Why is Kelly's criterion a suitable objective to optimise? In a betting setup with round $N$. Starting from $1$, we bet with proportion $f$ at each round. It wins with probability $p$ and gains $bf$ dollars, or loses everything with probability $1-p$. The Kelly's Criterion says
$$ f^*=\frac{bp-(1-p)}{b} $$
is the optimal allocation. Let $X_N$ denote the total assets at the round $N$. Then
$$ \mathbb{E}[X_N]=\sum_{k=1}^N [p(1+bf)]^K[(1-p)(1-f)]^{N-k}.$$
Kelly's Criterion optimizes the $\mathbb{E}[\log X_N]$ instead of $\mathbb{E}[X_N]$ directly. What's the motivation for doing that? In general, $\arg\max\mathbb{E}[X_N]\neq \arg\max \mathbb{E}[\log X_N]$, right?
 A: A handwaving explanation:
Suppose you had to bet a constant fraction $f$ of your wealth each time, and I told you what the outcome of the next three bets were going to be: in two cases you would gain the amount staked and one case you would lose the amount staked.
I did not tell you the order of these outcomes, but it does not really matter since the ratio of your final wealth to initial wealth was $(1+f)^2(1-f)$,  and this would be maximised when $f^*=\frac13$. Consider why this is better than using $f=1$, which guarantees bankruptcy in this scenario, or some intermediate $f$.
You would have got the same $f^*$ maximising $2\log(1+f)+\log(1-f)$, i.e. the frequency-weighted sum of the logarithms of the outcomes.
Something similar happen with probabilities close to long-run frequencies: you want to maximise the probability-weighted product i.e. $(1+bf)^p(1-f)^q$, and you will get the same $f^*$ if you maximise the frequency-weighted sum of the logarithms of the outcomes $p\log(1+bf) + q\log(1-f)$, i.e. maximise the expectation of the logarithms.  It is easier to work with the (arithmetic) expectation of the logarithms than with what you might perhaps call the geometric expectation of the outcomes themselves.
Betting more than this can appear to increase your expected gain, but at the cost of being much more likely to lose money.  Maximising the expected gain in the probability scenario happens with $f=1$ but the probability of making anything and not going bankrupt is $p^n$ , which approaches $0$ exponentially quickly; the likely alternative is not desirable. In effect you are betting against the law of large numbers, and are compensated for doing so in the rare occasions when it takes a long time to produce frequencies close to the probabilities.
