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2 of 3 (8 Magic # system) - Sad Reality

Thoth's avatar - binary
Thoth

I have seen many groupings of "Magic" eight numbers that when played boxed are guaranteed to get you 2 of the 3 winning pick 3 number as long as the combo drawn is not a double or triple digit combination.  I have also seen where people have stated that your odds of winning boxed with these combos are 1 in 10 or even 1 in 8 when a nomatch number is drawn.

Here is the sad reality when playing the "magic 8 #" systems......

True Odds:
Yes - any of the magic 8 number sets will guarantee you 2 of 3 in boxed form when a nomatch is drawn, HOWEVER - you are still ONLY playing 8 of the 120 NoMatch Boxed combinations that are possible when a nomatch number is drawn.  This means your odds are 8 in 120 (1 in 15 reduced) or 6.67%.  This is far from 1 in 10 (10%) or 1 in 8 (12.5%).

Probability:
In the long run - you can expect any of the guaranteed magic 8 number sets to each hit only around 4.8% of the time. Heres why:

As stated above the true probabilty is 6.67%, but this percentage occurs only when a nomatch is drawn and that is 72% of the time.  So 6.67% X 72% gives you the overall probability of .048 or 4.8% for any givin drawing.

Another way to think of it is that the 8 boxed combos also represent 48 straight numbers (each of the 8 boxed play 6 ways straight), so thats 48 in 1000 or still 4.8%.

Results:
I have back tested 3 of the magic 8 number sets in 10,053 Ohio Pick 3 games. Here are the results that verify the 4.8% probability:

045

123

169

178

268

279

367

389

039

059

069

124

178

278

356

478

012

148

024

028

359

367

567

679

 489 Total Hits 427 Total Hits 505 Total Hits

Conclusion:
The overall probability of 4.8% is shown to be accurate - Now how "Magic" or "Lucky" is the system really?  Additionally, if you try to calculate the 3rd digit and/or replace digits in any of the 8 combo sets, you will lose the mathematical arrangement that is guaranteeing you 2 out of three!

 

 

BobP's avatar - bobp avatar.png
BobP

There are odds for every possible event in the game.  It cannot be denied that when you have a three different digit number on your ticket with two of the winning digits, there are only 8 other possible digits to complete the number. Therefore your odds of this ticket winning when all different digits are drawn is 1 in 8.  The idea is to wait until all different digits are due.  You don't have to play in every draw you know.   I never expect a generic set of 2in8 to do well because it isn't connected to the game.  For some reason hot numbers remain hot and cold numbers tend to remain cold for the life of the ballset collection.  It makes sense to build the wheel around one or more hot numbers in the game until they destroy the ballset collection and replace them with new balls.  This effect may not exist in a computerized game, Ohio may not be the best choice of a test state.

BobP


Thoth's avatar - binary
Thoth

There are odds for every possible event in the game.  It cannot be denied that when you have a three different digit number on your ticket with two of the winning digits, there are only 8 other possible digits to complete the number. Therefore your odds of this ticket winning when all different digits are drawn is 1 in 8.  The idea is to wait until all different digits are due.  You don't have to play in every draw you know.   I never expect a generic set of 2in8 to do well because it isn't connected to the game.  For some reason hot numbers remain hot and cold numbers tend to remain cold for the life of the ballset collection.  It makes sense to build the wheel around one or more hot numbers in the game until they destroy the ballset collection and replace them with new balls.  This effect may not exist in a computerized game, Ohio may not be the best choice of a test state.

BobP


I still have to stand by my calculated odds, though it does SEEM like the odds should be 1 in 8 they simply cannot be.

 

You get the same situation when looking at doubles and triples.  Every time you see a double digit pair in a pick 3 drawing you might think that the last non-repeating digit has (or had) a 10% chance of creating a triple (55X  5X5  X55) but there’s no way, not with 270 doubles and only 10 triples in the game.  In 10,000 games you will see about 100 triples and 2700 doubles.  If the triple had a 10% chance whenever a double was drawn you would have around 270 triples in 10,000 games instead of 100.

