What is the impact of preprogrammed velocities on roulette probabilities?

[borson]

Hey guys! first off, thank you all for your attention!.
Okay I am a bit naïve on this, and also this question may seem a bit strange. But here it goes.
Let's say that a mechanic roulette throws the ball at different preprogrammed velocities.
Each velocity will lead to a determined final time, that is, the time at which the ball leaves the stator to fall into the wheel. These different final times are recorded and the data gathered.
we see then, that there are 2 possible time intervals. The data also shows that the likelihood of one is slightly higher than the another's.
Let's say that the likelihood of the former interval is 30%. It is also remarkable to mention, that that roulette has a bias, so the ball leaves the stator always at the same place, that is, within the same space interval.
Next, in real game we'll clock/time the rotor to find out its speed, and thus we'll figure out which numbers are going to be in the space interval during that time interval. In overall, we'd bet on 12 numbers each spin.
So here my question comes. If the frecuency (probability) at which that time interval turns out is 30%, it means that at least I should win 30% of the times. But... the likelihood of betting on 12 numbers in a European roulette is 12/37=32%!. Does this mean that by using this method my chances of winning are only decreasing? or I should combine both probabilities? Or what does it mean?

Thank you all!

 

[scottdave]

I did not follow something:

...we see then, that there are 2 possible time intervals. The data also shows that the likelihood of one is slightly higher than the another's.
Let's say that the likelihood of the former interval is 30%.

So if there are just 2 time intervals, and one is slightly more likely than the other, I would expect to see percentages such as 48% and 52%. If one is 30%, then what is the other one: 70%? That is hardly slightly more likely.

You are correct about there being 37 spaces on a European wheel. Each number pays 35 to 1, if I remember correctly. Without any advance knowledge, you will lose in the long run. It seems like you are trying to come up with a method to predict that some numbers come up more often? If you could do that, then it may be possible to offset the house advantage, but I don't see it in the way you have stated it.

 

[scottdave]

But, let us suppose that you determine that 12 of the 37 numbers will come up 30% of the time, rather than the normal 32%. So that means 70% of the time, the other 25 numbers are coming up. What if you bet on those 25 numbers? Say you bet 1 chip on each of those 25 numbers, for every spin. What is going to happen in the long run?
If you divide it up, so that 30% of the time, there is equal likelihood of the 12 numbers (not bet) coming up - you lose in those situations.
In the other 70% of the time, say that each of those 25 numbers comes up with equal chance, so in those situations, you collect 36 chips, but you bet 25, for a gain of 11.

In 10 hypothetical spins, you have bet a total of 250. You collect (36 x 7 = 252). So that might be on to something. It would take a long time to win much, though. And you could go through long streaks of not winning any, so you'd need some deep pockets to start. Remember this is expected value (over a long time, not 10 or 20 spins). Try flipping a coin for a bunch of times. On average it is about 50% heads and 50% tails, but you can have runs of 7 or more heads in a row, for example, event though on average it is 50%.

 

[borson]

I did not follow something:

So if there are just 2 time intervals, and one is slightly more likely than the other, I would expect to see percentages such as 48% and 52%. If one is 30%, then what is the other one: 70%? That is hardly slightly more likely.

There are two time intervals which are the most likely/are turned out the most. But there are other possible "final times" which are out of these intervals, but that turn out less often. In overall, that time interval is the most often turned out (with a 30%).
So the thing is... if I timed the velocity of the rotor, and with it I predicted which numbers were within the space interval, during that time interval, I would hope to get a profit. But the problem is that, that time interval (I would be making the prediction for*) only happens 30% of the times. Whilst betting on 12 numbers, without any prediction method, I should at least have a 32% of likelihood. So it seems that I would even have less chances with my method? How do I interpret it?

* The prediction would be made by simply multiplying the rotor's velocity by the final time (at which the ball is supposed to fall). In that way I'd get the space traveled by the rotor and thus its position at that time. If the time interval is, let's say, 22'6s to 23'3s (which means that between both final times, are found a 30% of the final times that turn out ) I would calculate the space traveled in 22'6 seconds, and then all of the numbers that would get in the space interval* during the next 0'7 seconds.

*The space interval is an area of that specific roulette at which the ball always fall.

Thanks!

 

[scottdave]

So 12 spaces out of 37 is 32.43% of the spaces. With a fair wheel, you would expect to collect (in the long run) 11.35 chips per 12 chip bet. Since you have determined that it actually lands on these 12 spaces less than 32.43% of the time, then you should stay away from those 12 numbers. Look at the other time interval (number intervals) in the same way. If the wheel is indeed operating this way, you might find an interval of only 2 or 3 numbers which could be to your advantage, for example.