Answers

*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.