 

In the same way, if a set of magic 8 #'s had a 1 in 8 chance of winning every time a nomatch was drawn - then in 10,000 games we should have about 7200 no matches with 1/8 of those being from that 8# group.  This equates to 900 hits (.125*7200=900).  That’s almost twice as many as we actually have for any one of the groups.

 

I am sure it all has to do with ball order or how the numbers are drawn.  Ohio still uses a mechanical drawing system and is a very good state to test in.  Just for the heck of it i will test the same number sets throughout the Michigan Daily 3 history.  I'd bet money we will see close to an overall 4.8% hit rate there too.

 

Thoth's avatar - binary
Thoth
Performance of Magic 8# System for Michigan Daily 3 throughout 9,732 consecutive drawings.

045
123
169
178
268
279
367
389

039

059

069

124

178

278

356

478

012
148
024
028
359
367
567
679

463 Total Hits

4.76%

 474 Total Hits

4.87%

 431 Total Hits

4.43%

 

The total of NoMatch numbers that were drawn within the 9732 games was 6942.  IF everytime a NoMatch number is drawn and a Magic 8 group has a 1 in 8 chance of winning, then each group should win 1/8 of the total NoMatch games or (1 divided by 8 = 12.5%).  This does not appear to be the case.....

Magic 8 group #1 won 463 times in 6942 NoMatch games: 463/6942=6.67%

Magic 8 group #2 won 474 times in 6942 NoMatch games: 474/6942=6.83%

Magic 8 group #3 won 431 times in 6942 NoMatch games: 431/6942=6.21%

As shown above - they are only winning about as much as the true odds of 1 in 15 or 6.67%. allows them to.

BobP's avatar - bobp avatar.png
BobP

There are odds for every possible event in the game.  It cannot be denied that when you have a three different digit number on your ticket with two of the winning digits, there are only 8 other possible digits to complete the number. Therefore your odds of this ticket winning when all different digits are drawn is 1 in 8.  The idea is to wait until all different digits are due.  You don't have to play in every draw you know.   I never expect a generic set of 2in8 to do well because it isn't connected to the game.  For some reason hot numbers remain hot and cold numbers tend to remain cold for the life of the ballset collection.  It makes sense to build the wheel around one or more hot numbers in the game until they destroy the ballset collection and replace them with new balls.  This effect may not exist in a computerized game, Ohio may not be the best choice of a test state.

BobP


I still have to stand by my calculated odds, though it does SEEM like the odds should be 1 in 8 they simply cannot be.

 

You get the same situation when looking at doubles and triples.  Every time you see a double digit pair in a pick 3 drawing you might think that the last non-repeating digit has (or had) a 10% chance of creating a triple (55X  5X5  X55) but there’s no way, not with 270 doubles and only 10 triples in the game.  In 10,000 games you will see about 100 triples and 2700 doubles.  If the triple had a 10% chance whenever a double was drawn you would have around 270 triples in 10,000 games instead of 100.

 

In the same way, if a set of magic 8 #'s had a 1 in 8 chance of winning every time a nomatch was drawn - then in 10,000 games we should have about 7200 no matches with 1/8 of those being from that 8# group.  This equates to 900 hits (.125*7200=900).  That’s almost twice as many as we actually have for any one of the groups.

 

I am sure it all has to do with ball order or how the numbers are drawn.  Ohio still uses a mechanical drawing system and is a very good state to test in.  Just for the heck of it i will test the same number sets throughout the Michigan Daily 3 history.  I'd bet money we will see close to an overall 4.8% hit rate there too.

 

When you conduct such tests please supply the test set numbers as I have no idea whether your set is a true 2in8 or say a 90% 2in8 set.  Maybe not all six boxed combinations are being checked for.  I have no idea what might be going wrong, but again there are only 8 remaining digits and the winning digit has to be one of them so if that isn't 1in8 odds . . .

As with the Birthday Paradox and gambler's faliacy what seems correct may not be.
Sometimes the virtue of playing all the numbers in a limited number of lines forces a certain outcome then taking a set of 8 random three different digit numbers.  Are you sure your test is splitting out the all different digit numbers from the doubles and triples and testing all 8 numbers against them?  Sorry to ask, but all too often we see what we want to see.  BobP

powerplayer's avatar - Lottery-022.jpg
powerplayer

Very interesting..................I have a question for this % magic 8 #??

