callipygian
Well-Known Member
So the title formula is one that has oft been quoted to me. I'm trying to figure out where this comes from.
I'll try to standardize all the notation such that:
B = bankroll
f = fraction of bankroll bet
p(i) = probability of ith outcome
y(i) = payout of ith outcome
r = rate of growth
E = expected value
Given a game with multiple outcomes, the bankroll grows as:
B' = B*Product{i = 1 to n, (1+f*y(i))^p(i)}
This leads to an aggregate rate of growth of:
ln(1+r) = B'/B = Sum{i = 1 to n, p(i)*ln(1+f*y(i))}
Maximizing ln(1+r) with respect to f involves the differentiation with respect to f, or
d[ln(1+r)]/df = Sum{i = 1 to n, p(i)*y(i)/(1+f*y(i))} = 0
In the case of n = 2 (one win payout and one loss payout) this is easy to solve and ends up being:
f(derivative) = -[p(1)*y(1)+p(2)*y(2)]/[y(1)*y(2)*(p(1)+p(2))]
Or for p(1) + p(2) = 1 [no push] and y(2) = -1,
f(derivative) = E / y(1) = advantage / payout
Up to this point, this all makes sense to me. However, this formula is somewhat different than the formula given, which is
f(title) = advantage / variance
Expanded, would be:
f(title) = Sum {i = 1 to n, p(i)*y(i)}/Sum {i = 1 to n, p(i)*(y(i)-E)^2}
or for n = 2:
f(title) = [p(1)*y(1)+p(2)*y(2)]/[p(1)*(y(1)-(p(1)*y(1)+p(2)*y(2)))^2 + p(2)*(y(2)-(p(1)*y(1)+p(2)*y(2)))^2]
(In case the comparison is not obvious, f(title) has y(1)^2 and y(2)^2terms in the denominator while f(derivative) does not ... and no they don't cancel, I checked)
I did find a page written by Stanford Wong which suggests that variance is actually an approximation, that the exact solution is actually:
f(Wong) = Sum {i = 1 to n, p(i)*y(i)}/Sum {i = 1 to n, p(i)*y(i)^2}
In the case where E << y(i), which is true for blackjack, then f(Wong) = f(title). So that part makes sense to me. But I still can't reconcile f(Wong) or f(title) with f(derivative).
Can anyone help?
I'll try to standardize all the notation such that:
B = bankroll
f = fraction of bankroll bet
p(i) = probability of ith outcome
y(i) = payout of ith outcome
r = rate of growth
E = expected value
Given a game with multiple outcomes, the bankroll grows as:
B' = B*Product{i = 1 to n, (1+f*y(i))^p(i)}
This leads to an aggregate rate of growth of:
ln(1+r) = B'/B = Sum{i = 1 to n, p(i)*ln(1+f*y(i))}
Maximizing ln(1+r) with respect to f involves the differentiation with respect to f, or
d[ln(1+r)]/df = Sum{i = 1 to n, p(i)*y(i)/(1+f*y(i))} = 0
In the case of n = 2 (one win payout and one loss payout) this is easy to solve and ends up being:
f(derivative) = -[p(1)*y(1)+p(2)*y(2)]/[y(1)*y(2)*(p(1)+p(2))]
Or for p(1) + p(2) = 1 [no push] and y(2) = -1,
f(derivative) = E / y(1) = advantage / payout
Up to this point, this all makes sense to me. However, this formula is somewhat different than the formula given, which is
f(title) = advantage / variance
Expanded, would be:
f(title) = Sum {i = 1 to n, p(i)*y(i)}/Sum {i = 1 to n, p(i)*(y(i)-E)^2}
or for n = 2:
f(title) = [p(1)*y(1)+p(2)*y(2)]/[p(1)*(y(1)-(p(1)*y(1)+p(2)*y(2)))^2 + p(2)*(y(2)-(p(1)*y(1)+p(2)*y(2)))^2]
(In case the comparison is not obvious, f(title) has y(1)^2 and y(2)^2terms in the denominator while f(derivative) does not ... and no they don't cancel, I checked)
I did find a page written by Stanford Wong which suggests that variance is actually an approximation, that the exact solution is actually:
f(Wong) = Sum {i = 1 to n, p(i)*y(i)}/Sum {i = 1 to n, p(i)*y(i)^2}
In the case where E << y(i), which is true for blackjack, then f(Wong) = f(title). So that part makes sense to me. But I still can't reconcile f(Wong) or f(title) with f(derivative).
Can anyone help?
