Ok so here’s the rules
- I just bet on red every time
- I start with 1 dollar
- every time I lose, I triple my previous bet
- every time I win I restart
I’m going to simulate 10 games
- Game 1 - Bet $1 Lose
- Game 2 - Bet $3 Lose
- Game 3 - Bet $9 Win $18
- Game 4 - Bet $1 Lose
- Game 5 - Bet $3 Lose
- Game 6 - Bet $9 Win $18
- Game 7 - Bet $1 Lose
- Game 8 - Bet $3 Lose
- Game 9 - Bet $9 Lose
- Game 10 - Bet $18 Win $36
In this simulation I’m losing at a rate of 70%. In reality the lose rate is closer to 52%. I put in $54 but I’m walking away with $72, basically leaving the building with $18.
Another example. Let’s pretend I walk in with $100,000 to bet with. I lose my first 10 games and win the 11th.
- 1 lose
- 3 lose
- 9 lose
- 27 lose
- 81 lose
- 243 lose
- 729 lose
- 2187 lose
- 6561 lose
- 19683 lose
- 59049 win $118098
$88573 spent out of pocket, $118098 won
Walk out with roughly $29525.
I get most casinos won’t let you be that high but it’s a pretty extreme example anyway, the likelyhood of losing 10/11 games on 48% odds is really unlikely.
So help me out here, what am I missing?
Unless your bank roll is infinite, you always lose in the average case. My math was just an example to show the point with concrete numbers.
In truth it is trivial to prove that there is no winning strategy in roulette. If a strategy is just a series of bets, then the expected value is the sum of the expected value of the bets. Every bet in roulette has a negative expected value. Therefore, every strategy has a negative expected value as well. I’m not saying anything ground-breaking, you can read a better write-up of this idea in the wikipedia article.
If you don’t think that’s true, you are welcome to show your math which proves a positive expected value. Otherwise, saying I’m “completely wrong” means nothing.