let me begin by saying i do not use martingale
this is purely to satisfy my mathematical curiosity
what is the formula to calculate the ROR using basic martingale
assuming a min bet of 'a' and max bet 'b'
assuming infinite BR
what is the probability of reaching the max bet without ever winning a hand?
and as the max bet approaches infinity what number does your risk of ruin approach and at what rate does it approach that number?
Running into the max bet is an eventuality, not just a "probability."
Thus, each "session" (unless there is a specific "win goal") must necessarily run into a situation where, in order to continue the bet sequence, a bet greater than the table limit is required. IOW, you lose BIG. In fact, the odds say that single loss will exceed the sum of ALL previous wins -- i.e., the very definition of a negative expection game.
Technically, with an infinite bankroll, you can never be "ruined," but you can and WILL be in negative territory virtually all the time.
what is the probability of reaching the max bet without ever winning a hand?
First, you need to calculate the number losses need to ruin. First you divide the max bet by the min bet, and then add 1 to the floor of log based 2. For example, if the min and max bets are 10 and 5000:
5000 / 10 = 500
floor(log2(500)) + 1 = 9
Specifically, your bets will be 10, 20, 40, 80, 160, 320, 640, 1280, and 2560. The next bet in the sequence, 5120 would exceed the max bet.
Next you need to calculated the probability of losing that many bets in a row starting with the next bet. The formula is the probability of a loss raised the number losses needed. Suppose we have an even money bets where the probablity of winning or losing is 50%. Continuing our previous example, the probability of losing the next 9 hands is 50% raised to the 9th power = 0.20%.
Now, we have to determine how many progression sequences we are going to make. Suppose we decide on a 100 sequences. The probability of not going ruin over those hundred sequence is the probability of not starting a ruin sequence raised to 100th power. Using our previous example, we get (1-(50%^9))^100 = 82.24%. Therefore, the probability of ruin is 17.76%.
Following are the risk of ruin for various number of sequences:
200: 32.36%
500: 62.38%
1000: 85.84%
5000: 99.99%
10000: 100.00%
and as the max bet approaches infinity what number does your risk of ruin approach and at what rate does it approach that number?
The risk is 100% at all max bets less than infinity. There have been heated discussions about what happens at infinity, so I won't get into that. You can an idea of the speed based on my example.
I took a dif eq class a while back and I remember how to determine what will happen as n approaches infinity I just don't know how to set up the series
This my best guess. (2n)/(.5^n) this series would diverge but what does that mean?
My other guess is the inverse situation in which case the series converges but once again I don't know what that means to the bank roll
Thanks for the answer to my first question panama rick
...is that you can never win regardless of how you play.
If you start with an infinite BR, then nothing you can do will ever possibly increase your BR, so there is no chance of ever winning.
To have any possibility of winning (increasing your BR), your starting BR has to be a finite number, and the ROR using a Martingale system for any finite BR is 100%.
Also, there is the problem that you could never bet against any entity that has an infinite BR, so your max bet will always necessarily be limited, meaning that your effective BR will always be finite and your Martingale ROR will therefore always be 100%.
The other problem is that you can never actually have an infinite BR, so the whole question is really N/A, at least for the realm of being that we all like to refer to as "reality".
