Searching for a winning strategy for a fair betting game

Consider a fair betting game, where a player bets on a number between $$0$$ and $$100$$. The numbers are then drawn out of a wizard's hat randomly.

The payoff for a successful bet is equal to $$1 / \Pr(win)$$ . For example, if a player successfully bets $$1.0000\dots$$ coins that a number will be greater than $$50$$, then the player receives a $$2.0000\dots$$ coin payout. If the player successfully bets that the number will be less than $$20$$, then the player will receive a $$5.0000$$ coin payout (coins are allowed to be irrational numbers).

My question is if there are any winning strategies such that the expected value of the bets increases over time?

Progress so far:

Recently I have come to learn of martingale sequences, and that as a result, any sort of doubling up strategies on a $$50:50$$ bet ultimately will not result in an increase in the expected value over time.

It seems intuitive that, regardless of the betting odds, the principle that the expected value should not change over time should hold.

Nevertheless, I would like to consider the case of going all in on a $$99:1$$ bet. That is to say, a player repeatedly bets everything they have, earing $$1 / (99/100) = 1.0101\dots$$.

Here I make some crude assumptions:

In this example, the player apparently has a $$1/100$$ chance of losing everything, however due to exponential growth, the player has already more than doubled their profit on the $$69^{th}$$ bet (nice!).

It would then seem to follow, the the player could beat the game by putting half of their winnings aside every $$69$$ bets.

Could this be considered a winning strategy?

Bonus: We further notice that the by placing bets with increasingly higher odds of success & more often, the limit of $$\lim_{n\to\infty} (1 + \frac{1}{n})^n$$ approaches $$e$$. Can this be interpreted to mean that an optimal betting strategy would be place bets with the highest odds possible?

• Are 0 and/or 100 included?????
– zoli
Commented Aug 26, 2019 at 2:11
• If zero is not included then the probability of winning is $\frac{99}{100}$. In case of success you earn $\frac{1}{\frac{99}{100}}$. The expectation is the product of the two numbers above, that is, $1$.
– zoli
Commented Aug 26, 2019 at 2:37
• If zero is incl. then the result is the same.
– zoli
Commented Aug 26, 2019 at 2:38

NO If each bet has an expectation of $$0$$, any series of bets will also have an expectation of $$0$$. You can formulate strategies that give a high probability of a gain, but that will be balanced by a small probability of a large loss.