Thunder said:
How do you figure a 65% chance? I'm guessing your odds of winning on any spin in that scenario is 48%. IF you lose the first spin, your odds of winning the next spin is still 48% Your odds of losing 2 in a row would be 27%
Oh it's kindof like I could bet $150 on any 2 of the 3 columns with a 2-1 payoff.
I'll just switch to a single-zero game now lol.
If I do that I have a 24/37 chance of winning. (64.9% chance). Or I could bet $75 on any 1 dozen. If I win, I won my $150. If I lose, I bet $112.50 on any one dozen. If I win, I win $225 and am $150 ahead and quit. If I lose again, I still have $112.50 left and could bet $12.50 on any 9 numbers and still win my $150.
So, if you multiply it out, I have increased my chances of winning from originally 64.9% to 65.5% chance.
The point is, that with identical orig $roll, I have increased my chances of winning the same amount, of achieving my goal, only by changing how I choose to bet it. The house edge of the game hasn't changed, only how long my avg bet has been exposed to it.
Obviously, the given 1 hr time constraint enters into it too. It allows me to play 3 spins, at most, in this case rather than being limited to only 1 spin.
With an entire hour to play with, it wouldn't surprise me that this could be improved upon further.
But, hey, there you go, if you have time for 3 spins at single-zero roulette, and want to win 50% of starting roll 65.5% of the time, why not do it if that risk is acceptable to you lol?
If 3 spins takes 10 minutes, and you only have 10 minutes, if you can figure out a higher chance of achieving 50% of starting roll in 10 minutes, then definitely do that lol.