A series of self-contradictory non-sequiturs

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Self-sufficiency is the road to poverty

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games of production,
games of consumption,
in the end a game is a game

why be picky?

retirement is difficult, and so it is that meaningful play isn't all that much easier to get right than meaningful work.

I feel like that's one of the central lessons of The Culture. In a utopia, Iain Banks shows us how in a limitless and needless world, the problems are very much the same.

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Okay, fine, maybe meaningful play is not all that much easier than meaningful work, but it IS easier. The question is whether that is due to others hacking your sense of aesthetics and undermining your free will and things like that. To be truly happy in consumption, you have to find a group of mutually serving people where everyone contributes some atom of meaning, but at that point it's probably possible to reorganize it into a commercial enterprise anyway.

Point being, I sincerely believe the best things in life are free, or more than pay for themselves.

General form for the return R:

$$R = \Pi_i (1+xa_i)^{p_i}$$

Optimal conditions:

$$\partial_x R = \sum_i \frac{p_i a_i}{1+xa_i}\Pi_j ((1+xa_j)^{p_j} = R \sum_i \frac{p_i a_i}{1+xa_i} = 0$$

Specializing to previous case, where i runs from 1 to 2:

$$\frac{p_1 a_1}{1+xa_1} + \frac{p_2 a_2}{1+xa_2} = 0$$

$$(p_1 a_1)(1+xa_2) + (p_2 a_2)(1+xa_1) = 0$$

$$p_1 a_1+ x(p_1+p_2)a_1a_2 + p_2 a_2 = 0$$

$$x = -\frac{p_1 a_1+p_2 a_2}{a_1a_2} = -\frac{\left<a\right>}{a_1a_2}$$

Notice that when $p_1\rightarrow 1$, $x \rightarrow -\frac{1}{a_2}$, so as disaster becomes ever more unlikely, you would bet a proportion of your wealth up to the loss ratio. This is a reflection of the Kelly criterion's tendency to never allow anything to go to zero, under any circumstance.

We'll consider why this is undesirable in some future post. For now, Let's make most use of the formulation, and try to find good ways of summarizing win/loss ratios and frequencies to fit into the form above.

The terms for each outcome ($\frac{p_i a_i}{1+xa_i} =\frac{p_i}{a_i^{-1}+x}$)  sum to zero for the optimal $x$. Since $x > 0$, These terms are either monotonically increasing or decreasing with $x$ depending on the sign of $a_i$.

The question then is how one would represent the two groups of monotonically increasing and decreasing terms so has to figure out which $x$ they net out at. This will be covered in the next post.

Suppose you have a bet which loses money most of the time, but wins a massive amount now and then, how much money should you put on it?

Let's say the 1 time you win, you win $a for each dollar you bet, and the N times you lose, you lose $b for each dollar you bet. By the Kelly criterion, the geometric average rate of gain if you bet $x$ of your wealth would be

$$R = (1+xa)(1-xb)^n$$

Setting $\partial_x R = 0$, you get

$$x = \frac{a - nb}{(n+1)ab} = \frac{\left<\mathrm{arithmetic\ gain}\right>}{ab}$$

Suppose you are asked to flip a coin, and heads you win $3, and tails you lose $1---then $n=1, a=2,b=1$, and therefore $x = \frac{1}{2 \cdot 2} = \frac{1}{4}$, i.e., you should bet 25% of your wealth.

If you have a lottery ticket that has a 1 out of 5,000 chance of winning $10,000 that costs $1, and you are only allowed to buy one number, then $n=4999,a=9999, b=1$, and
$x = \frac{5000}{49999 \cdot 9999} \approx 10000^{-1}$ and you should only bet 0.01% of your wealth at a time.

Conversely, if you were selling a lottery ticket that had 1 out of 10,000 chance of winning $5,000 that cost $1, $n = 9999, a = -4999, b = -1$, and $x = \frac{5000}{9999\cdot4999} \approx 10000^{-1}$ and you should be trying to have about 0.01% of your wealth at stake.

Related: Do not play the lottery unless you are a millionaire