Betting on a coin flip

Question

SETUP

I am allowed to make a bet on the odds of a coin flip. The coin is a normal metal coin, is not weighted in any particular direction, and I can bet as much or as little as I'd like, and I can bet on either side of the coin. For example, I can bet 100$ that the coin will land heads side up, and if the coin lands head side up, I would win. I win all bets 1:1.

QUESTION

Is there a way to ultimately make money on this coin flip, knowing that in the long run, the results will tend towards 50% for heads and for tails? Is there a way for me to take advantage of that fact? I thought I had a method, but after putting together a simple script and running it a couple ten thousand times, I ended up breaking even on the results. I got my numbers from random.org so the numbers are as close to "true" random as I can get.

Answer 1

The following strategy guarantees that you win 1\$:

Bet 1\$ on any outcome. Each time you lose, bet twice as much (i.e., 2\$, 4\$, 8\$, etc.) in the next round (on any outcome). Stop when you win.

Howeve, this requires that you have infinite money to play with, infinite time, and the bank allows arbitraily large bets.

Answer 2

There is one strategy but it has some requirements that are not likely in real life.

Bet \$1 on the first throw. If you win then walk away with a profit of \$1.

If you lose then bet \$2 on the next throw. Again, if you win then walk away with a profit of \$1.

If you lose then bet \$4 ...

Here are the problems: you need an unlimited budget. If you were unlucky enough to lose the first 20 throws then you would be betting more than \$1,000,000 just to win \$1. Also, you need an opponent who is willing to accept these bets and allow you to stop at any point.

I think that neither condition is likely to be satisfied in real life.

Knowledge of the distribution (split) of possible outcomes would be useful if it did not match the returns. E.g. if you are playing a game with two dice and your opponent believed (as some do) that the chance of at least one 6 was $\frac{1}{3}$ then you might be able exploit it.