The 'repeat window' - Math vs Reality

[Ricky]

Mathematically the repeat can appear at spin 38, but in reality the limit is spin 29.

[Ricky]

Unfortunately, 76% is still quite a large proportion of the cycle and when adding in table limits, this is still difficult to create an advantage from.

There are two paths to solve this problem. You have assumed one path - progression.
The other path will get you there with much less risk and more successful cycles then losses.

cheers,
Ricky

[Steve]

You need to take into account he was full of shit.

[Ricky]

We need to make the 'repeat window' smaller so we're only betting in a small section of the cycle.

Or somehow overcome the lost spins using some other means. What if it did not cost you anything to see the next number or you invented a 'time machine'?

Cheers,
Ricky

[Ricky]

You need to take into account he was full of shit.

You mean 'Dyskexlic'? Never met the guy. But he may not have found the ideal solution but there is another way to solve his problem!

Cheers,
Ricky

[Ricky]

The repeat window is 50/112 = 44% of the cycle.

Are you using the same number of spin cycles to get your results?

I would have thought if you doubled your number set you should double your spin cycle to get an equivalent max uniques before a repeat. This is due to the more combinations in the larger spin set we have not exhausted all possible paths. This may be a reason why your ratios are dropping. You may also find if you double the 37 spin cycles you may get more than 29 uniques.

Cheers,
Ricky

[Ricky]

Take the result of 2 spins, for the first spin record if it's red or black, then for the second spin record the straight number.

Ignore my previous responses . Now I see how you are creating your 74 options from the 37 numbers so your analysis is correct. I should have read the entire thread before responding, I thought your were using a wheel with 74 numbers.

Cheers,
Ricky

[Ricky]

Red hot, for example in a roulette of 74 numbers R0......R36 and B0....B36 if looking for the repetition of a number land zero in 1st spin (we need two spins to make each number of our roulette of 74 numbers), Do we spin 1 more time?.

Actually, if you include the Zero, would you not have RBG as the color? So there will be more than 74 options. This is where the house edge comes and your stats then become skewed as some results will be as you say G0 - G36

[redhot]

Actually, if you include the Zero, would you not have RBG as the color? So there will be more than 74 options. This is where the house edge comes and your stats then become skewed as some results will be as you say G0 - G36

I'd ignore the zero for now, we need to create a method that can beat non-zero roulette first (1-36).

Once that is established, then think about how to handle zero.

[quos]

Hi Redhot,the best method so far may be that of a roulette of 1369 numbers.

two spins to make each number of our roulette of 1369 numbers.

0-0, 0-1, 0-2, 0-3.......until 36-36

Our repeat window is 13,8% (189 numbers of the 1369)

Then I would start writing down numbers every two spins. Each time in the first spin, a number appears that previously appeared in another first spin, we will bet on its second turn.

Example:

sequence:

2-15
3-18
25-35
14-23
.
.
.
2-  Here first spin corresponds to the first spin of a number that has already appeared. then we would bet for the number 15.

we would have to follow a progression that, in the event of a hit, leaves at least one unit up.
The question is that I do not know how far the progression could go and if it would reach the limits of the roulette.

My formula for calculating the bet is:

X = (1+Y) / (36-P)

X = bet to make in each Straight
Y = accumulated losses within the micro-game
P = Number of Straight to play in a certain spin

Thanks for continuing to explain Redhot!

[quos]

Up

[Blueprint]

Looks complicated.

[quos]

Redhot, could you share your opinion about the last post please?
It's seem very interesting.
Thanks!!!

[redhot]

Redhot, could you share your opinion about the last post please?
It's seem very interesting.
Thanks!!!

Your method looks interesting have you tested it? How does it play out?

I don't have the answer, I'm not here to say what's right or wrong. I'm simply sharing my thoughts to hopefully inspire other ideas.

[Orca]

Mathematically, roulette can't be beaten. This is a fact.

However, this is under the assumption that roulette will behave in a mathematically expected way defined by probability. If it doesn't, the above statement no longer holds true.

If we could find a situation where roulette results do not conform to the mathematical expectation, we may have a chance to beat the game.

A simple example is a biased wheel, certain numbers/sectors hit above the mathematical expectation. Betting on these numbers gives an advantage.

What an interesting proposition? Incorrect but interesting all the same. You are clearly mistaken about the math.