This is the mathematical proofing i did in another post - you only need to concentrate on the last two parts bolded in blue , although you need to do the algebra and calculus to get there through reduction. It's not even a zero-sum system but a negative-sum system.
So the mathematical proof that these type of "staking systems" don't work, will never work and have you attending GA meetings twice a week reciting the 12 steps or worse staying in a cardboard box and looking for safe-spots at night.....is such
b_k = bet value at the kth level
p_k = probability that series terminates with a win at the kth level, having been preceded by k-1 losses in a row
n - 1 = greatest number of losses in a row that can be sustained
e = player's expectation
e = + p_1b_1 + p_2(b_2 - b_1) + p_3(b_3 - b_2 - b_1) +...
+ p_n(b_n - b_(n-1) - ... - b_1)
+ (1 - p_1 - p_2-...
- p_n)(-b_n - b_(n-1) - ... - b_1)
The terms on the first line represent products of the probability that the series will terminate with a win at each successive level times the net profit at that level. The term on the third line gives the product of the probability that the series ends in failure at the nth level times the net loss.
regroup the terms:
e = 2p_1b_1 + (2p_2 + p_1)b_2 + (2p_3 + p_2 + p_1)b_3 + ...
+ (2p_n + p_(n-1) + ... + p_2 + p_1)b_n
- (b_1 + b_2 + ... + b_(n-1) + b_n)
p_k = (1 - p)^(k-1)p, where p is the probability of a win on any individual play and 1 - p is the probability of a loss.
substituting:
e = [2p]b_1 + [2p(1 - p) + p]b_2 + [2p(1 - p)^2 + p(1 - p) + p]b_3 + ...
+ [2p(1 - p)^(n-1) + p(1 - p)^(n-2) + ... + p(1 - p)^2 + p(1 - p)^1
+ p(1 - p)^0]b_n - (b_1 + b_2 + ... b_(n-1) + b_n)
factor out p so that the kth term is rewritten as:
p[(1 - p)^(k-1) + (1 - p)^(k-1) + (1 - p)^(k-2) + ... + (1 - p)^2
+ (1 - p)^1 + (1 - p)^0]b_k
use the formula for the sum of a geometric series and rewrite the kth term:
p[(1 - p)^(k-1) + (((1 - p)^k - 1)/((1 - p) - 1))]b_k
= [(2p - 1)(1 - p)^(k-1) + 1]b_k
e = sum_{k = 1}^{n} [2p - 1)(1 - p)^(k-1) + 1]b_k - sum_{k = 1}^{n} b_k
e = sum_{k = 1}^{n} [2p - 1)(1 - p)^(k-1)b_k + sum_{k = 1}^{n} b_k - sum_{k = 1}^{n} b_k
cancel the last two summations and factor out (2p - 1):
e = (2p - 1) sum_{k = 1}^{n} (1 - p)^(k-1)b_k
Since (1 - p) is positive and b_k is positive, the summation is positive. Therefore, the sign of e depends on the sign of (2p - 1). In an even-payoff unfair game, p < .5 and (2p - 1) is negative. In a fair game, p = .5 and (2p - 1) = 0.
Simplified these type of staking systems don't work and never will - there used to be an old saying that over a series of bets, if that series does not show a long term profit to either level or proportionate stakes (betting to win a fixed amount determined by the probability/odds of bet struck) - any other type of staking system is doomed - im excluding the Kelly Criterion method here which works through +EV and can be used full or fractional , only reason im excluding it is you need a probability distribution pre-race so that you can calculate "edge - most punters don't use probability distributions.