### Author Topic: How many repeats before 18/19 uniques?  (Read 18128 times)

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#### random

##### How many repeats before 18/19 uniques?
« on: August 09, 2015, 11:50:52 PM »
I count 19 unique numbers. Then i check how many spins it took to get them. The difference is the number of repeats.
« Last Edit: August 09, 2015, 11:55:11 PM by random »

#### Reyth

##### Re: How many repeats before 18/19 uniques?
« Reply #1 on: August 10, 2015, 12:11:21 AM »
Yes, I notice that when repeats are numbers which MISS my bets, they can be the cause of a serious downfall.

#### random

##### Re: How many repeats before 18/19 uniques?
« Reply #2 on: August 10, 2015, 12:19:29 AM »
random.org examples

#### random

##### Re: How many repeats before 18/19 uniques?
« Reply #3 on: August 10, 2015, 12:21:36 AM »
and some more...

#### Reyth

##### Re: How many repeats before 18/19 uniques?
« Reply #4 on: August 10, 2015, 12:37:22 AM »
I guess I will see what the upper range of repeats is per 37 spins.  I already know that 37 is the highest number of uniques which means the lowest number of repeats is 0.  But let's see what the highest number of repeats is...

The max number of repeaters is 36 and here is a sample output of 35:

23  17  23  15  31  9  9  36 3 31  20  2  15 19  33  6  28  24  5  27  19  19  19  28  2  5  3  36  33  15 9  0  27  17  20  31  0

So here our unique numbers are 24 and 6.  37 repeats do not seem to occur.

Code: [Select]
`1 OPEN "Max_Repeaters.txt" FOR OUTPUT AS #15 DIM sh(37), rs(37)10 RANDOMIZE TIMER20 r = INT(RND * 37): sp = sp + 1: IF sp <= 37 THEN sh(sp) = r: GOTO 2030 GOSUB 100 ' spin history is full, shuffle history40 FOR i = 1 TO 37: FOR j = 1 TO 37  50 IF sh(i) = sh(j) AND i <> j THEN rs(i) = 1: rs(j) = 160 NEXT j, i62 rc = 0: FOR i = 1 TO 37 64 IF rs(i) = 1 THEN rc = rc + 166 NEXT i68 FOR i = 1 TO 37: rs(i) = 0: NEXT i 'flush repeat counter array70 IF rc > mr THEN mr = rc: GOSUB 20080 LOCATE 1, 1: PRINT mr90 GOTO 20100 'shuffle history110 FOR i = 1 TO 36120 sh(i) = sh(i + 1): NEXT i130 sh(37) = r: RETURN200 FOR i = 1 TO 37: PRINT #1, sh(i);: NEXT i: PRINT #1,: PRINT #1,: RETURN`
I could work out the percentage chances of getting X # of repeats, very easily.  Since inside numbers pay very well, this could be a very powerful strategy where we could bet when X # of repeats have occurred in the last 37 spins because we would have Y% chance of hitting if we bet only the unique numbers.  If Y% is less than Z then we would not bet.

What I am hoping is that the loss distribution data will show an improvement over simply betting cold numbers every 37 spins.  The only way this could be true is because of the unique configuration caused by repeats as opposed to simply hits; i.e. the Law of the Third has the categories of "repeaters" and "hits" and normally these are counted as the same when determining cold numbers (they are just numbers that have hit).

Previously I have examined the last 37 spins and found that simply betting all numbers that have not appeared in the last 37 spins will generate a hit within 27 spins.  The idea is that when we specifically track REPEATERS that our statistics will improve and we will expect a hit in LESS THAN 27 spins.

Waiting for one of the repeating categories to hit 16M...

Ok 16M was going to take almost 1.5 hours and I am running late in my moving.  So I cut it off at 1M.

The total number of spins is: 7,484,128

We can observe the peak percentage of repeats at 23 and then a decline from 24 to 36.  Obviously our focus needs to be on 24 and greater:

(24)981819 0.1311868263076206
(25)860067 0.1149187988233232
(26)667228 0.08915240359331107
(27)457160 0.06108393656548899
(28)275391 0.03679667156948679
(29)143307 0.01914812253344678

(30)64511   0.008619708267950521
(31)24552   0.003280542502747147
(32)7760     0.001036860940913891
(33)1934     0.0002584135386246734
(34)364       0.00004863626063049697
(35)37         0.000004943795723429637
(36)2           0.0000002672322012664668

Anything 30 and below we will need to spin over 100 times on average to even see and so those don't look like very good triggers.

29 seems to look the best as far as balance between frequency and success chances.  We should see this around every 50-60 spins.

REPEAT COUNTING PROCEDURE:

1) Write out the entire 37 spin sequence on a piece of paper
2) Place a check mark next to EACH number that repeats
3) Count your check marks.  If they are 29 or above, you have a bet.

Obviously I can write a program that will do this automatically.

