Roulette Tool Edge Cases

[FreeRoulette]

I want to make a tool that tracks a bunch of different edge cases at 90 or 95 percent.

For example, the repeats, say 20 numbers spin with no repeat (I will figure out the percent), but lets say that has a 95% chance to win by playing those 20 next spin. The program would trigger telling the player to make the bet.

Now, that could take a long time, so I want your ideas on other cases. If we get enough cases, the we might be able to play at 90 win rate or higher without a progression.

I am open to any ideas.

[sugtips]

18 red in a row

[FreeRoulette]

18 red in a row

Okay 90% on all even and 2:1 for 'in a row'

[Joe]

For example, the repeats, say 20 numbers spin with no repeat (I will figure out the percent), but lets say that has a 95% chance to win by playing those 20 next spin.

FreeRoulette, how can playing 20 numbers give you a 95% chance to win? The chance any group of numbers has to win never changes depending on what came before, in this case it's 20/37 = 54%.

People keep making the same big mistake; they think that because a sequence of spins has some probability as a sequence, then it must apply to the next spin. This wrong, big time. So many systems are based on this fallacious idea.

[FreeRoulette]

FreeRoulette, how can playing 20 numbers give you a 95% chance to win? The chance any group of numbers has to win never changes depending on what came before, in this case it's 20/37 = 54%.

People keep making the same big mistake; they think that because a sequence of spins has some probability as a sequence, then it must apply to the next spin. This wrong, big time. So many systems are based on this fallacious idea.

Have you tested these 2 cases. How many times in a million trials does the next spin hit.

Case 1. pick 20 random numbers.

Case 2. Wait for 20 unique numbers in a row, then bet those numbers.

[Mister Eko]

The chances/odds never change in roulette, so fold up this idea rather.

[FreeRoulette]

The chances/odds never change in roulette, so fold up this idea rather.

Have you ever seen 38 sevens in a row?

To put it another way, 2 spins for red/black.

RR
RB
BB
BR

You have a one in 4 chance of picking the right combination excluding zeros

[ati]

Sorry, but Joe is right. Betting after a rare event of 20 unique numbers will not win any more than betting randomly.

Or betting after the rare event of 18 reds in a row. The rare event already happened, and the odds of winning in the next spin remain 50/50.

Betting for the rare event won't help either. Because it's a rare event, so you would lose a lot before the rare event finally happens.

Without going into the math, simply check the below 30 million spin results to calculate what would happen.

It's a total coincidence that in this set of 30 million spins you would profit 92 units. But let's check for 21 unique. It happened 165 times, the 22nd spin is the repeat. So 165 times you would win 15 units. But a total of 128 times there would be no repeat on the 22nd spin, so you would lose 128 times 21 units.
165*15=2475
128*21=2688
Total loss = 213 units. So you can see that there is zero advantage of waiting for 21 unique in a row. Not the mention the time wasted.

[reveal]2    9799
3    19199
4    26712
5    33099
6    36817
7    38117
8    37424
9    34613
10    30241
11    25616
12    20579
13    15855
14    11481
15    8103
16    5429
17    3428
18    2057
19    1187
20    665
21    372
22    165
23    79
24    35
25    10
26    3
27    0
28    1[/reveal]

[Joe]

Have you tested these 2 cases. How many times in a million trials does the next spin hit.

Case 1. pick 20 random numbers.

Case 2. Wait for 20 unique numbers in a row, then bet those numbers.

More to the point, have you tested the two cases? 

There shouldn't be any need to test them because it's a matter of logic. But just to satisfy you, here are the results of a simulation. Because 20 unique numbers in a row is a rare event I simulated 1000,000 sessions of 20 spins.

Code Select Expand

Number of 20 spin sessions with all unique #s : 1629
Number of wins directly after 20 uniques : 866
Number of wins directly after random #s : 890
Probability of getting a hit after 20 uniques : 0.532
Probability of getting a hit after random pick : 0.546

The difference isn't significant, here is another run :

Code Select Expand

Number of 20 spin sessions with all unique #s : 1728
Number of wins directly after 20 uniques : 937
Number of wins directly after random #s : 923
Probability of getting a hit after 20 uniques : 0.542
Probability of getting a hit after random pick : 0.534


and another : 

Code Select Expand

Number of 20 spin sessions with all unique #s : 1661
Number of wins directly after 20 uniques : 862
Number of wins directly after random #s : 914
Probability of getting a hit after 20 uniques : 0.519
Probability of getting a hit after random pick : 0.550

[Mister Eko]

Have you ever seen 38 sevens in a row?

