Imagine betting on a fair coin toss. Actually betting on multiple coin tosses. You obviously have 50% probability to win each toss. We have a bankroll of 1000 units. We can start/stop betting and increase/decrease our bets freely.
What we want to do is this: The average won bet to be higher than the average lost bet.
How could we achieve this?
Note 1: We don't need to be in profit. For example having 5 lost 1 unit bets and 2 won 2 units bets fulfils our requirement, yet we are not in profit.
Note 2: If you think it can't be done, please provide a mathematical proof it is impossible.
Bet \$1 each time until you lose. You have now won \$1 several times, say you have won $k$ times, and lost \$1 exactly once. For each subsequent turn, bet the smallest amount possible so that if you were to win the bet, your average wins would exceed your average losses. Here's what this looks like:
If $k=0$, then you should bet \$2 every time until you win. When this happens, your average win will be \$2, and your average loss will be less than \$2.
If $k=1$, you should first bet \$2, then \$3 if you lose, then \$4 if you lose that, and so on, incrementing your bet by \$1 each time. If the final bet you make is \$$n$, then your average wins will be $(1+n)/2$, while your average losses will be the average of \$1, \$2, ... , \$$(n-1)$, which is $n/2$.
For all other values of $k$, the series of bets you should make does not have a simple description. The strategy is summarized in the below table.
The only way this strategy can fail is if you run out of money to make the next bet. When $k=0$, you would need to lose $500$ times in a row for that to happen, which is extremely unlikely. For $k=1$, you must avoid losing $43$ times in a row, still very unlikely. When $k=9$, you lose if you lose $6$ times in a row. All in all, the probability of winning for this strategy works out to be $99.981\%$.
\begin{array}{r|l} k & \hspace{5.5cm}\text{Series of bets} \\\hline 2 & 2, 3, 5, 7, 9, 12, 15, 19, 23, 27, 32, 37, 43, 49, 55, 62, 69, 77, 85, 93, 102, 111 \\ 3 & 2, 4, 7, 12, 18, 27, 38, 52, 69, 90, 114, 142, 175, 212 \\ 4 & 2, 4, 8, 15, 27, 44, 69, 103, 148, 207, 282 \\ 5 & 2, 5, 12, 26, 51, 93, 158, 257, 399 \\ 6 & 2, 5, 13, 31, 67, 133, 247, 431 \\ 7 & 2, 6, 18, 48, 114, 246, 491 \\ 8 & 2, 6, 20, 58, 149, 347 \\ 9 & 2, 7, 25, 79, 220, 548 \\ 10 & 2, 7, 27, 92, 274 \\ 11 & 2, 8, 34, 125, 398 \\ 12 & 2, 8, 36, 141, 477 \\ 13 & 2, 9, 44, 184, 660 \\ 14 & 2, 9, 47, 208 \\ 15 & 2, 10, 55, 258 \\ 16 & 2, 10, 58, 286 \\ 17 & 2, 11, 68, 353 \\ 18 & 2, 11, 71, 386 \\ 19 & 2, 12, 82, 467 \\ 20 & 2, 12, 86, 511 \\ 21 & 2, 13, 97, 601 \\ 22 & 2, 13, 101, 651 \\ 23 & 2, 14, 114, 764 \\ 24 & 2, 14, 118, 820 \\ \end{array}