reset. If you look at the chart above - you will see how 10-15 in a row without a win on the "last" spot would destroy a bankroll. We can avoid this. Example #1 - The Red average bankroll line is around -100 (or -1 unit) In total there weren't many sessions since waiting for the last dozen to appear takes time. 96 units were placed in total. Example #2 - The Red average bankroll line is around +250 (or +2.5 units) compared to -1 unit. In total there were many more sessions played compared to the first example. 148 units were placed in total compared to 96 units in Example #1 These were the same numbers. So your own testing and results will tell you the same thing - Never continue to bet until the "last" of something wins. It's a 100% way to lose. Once you have identified this cold/sleeper bet, avoid it - once it appears you can feel free to add it back into the mix - it more than likely won't be the same long term sleeper again depending on the bet location. But that's for another thread and involves non-repeating rare patterns. Because..... once you know what NOT going to happen, you can certainly predict what IS going to happen. Thanks for reading and have a nice day.