# Probability Question - Calculating Multiple Conditionally Dependent Outcomes

Say I participate in a betting system. I can bet 1, 2, or 4 dollars with someone on an event that has precisely 50/50 chances. The outcomes of each event do not affect the outcomes of others. Now, I want to cheat the system. The only thing I can come up with is: 1) Bet with 1. If win, collect and stop for now. If lose, move on to next. 2) Bet with 2. If win, collect and stop for now. If lose, move on to next. 3) Bet with 4. If win, collect and stop for now. If lose, ...

And that's my problem. What do I do after that to make my money back? I have a 3/4 chance of making a net profit of 1, but what do I do if I lose? I'm now down 7, and in this context, that's a lot to make up. Simply running four times won't help, because my probable gain will be 3, but my probable loss will be 7, leaving me down 4. Is there an improvement to be made, or is the lesson "Don't gamble, kids?"

• In a fair game, your expected net gain is $0$. The more you bet, the higher the variance in your net gain, but it does not affect the expectation – Henry Apr 16 '16 at 0:14
• What I'm doing here, however, is an attempt to cheat the fair game using increasingly high values. A strategy I've developed in things like this is to start with a low amount, and continuously double it until I've won. Almost certainly, I will eventually make out with a sum of winnings. What I'm wondering is whether there is a way to adapt it to a system with a ceiling. – Siris Apr 16 '16 at 2:10
• You cannot "cheat the fair game" unless you have an infinite amount of money (in which case winning a finite extra amount makes no difference to your wealth). Your expected net gain at any particular time will be $0$, no matter what strategy you follow, though there are strategies which give you a high chance of winning a little balanced by a low probability of losing a lot, and other strategies which give you a low chance of winning a lot balanced by a high chance of losing a little. – Henry Apr 16 '16 at 12:28

It is called the martingale betting system and the gambler's expected value remain zero because "the small probability that he will suffer a catastrophic loss exactly balances with his small gain when he wins". That is given its a $50/50$-game.