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?
Another way of thinking about it is betting your entire bankroll for 99.9…% certainty that you will win $1.
Say you go into the casino with $1000.
Bet:
But what are the chances of getting 6 or 7 losses in a row? 1 in 64, or 128 respectively, actually worse because roulette wheels aren’t 50/50, they’re 18/19 (18 wins and 19 losses in 37 plays on average) or worse. So losing 6 times in a row will happen 1 in 54 plays, 7 losses is 1 in 106.
Google says roulette wheels spin 55 times per hour so with your strategy you will lose your bank roll in about one hour assuming your starting bet is 0.1% of your bank roll.