........ I am interested in both the probabilities and the loss runs, now that you have defined it. So you mean to say, that at 87.7779% chance to win, I need to get (on average) 300,000 loss runs? Meaning, 300,000 martingale sequences before I hit a 6 loss streak?
Not quite. The expected number of sequences until you
exactly 5 losses in a row is 41772.04. The expected number of loss runs until you hit
more than 5 losses in a row 3e05.
Keep in mind that the variance is large. If the expected number of loss runs is 1/3e05, then 95 times out of a hundred that you experiment, you'll hit a hit loss streak of 6 or more at between 7,595 and 1,106,661 sequences.
So, a loss run is any martingale sequence? Not including just 1 loss. That means at least 2 losses in a row? Or does loss run mean any sequence of losses, including a single loss?
"Losses in a row" can be any positive integer from zero up: 0, 1, 2, 3 .... For example loss, loss, win, loss, win, win is 2 losses in a row, followed by a one loss sequences, followed by a zero loss sequence.
According to the same formulas (and even other formulas from other websites) the probabilities of a 7 loss streak or run at this chance to win is extremely low, which I took and interpreted as, the 6th roll in these two sequences are most probably the last to lose, the next one (the 7th) will most likely win.
That was true with the first time. On this second time, I still don't know yet if it will win since I have not yet made the 7th roll. However, since the average number of loss runs until the next greater than 6 run of losses in a row is about 2.4 million, and I've only rolled 1.3 million times, I think I'm not going to lose the next roll.
I also understand that, they keep saying all rolls are independent and dice have no memory, and assuming that, I still have an 87.7779% chance the next roll will win.
They do keep saying that and it is true. I absolutely guarantee you that as long as the site is fair you will have an 87.7779% chance to win every time you play the 87.7779% game. I also think you wont lose the next roll, in fact I'm 87.7779% sure you won't lose. More info on memoryless probability distributions:
http://en.wikipedia.org/wiki/Memorylessness@ infested999
If I have bankroll of 0.512 btc and I martingale with min bet 0.001 with the only purpose of doubling that 0.512 ... <snip>
I realised that the probability of having exactly doubled your money in exactly 512 rolls is extremely low - less than 0.25%. This is because as well as calculating the probability of 9 losses in a row, you need to also include the possibility of the last roll being a loss and only having 0.511 btc. Also, you need to include the probability of the last two rolls being a loss, the last three rolls and so on to the last 8 rolls being a loss.
Most people would then keep playing past 512 rolls until a win, but then the original probabilities calculated are incorrect. However if you make your calculations based on loss runs this is no longer a problem.
Would "how much bank do I need to have a 90% chance of a positive balance in 512 rolls" or something similar be a better question? Or maybe "I want to win
x btc on probability
p and before
y loss runs, how much bank do I need?".
