The perfect gambler - would it work?

Question

Let's say you're odds of winning a game are 48%, and you have 100 coins. Every time you win, you double your bet. For example if you bet 10 coins and win, you would have 20 coins. If you stick with a certain strategy, I am thinking you win 1 coin every round. Here's how it works:

• Whenever you lose, you double the amount you bet previously
• Whenever you win, you restart your bet to 1 coin.

So if you start with 1, and lose on a 1, 2, 4, but win on an 8, you just made 1 coin. Could you restart the cycle and keep winning 1 coin?

There has to be a flaw in my logic, could someone find it?

Answer 1

The flaw is that if you lose seven in a row you don't have enough coins to double your bet again. It is a losing bet, so the more you play the more certain you are to lose. No betting strategy can fix that.

Answer 2

This is a gambling system called the "Martingale System" and it has a reasonable chance of working in the short term, but the issue is that in the long term you will always lose.

if you just bet one coin at a time, the expected value is 0.96 (0.48*2), and you can double that as much as you want, but the odds are stacked against you, and in the long run you WILL lose it all.

Answer 3

I just found it - it's called a martingale betting system. https://en.wikipedia.org/wiki/Martingale_(betting_system)