To put it another way, 2 spins for red/black.

RR
RB
BB
BR

You have a one in 4 chance of picking the right combination excluding zeros

Appearing seven has 1:37 chance in every spin.

[FreeRoulette]

I see where you are coming from that a number has 1/38 chance and that the table has no memory.

Still I can't get over the idea that in 38 spins usually around 24 numbers appear. The further away from that, the less likely it is to happen. Can't we take advantage of that somehow? Like if you spin until all numbers appear, it usually takes less than a couple hundred spins. If you go a million trials, I highly doubt it would go a million spins without all the numbers showing up. Even though the wheel has no memory.

[FreeRoulette]

Just to add one more thing.

For argument sake lets say you tested 100, I know that is small, 100 million would be better, but anyway. The results show that only one time, red came up 15 times. So during play, red comes up 14 times and you will play the next spin.

Math says that red has an18/38 chance, but it also says it has a 1 in a hundred chance to happen.

[FreeRoulette]

Sorry, but Joe is right. Betting after a rare event of 20 unique numbers will not win any more than betting randomly.

Or betting after the rare event of 18 reds in a row. The rare event already happened, and the odds of winning in the next spin remain 50/50.

Betting for the rare event won't help either. Because it's a rare event, so you would lose a lot before the rare event finally happens.

Without going into the math, simply check the below 30 million spin results to calculate what would happen.

It's a total coincidence that in this set of 30 million spins you would profit 92 units. But let's check for 21 unique. It happened 165 times, the 22nd spin is the repeat. So 165 times you would win 15 units. But a total of 128 times there would be no repeat on the 22nd spin, so you would lose 128 times 21 units.
165*15=2475
128*21=2688
Total loss = 213 units. So you can see that there is zero advantage of waiting for 21 unique in a row. Not the mention the time wasted.

[reveal]2    9799
3    19199
4    26712
5    33099
6    36817
7    38117
8    37424
9    34613
10    30241
11    25616
12    20579
13    15855
14    11481
15    8103
16    5429
17    3428
18    2057
19    1187
20    665
21    372
22    165
23    79
24    35
25    10
26    3
27    0
28    1[/reveal]

28 uniques happened only once. 29 uniques never happened, so if 28 uniques happened, would you want to bet on those 28 or 28 random numbers that include all the numbers that did not come up?

[ati]

28 uniques happened only once. 29 uniques never happened, so if 28 uniques happened, would you want to bet on those 28 or 28 random numbers that include all the numbers that did not come up?

No, because I would still have 28/37 chance to win. And it isn't even worth to talk about it, since it's so rare, I wouldn't want to wait months or years to make a single bet.

The results show that only one time, red came up 15 times. So during play, red comes up 14 times and you will play the next spin.

Math says that red has an18/38 chance, but it also says it has a 1 in a hundred chance to happen.

I know it can be a bit hard to understand, but there are different odds for the next spin and the next series of spins.

You talk about the odds of the series, but you can only bet on the next spin.

Let's look at simple sequence of two EC outcomes. There are four possible ways, RR, RB, BR, BB. So two reds in a row has 25% chance to happen. Or 1/4.
But note, this is before the first spin!
Now after the first red comes out, do we still have 25% chance to see RR? No!
We already have a red, so it cannot be BR or BB. It can be ether RR or RB, so two possibilities remain at 50/50 chance.

This simple example applies to longer series as well, like 14 reds in a row, or 20 unique straight in a row.

[FreeRoulette]

No, because I would still have 28/37 chance to win. And it isn't even worth to talk about it, since it's so rare, I wouldn't want to wait months or years to make a single bet.

I know it can be a bit hard to understand, but there are different odds for the next spin and the next series of spins.

You talk about the odds of the series, but you can only bet on the next spin.

Let's look at simple sequence of two EC outcomes. There are four possible ways, RR, RB, BR, BB. So two reds in a row has 25% chance to happen. Or 1/4.
But note, this is before the first spin!
Now after the first red comes out, do we still have 25% chance to see RR? No!
We already have a red, so it cannot be BR or BB. It can be ether RR or RB, so two possibilities remain at 50/50 chance.

This simple example applies to longer series as well, like 14 reds in a row, or 20 unique straight in a row.

What you said makes perfect sense RB has a 25% chance and after the first color, the second color has a 50% chance. So you convinced me about how the sequence works. But how does it work over time? Lets say RB has a 25% chance to come up, 1 in 4. Lets say 10 trials came up RR BB , the longer it goes without a RB, wouldn't RB have a better chance to come up in order to stay with the odd of 1 in 4?