What would be the result of this coin tossing betting strategy?

Question

I am currently playing an online game where you can bet against other players with pixel money.

The probability for win and loss is 50%. There is no (pixel money) fee or anything that would change the expected value to anything else than 0$ in the long run.

B-but I'd like to profit from it so I thought of a simple (!) strategy, but I am not proficient enough in math to know 'how' to check, if it is a winning, losing or neutral strategy.

I'd appreciate the thoughts of you!

The bankroll would be 100€.

1)

I'd bet 1€.

If I lose my bankroll would become 99€ and I'd start a new bet with 1€.

If I'd win, I wouldn't add the won dollar to my bankroll, but bet 2€. (bankroll still at 99€, 1€ of original bet and the 1€ won)

2)

If this 2nd bet is a loss, I'd start with an 1€ bet (and 99€ bankroll) again.

If the 2nd bet is a win (4€ now), I'd put 1€ into my bankroll (now 100€) and bet the other 3€.

3)

If the 3rd bet is a loss, I'd start with 1$ again (bankroll 100€).

If the 3rd bet is a win (6€ now), I'd put 2€ into my bankroll (now 102€) and bet the other 4€.

4)

If the 4th bet is a loss, I'd start with 1$ again (bankroll 102€).

If the 4th bet is a win (8€ now), I'd put 3€ into my bankroll (now 105€) and bet the other 5€.

5)

If the 5th bet is a loss, I'd start with 1€ again (bankroll 105€).

If the 5th bet is a win (10€ now), I'd put 4€ into my bankroll (now 109€) and bet the other 6€.

.... ....

and so on

I am not sure how I could explain my (simple) thought process better in english or math.

A win streak of 5 would make be enough to cover 9 losses.

But on the other side I'd need need at least 2 wins to get break even, 3 wins for profit. Everytime I only win once, I'd bleed money.

So I wonder what would be the expected value after 100 / 1000 tosses.

I'd appreciate any thoughts on this!

Have a nice day!

Answer 1

Tracking what happens up until your first loss, if it occurs on the first or second toss (probability $\frac34$) you have a bankroll of $99.$ If it's the third toss (probability $\frac18$) your bankroll is $100,$ if it's the fourth toss (probability $\frac1{16}$) your bankroll is $102$ (you seem to have made a small mistake here, $100 + 2 = 102,$ not $103$).

Your expected bankroll is $$ \frac34 (99) + \frac18 (100) + \frac1{16}(102) + \frac1{32}(105) + \frac1{64}(109) + \cdots = 99 + \frac18 + \frac{3}{16} + \frac{6}{32} + \frac{10}{64} + \cdots. $$

If you work this out as an infinite sum, it's $100.$ Since you want to limit play to $100$ or $1000$ tosses, however, you have only a finite sum, with a larger-than usual last term (because you have to put all your winnings in the bankroll) in place of the infinite tail. If you work this out carefully you should find that it still comes out to $100,$ since all your bets (regardless of timing or size) have zero expected value.

Answer 2

I haven't followed your betting scheme in detail but know it won't win in the long run. You have invented some version of the gambler's ruin strategy: https://en.wikipedia.org/wiki/Gambler%27s_ruin

The problem is that you will sooner or later encounter a run of losses long enough to eat up any fixed starting stake. For example, since $2^{10} \approx 1000$ you will lose $10$ times in a row about once in every thousand plays.

That may seem unlikely to you, but it will happen. When it doesn't you accumulate small wins. When it does you lose all the small wins at one time and have no more stake to play with.

As long as you play with just pixel money you can try this out, although it will take a while. Or you could write a small computer program which will play many games much faster.