The one genuinely useful piece of mathematics a Balloon player can carry into a session is the expected value of continuing versus banking. The expected value ($EV$) of each additional step of inflation is:
$$EV = P(\text{success}) \times (S + \Delta S) - S$$
Here $P(\text{success})$ is the probability that the balloon does not burst on the current step, $S$ is the accumulated win available right now, and $\Delta S$ is the increment gained from the next fraction of a second of holding. Because $P(\text{success})$ falls continuously as the balloon expands, an inflection point arrives at which $EV$ turns negative, and every step past that point destroys value.
The same logic can be illustrated on the classic bounded balloon task, where the burst threshold $M$ is unknown but assumed uniform on a finite range. If you have already pumped $i$ times and the balloon can take at most 20 pumps, the chance it bursts on the next attempt is $1/(21-i)$, and the expected value of pumping once more is:
$$EV_{\text{pump}}(i) = \frac{20-i}{21-i} \times (i+1)$$
Compare that against the immediate payoff of $i$ for cashing in. Plotting both curves shows the crossover, the break-even point at 10 pumps, after which risking the accumulated amount is no longer rational. This is the analytical skeleton behind every "cash out early" recommendation you will read. Not superstition, but the point where the growth of the prize stops compensating for the growth of the burst probability.
In a commercial crash game the picture is harsher than in the bounded task, because the house edge shifts the crossover earlier and no amount of observation reveals the hidden threshold in advance. The RNG draws a fresh, independent burst point every round.