So the procedure is to track every spin and examine the rolling 37 spin snapshot.  When there is at least 29 repeating numbers, bet the remaining numbers that have not yet shown up.  The chances of success in one spin is:

143307/242467=.5910371308260506

The max consecutive loss we should see on this bet is 17 times in a row.  This is definitely an improvement over 27 spins.  Let's look at the other bets:

30: .6505748285599032  Max Loss=16
31: .7085918785534936  Max Loss=13

Now for the progressions.  If each of the 29 numbers were to represent a single number repeated once (they won't ever be only once) the total numbers bet would be 15, so that's the minimum bet:

Bet 1 on 15 spots for 2 spins. Minimum Profit: 6
Bet 2 on 15 spots for 1 spins. Minimum Profit: 12
Bet 3 on 15 spots for 1 spins. Minimum Profit: 3
Bet 6 on 15 spots for 1 spins. Minimum Profit: 21
Bet 10 on 15 spots for 1 spins. Minimum Profit: 15
Bet 17 on 15 spots for 1 spins. Minimum Profit: 12
Bet 29 on 15 spots for 1 spins. Minimum Profit: 9
Bet 50 on 15 spots for 1 spins. Minimum Profit: 15
Bet 86 on 15 spots for 1 spins. Minimum Profit: 21
Bet 147 on 15 spots for 1 spins. Minimum Profit: 12
Bet 252 on 15 spots for 1 spins. Minimum Profit: 12
Bet 432 on 15 spots for 1 spins. Minimum Profit: 12
Bet 741 on 15 spots for 1 spins. Minimum Profit: 21
Bet 1270 on 15 spots for 1 spins. Minimum Profit: 15
Bet 2177 on 15 spots for 1 spins. Minimum Profit: 12
Bet 3732 on 15 spots for 1 spins. Minimum Profit: 12 <=== 17th bet

The maximum numbers bet will be 20, otherwise the bet amount goes over 1M units:

Bet 1 on 20 spots for 1 spins. Minimum Profit: 16
Bet 2 on 20 spots for 1 spins. Minimum Profit: 12
Bet 4 on 20 spots for 1 spins. Minimum Profit: 4
Bet 9 on 20 spots for 1 spins. Minimum Profit: 4
Bet 21 on 20 spots for 1 spins. Minimum Profit: 16
Bet 47 on 20 spots for 1 spins. Minimum Profit: 12
Bet 106 on 20 spots for 1 spins. Minimum Profit: 16
Bet 238 on 20 spots for 1 spins. Minimum Profit: 8
Bet 536 on 20 spots for 1 spins. Minimum Profit: 16
Bet 1206 on 20 spots for 1 spins. Minimum Profit: 16
Bet 2713 on 20 spots for 1 spins. Minimum Profit: 8
Bet 6104 on 20 spots for 1 spins. Minimum Profit: 4
Bet 13734 on 20 spots for 1 spins. Minimum Profit: 4 <=== 13th bet

The maximum number of numbers for the 29 repeaters is 17:

Bet 1 on 17 spots for 2 spins. Minimum Profit: 2
Bet 2 on 17 spots for 1 spins. Minimum Profit: 4
Bet 4 on 17 spots for 1 spins. Minimum Profit: 8
Bet 8 on 17 spots for 1 spins. Minimum Profit: 16
Bet 15 on 17 spots for 1 spins. Minimum Profit: 13
Bet 28 on 17 spots for 1 spins. Minimum Profit: 5
Bet 53 on 17 spots for 1 spins. Minimum Profit: 4
Bet 101 on 17 spots for 1 spins. Minimum Profit: 15
Bet 191 on 17 spots for 1 spins. Minimum Profit: 8
Bet 362 on 17 spots for 1 spins. Minimum Profit: 10
Bet 686 on 17 spots for 1 spins. Minimum Profit: 12
Bet 1300 on 17 spots for 1 spins. Minimum Profit: 16
Bet 2463 on 17 spots for 1 spins. Minimum Profit: 13
Bet 4667 on 17 spots for 1 spins. Minimum Profit: 18
Bet 8842 on 17 spots for 1 spins. Minimum Profit: 4
Bet 16754 on 17 spots for 1 spins. Minimum Profit: 18 <=== 17th bet

So at a .01 table the bankroll required is 167.54*17=2,848.18

This is a definite improvement over the 4000+ required for 27 spins!

The reason why this works better than just betting the cold numbers is because a minimum of 29 repeats has taken place in the last 37 spins.  So tracking repeaters can improve the odds of hitting cold numbers!

Thanks for this post Random! : D
« Last Edit: August 10, 2015, 03:00:02 PM by Reyth »

#### scepticus

##### Re: How many repeats before 18/19 uniques?
« Reply #5 on: August 10, 2015, 01:32:46 PM »
If it is accepted that the maths in The Birthday Problem is valid  then the Probability is that there will be 2 the same in any 10  spins -  with a 95 % probability.
The question is - Can we profitably make use of this ?

#### Mike

##### Re: How many repeats before 18/19 uniques?
« Reply #6 on: August 10, 2015, 07:06:33 PM »
the Probability is that there will be 2 the same in any 10  spins -  with a 95 % probability.

