Roulette Mathematical Advantage

s2baker's answer is easy to demonstrate.

Essentially, the system consists of the fact that the bet will be won 12/25 of the time and lost 13/25 of the time. You ignore the middle dozen because there is no result. The bet is only resolved after a win or loss, and as the number of throws approaches infinity, the end result is you win 12/25th of the time and lose 13/25th of the time. The HA therefore is 4% (-1/25). You multiply it by two which is the value of the bet (2 units) to get 8%.

You can try the same on a 00 wheel. The system consists of the fact that your bet will be won 12/26 of the time and lost 14/26 of the time. The HA is -2/26 (-.087119298) You multiply by 2 to get an expected loss of -.15384 units.

Quote: boymimbo

s2baker's answer is easy to demonstrate.

Essentially, the system consists of the fact that the bet will be won 12/25 of the time and lost 13/25 of the time. You ignore the middle dozen because there is no result. The bet is only resolved after a win or loss, and as the number of throws approaches infinity, the end result is you win 12/25th of the time and lose 13/25th of the time. The HA therefore is 4% (-1/25). You multiply it by two which is the value of the bet (2 units) to get 8%.

You can try the same on a 00 wheel. The system consists of the fact that your bet will be won 12/26 of the time and lost 14/26 of the time. The HA is -2/26 (-.087119298) You multiply by 2 to get an expected loss of -.15384 units.

I think I got that for the double zero wheel as well. Since it did't come out to .16 I wanted to doublecheck.

Quote: MangoJ

Maybe then you might learn something about your system or the game of roulette itself.

You people don't understand how these system
inventors think. Colbster is absolutely convinced
he's right and you'll never talk him out of it. He
wants to believe he's beaten roulette so badly that
he's in complete denial. You see this over and
over on the roulette boards. Some of these guys
have been around for years and they never admit
defeat. Something is wrong with them, they seem
sane and reasonable, but there's a tiny screw loose
in there somewhere.

I agree Bob At AT&T I used to handle the nut calls for the clerks. From the people who knew the FBI was tapping their phones to the
Indians who knew we were blocking all circuits to India on the weekends. No arguement or logic will ever convince them they are WRONG !

Quote: colbster

The following is from a PM I sent regarding a theoretical 36 spins. Hope this may be clearer for some:

We can agree that 12 losses at -2 give us -24 and 12 wins at +1 give us +12, netting -12 before considering the repeats.

When we get repeats, the new 12-spin average is +1 (-1/2/2) which covers the -12 evenly, meaning we have EV of 0.

BUT, 1/3 of those twelve will move up to the next level which averages +2 (0/3/3). This new 1/3 of 12 adds 4 units to our equation, meaning we are at -24/16/12, or +4

BUT, 1/3 of those 4 move up to the next level which average +3 (1/4/4). This new 1/3 of 4 adds 1.333 units, meaning we are at -24/17.333/12, or +5.333

BUT, 1/3 of those 1.333 move up to the next level which average +4 (2/5/5). This new 1/3 of 1.333 adds 0.443 units, meaning we are at -24/17.776/12, or +5.776

BUT, 1/3 of those 0.443 units move up to the next level which average +5 (3/6/6). This new 1/3 of 0.443 adds 0.148 units, meaning we are at -24/17.924/12, or 5.924.

This continues on until the 2nd dozen (the repeats) approach, but don't reach, 18, meaning we expect slightly less than EV of 6 per 36 spins on a no-zero table. Obviously, 36 bets is too small for these fractions, but expanded over a larger quantity of bets with normal distribution, the math holds up. If we considered, 2187 spins instead of 36 spins, it is more evident, but the numbers remain exactly the same throughout.

The following is my response to that PM:

Quote: paraphrasing of me

So let's look at the repeats. 1/3 of the time, you'll lose a net of 1 (+1, then -2). 1/3 of the time, you'll win a net of 2 (+1, then +1). 1/3 of the time, you'll enter the string of repeats. Each of these 12 spins needs to average +1 to get you to breakeven over the 36 total spins, since you have 12*-2 and 12*+1 already calculated. so far, you're at .5/spin on the first 8 of these 12 spins. So you need to average +2 on the repeat sequence to get to even, and more than that to get ahead. The total expected value of the repeat sequence is 8, which is +2/spin. So all you do is get to even.

That's actually a pretty good check on the math. You aren't doing anything to affect the house edge of any spin, so you can't affect the house edge of the total game. The fact that the math shows that this system is break-even on a no-0 wheel is a pretty good proof that the math is correct.

Quote: colbster

The following is from a PM I sent regarding a theoretical 36 spins. Hope this may be clearer for some:

We can agree that 12 losses at -2 give us -24 and 12 wins at +1 give us +12, netting -12 before considering the repeats.

When we get repeats, the new 12-spin average is +1 (-1/2/2) which covers the -12 evenly, meaning we have EV of 0.

Correct, as you are not taking into account any of the green numbers, and you are not counting anything won by that first spin when it is in the "repeat" column.

Quote:

BUT, 1/3 of those twelve will move up to the next level which averages +2 (0/3/3).

Once again I point out, here is where you make your mistake. You already say that your second spin earns you +2 when it lands in the repeat column - but then you include that +2 again when showing the results of the third spin. Assuming the second column is the repeat, either your second spin numbers should be -1/0/2, or your third spin numbers should be -2/1/1.

Let's use your reasoning in a slightly different way. In this system, always bet 1 on black, and "repeat" each time black comes up (so your bet continues for however many spins black comes up, and ends when red (or green, but we'll ignore green for now) comes up)
First spin: 1/2 of the time, it is red, and your bet ends with -1, and 1/2 of the time, it is black, and you are at +1; by your reasoning, your EV is 0, which is correct for that one spin.
"BUT, 1/2 of the time, you move up to the next level; if the spin is red, you are at 0, and if it is black, you are at +1, so the average EV is +1, and we add (1/2 x 1) to the bet's overall EP, for a total of +1/2." Obviously, this is wrong.
"BUT, 1/4 of the time, you move up to the next level, where the bet's EP is +1 if red and +3 is black, for an average of +2, and we add (1/4 x 2) to the bet's overall EP, for a total of +1."
"BUT, 1/8 of the time, you move up to the next level, where the bet's EP is +2 if red and +4 if black, for an average of +3, and we add (1/8 x 3) to the bet's overall EP, for a total of +1 3/8."
If you keep going, the eventual overall EP using your reasoning is +3.
This can't possibly be right, as if you switch "red" with "black", the overall EP for that strategy would be +3 as well, so if you bet on red and black simultaneously, you should make money every time the same color comes up two or more times in a row. Needless to say, that is incorrect.