Thinking about the finite amount of money.
If your goal is to double your original amount. Each round you bet some amount, if you win you double it, if you lose you lose the entire bet. The probability of winning in a given round is p.
If you start at the max bid, you have p probability of winning in the first round, thus attaining your goal.
If you start at half the max bid, you can only bid once... if you win, with p probability, you only have 1.5x your original amount. So you start again. Probability of doubling your money is p^2, which, unless you always win, is strictly less than p.
If you start at 1/3 the max bid, you can bid twice.
This makes it complicated to figure out what happens.
> a) 1, next bid 1/3
-> b) p - 4/3, next bid 1/3
--> c) p - 5/3, next bid 1/3
---> d) p - 6/3, win
---> b) (1-p)
--> a) (1-p)
-> d) 1-p - 2/3, next bid 2/3
--> b) p - 4/3, next bid 1/3
--> x) (1-p) - none left.
Solving this sequence is kinda complicated and I'm lazy. After this it just gets worse.
a = p*b + (1-p)*d
b = p*c+ (1-p)*a
c = p*1 + (1-p)*b
d = p*b
Solving for a, b, c, d, gives us a value of a equivalent to the overall probability of winning.
Eliminate d:
a = p*b + (1-p)*(p*b)
Eliminate c:
b = p*(p+(1-p)*b) + (1-p)*a
b = p*(p+b-p*b) + (1-p)*a
b = p*p+p*b-p^2*b + (1-p)*a
b - p*b + p^2*b = p*p + (1-p)*a
b = (p*p+(1-p)*a)/(1 - p + p^2)
Eliminate b:
a = p*(p*p+(1-p)*a)/(1 - p + p^2) + (1-p)*p*(p*p+(1-p)*a)/(1 - p + p^2)
a = (p*(p*p+(1-p)*a) + (1-p)*p*(p*p+(1-p)*a))/(1 - p + p^2)
a - p*a + p^2*a = 2*p^3+2*a*p-3*a*p^2-p^4+a*p^3 -> I used Maple to simplify this for me, lol.
a - 3*p*a + 4*p^2*a - p^3*a = 2*p^3 - p^4
a = (2*p^3 - p^4) / (1 - 3*p + 4*p^2 - p^3)
This is not getting any simpler. But I can use the power of mathematical tools (Maple again) to plot a.
Turns out, if p < 0.5, a < p. That implies that it's better to just use the original 1-shot plan than to do it this way.
I would guess that the best chance you have of doubling your money is (p), no matter what fraction of the original you start with. Unless p > 0.5, in which case the game is rigged in your favour anyway, and the best strategy is to just bet a small amount repeatedly, and accumulate wealth over time.
If your goal is to double your original amount. Each round you bet some amount, if you win you double it, if you lose you lose the entire bet. The probability of winning in a given round is p.
If you start at the max bid, you have p probability of winning in the first round, thus attaining your goal.
If you start at half the max bid, you can only bid once... if you win, with p probability, you only have 1.5x your original amount. So you start again. Probability of doubling your money is p^2, which, unless you always win, is strictly less than p.
If you start at 1/3 the max bid, you can bid twice.
This makes it complicated to figure out what happens.
> a) 1, next bid 1/3
-> b) p - 4/3, next bid 1/3
--> c) p - 5/3, next bid 1/3
---> d) p - 6/3, win
---> b) (1-p)
--> a) (1-p)
-> d) 1-p - 2/3, next bid 2/3
--> b) p - 4/3, next bid 1/3
--> x) (1-p) - none left.
Solving this sequence is kinda complicated and I'm lazy. After this it just gets worse.
a = p*b + (1-p)*d
b = p*c+ (1-p)*a
c = p*1 + (1-p)*b
d = p*b
Solving for a, b, c, d, gives us a value of a equivalent to the overall probability of winning.
Eliminate d:
a = p*b + (1-p)*(p*b)
Eliminate c:
b = p*(p+(1-p)*b) + (1-p)*a
b = p*(p+b-p*b) + (1-p)*a
b = p*p+p*b-p^2*b + (1-p)*a
b - p*b + p^2*b = p*p + (1-p)*a
b = (p*p+(1-p)*a)/(1 - p + p^2)
Eliminate b:
a = p*(p*p+(1-p)*a)/(1 - p + p^2) + (1-p)*p*(p*p+(1-p)*a)/(1 - p + p^2)
a = (p*(p*p+(1-p)*a) + (1-p)*p*(p*p+(1-p)*a))/(1 - p + p^2)
a - p*a + p^2*a = 2*p^3+2*a*p-3*a*p^2-p^4+a*p^3 -> I used Maple to simplify this for me, lol.
a - 3*p*a + 4*p^2*a - p^3*a = 2*p^3 - p^4
a = (2*p^3 - p^4) / (1 - 3*p + 4*p^2 - p^3)
This is not getting any simpler. But I can use the power of mathematical tools (Maple again) to plot a.
Turns out, if p < 0.5, a < p. That implies that it's better to just use the original 1-shot plan than to do it this way.
I would guess that the best chance you have of doubling your money is (p), no matter what fraction of the original you start with. Unless p > 0.5, in which case the game is rigged in your favour anyway, and the best strategy is to just bet a small amount repeatedly, and accumulate wealth over time.