I beg to differ. The probability is just under 74% for at least one repeat in 10 spins.

#### scepticus

##### Re: How many repeats before 18/19 uniques?
« Reply #7 on: August 10, 2015, 07:52:24 PM »
Well I am not going to argue on this Mike. I was using a mathematician' s take on this. At least we agree that the probability can be determined fairly easily. and it is  around 74 % to 95%  which might be useful to progressive bettors.

#### random

##### Re: How many repeats before 18/19 uniques?
« Reply #8 on: August 10, 2015, 11:22:19 PM »
Thank you all for your efforts on this.
Maybe i did not explain myself correctly on this. I am not targeting repeats. I use the number of repeats as an indication of which side i should bet on. If not many repeats i bet the first 19 uniques, on the contrary i bet the unseen 18.

#### Reyth

##### Re: How many repeats before 18/19 uniques?
« Reply #9 on: August 10, 2015, 11:43:53 PM »
Very interesting!

I certainly agree with your analysis of many repeats, it will definitely improve your statistics with cold numbers!

The opposite picture is intriguing!  Obviously the lower the number of repeats the better.  Maybe 10 and under?

I think there should be some ranges of repeats where I would not bet (middling numbers like maybe 11-28).  I think focusing on the extremes (very low repeats, very high repeats) will perform best?

What numbers do you use?

#### palestis

##### Re: How many repeats before 18/19 uniques?
« Reply #10 on: August 11, 2015, 12:45:01 AM »
Thank you all for your efforts on this.
Maybe i did not explain myself correctly on this. I am not targeting repeats. I use the number of repeats as an indication of which side i should bet on. If not many repeats i bet the first 19 uniques, on the contrary i bet the unseen 18.
Observe 12 unique numbers in  a row. (No repeats among them).  Then see what happens in the next 6 spins.
1. The next 6 spins bring 6 new unique numbers. I find that extremely rare.
2. The next 6 spins bring unique numbers and some of them repeat among themselves without any repeat in the original 12. That is rare too.
Therefore after 12 unique numbers most likely at least one of them will repeat in the next 6 spins.
All you have to do is test to find out the probability of that happening.
The question is, who has the patience to wait for 12 unique numbers, bet them until one repeats and stop. Then wait patiently again for another 12.
« Last Edit: August 11, 2015, 12:46:44 AM by palestis »

#### random

##### Re: How many repeats before 18/19 uniques?
« Reply #11 on: August 11, 2015, 03:22:37 AM »
Thank you all for your efforts on this.
Maybe i did not explain myself correctly on this. I am not targeting repeats. I use the number of repeats as an indication of which side i should bet on. If not many repeats i bet the first 19 uniques, on the contrary i bet the unseen 18.
Observe 12 unique numbers in  a row. (No repeats among them).  Then see what happens in the next 6 spins.
1. The next 6 spins bring 6 new unique numbers. I find that extremely rare.
2. The next 6 spins bring unique numbers and some of them repeat among themselves without any repeat in the original 12. That is rare too.
Therefore after 12 unique numbers most likely at least one of them will repeat in the next 6 spins.
All you have to do is test to find out the probability of that happening.
The question is, who has the patience to wait for 12 unique numbers, bet them until one repeats and stop. Then wait patiently again for another 12.

For this type of approach the most significant number of spins is seven (statisticaly). Seven spins with seven unique numbers with no repeats back to back.

What is being proposed here is an even chance bet within some parameters.
Cheers

#### scepticus

##### Re: How many repeats before 18/19 uniques?
« Reply #12 on: August 11, 2015, 09:09:07 AM »
For this type of approach the most significant number of spins is seven (statisticaly). Seven spins with seven unique numbers with no repeats back to back.

What is being proposed here is an even chance bet within some parameters.
Cheers

so what is the significance  of the 8th spin  ?
" Some parameters " being What ?

#### Mike

##### Re: How many repeats before 18/19 uniques?
« Reply #13 on: August 11, 2015, 11:05:12 AM »
Therefore after 12 unique numbers most likely at least one of them will repeat in the next 6 spins.
All you have to do is test to find out the probability of that happening.

The problem with this idea is that it's basically no different from waiting for a run of say 10 reds in a row and then betting black. In order to catch a repeat you have to bet on each of the previous 12 numbers plus each of the subsequent numbers until one repeat. I haven't actually worked out the progression but WHEN none of the following 6 numbers repeat you are left with a big loss. You may win most of the time, but when the loss occurs it will make up for the previous profits gained. Just like a martingale, you may make a profit for a while... until disaster strikes.

#### Mike

##### Re: How many repeats before 18/19 uniques?
« Reply #14 on: August 11, 2015, 11:23:10 AM »
Here's the math:

You have observed 12 numbers with no repeat, so for the next 6 spins you bet on 12, 13,... up to 17 numbers. The chance of success (at least one repeat in the 6 bets ) is

1 - (25/37 x 24/37 x ....... x 20/37) = 99.87%

Great! but how much have you lost on those rare occasions when you DO lose?