What is the percentage of taking those #'s and playing them in all states? Does the % go up? I would think it would. I have done this before with 1 #. Like 12(wheeled the last digit) I played it day/night all states @ a .25. I ended up hitting on it 2 times. Both at night.

 

Let me know

 

Powerplayer

Thoth's avatar - binary
Thoth

Bob,

The test numbers are listed in my above posts, each individual 8# group represents a 2 out of 3 guarantee 100% of the time when a NoMatch is drawn and occasionally when a double is drawn - this has been checked and verified in both Ohio and Michigan.  The actual hits or total wins for all 3 digits is also 100% accurate.

The fact that when you look at your ticket and know that you are guaranteed 2 of 3 by playing those 8 boxed combos does not negate the fact that there are still 120 possible no match boxed numbers - anyone of which could be drawn. That means that there 112 more combinations besides the 8 that you are playing, let alone the probabilty for doubles and triples.

If you knew in ADVANCE that the drawing was going to be a NoMatch number and you also knew that the first 2 digits were going to be a 1 and then a 2 then you could play 1-2-X (X= the other 8 numbers) and have a 100% chance of winning or just play 1-2- and one other digit (besides 1 or 2) for a 1 in 8 chance of winning.  But the problem is that you do not know what two numbers will be drawn or what positions that they will be drawn in.  All of that affects the probability.

 

Thoth's avatar - binary
Thoth

Very interesting..................I have a question for this % magic 8 #??

What is the percentage of taking those #'s and playing them in all states? Does the % go up? I would think it would. I have done this before with 1 #. Like 12(wheeled the last digit) I played it day/night all states @ a .25. I ended up hitting on it 2 times. Both at night.

 

Let me know

 

Powerplayer

The overall percentage would stay the same.  Sure, you may have days where they win in multiple states at the same time, but in the long run they will still statistacally perform the same.  Their odds will never change.

Avatar
pacattack05

Maybe they're not so magical afterall!!!!!   

There are no absolutes....there are no perpetual motion machines........and wrestling is indeed fake. LOL 

I coulda been a contendah......

BobP's avatar - bobp avatar.png
BobP

Let's say you went to http://www.lotto-logix.com/txthouse/lottodojo.html  and got a free 2in8 set built around a hot number from the game . . . on second thought let's use your worst set shown here with 431 wins . . .

012
148
024
028
359
367
567
679


Let's say you invested $1.00 per number .50/.50 straight/box and either played in all 9723 draws or tracked the draws and only played in the 6942 all different digit draws over the roughly 13 year period.

Your cost to play would be . . .

9,723 x $8.00 = $77,784. or 6,942.00 x $8.00 = $55,536.

Figure 431 box wins at $75. each and 1/6th of the boxed wins to also be straight for an additional $450. each  playing at the no name place.

$75.00 x 431 = $32,325. + $450.00 x 72 = $45,072. = $77,397. total win.

So with your worst example set we virtually broke even playing in default mode against every drawing.  If we had played either of your better sets or had any skill ability at picking days all different draws are due say after two double days the system actually works.   Quid pro quo.  BobP

Cheers

Thoth's avatar - binary
Thoth

Let's say you went to http://www.lotto-logix.com/txthouse/lottodojo.html  and got a free 2in8 set built around a hot number from the game . . . on second thought let's use your worst set shown here with 431 wins . . .

012
148
024
028
359
367
567
679


Let's say you invested $1.00 per number .50/.50 straight/box and either played in all 9723 draws or tracked the draws and only played in the 6942 all different digit draws over the roughly 13 year period.

Your cost to play would be . . .

9,723 x $8.00 = $77,784. or 6,942.00 x $8.00 = $55,536.

Figure 431 box wins at $75. each and 1/6th of the boxed wins to also be straight for an additional $450. each  playing at the no name place.

$75.00 x 431 = $32,325. + $450.00 x 72 = $45,072. = $77,397. total win.

So with your worst example set we virtually broke even playing in default mode against every drawing.  If we had played either of your better sets or had any skill ability at picking days all different draws are due say after two double days the system actually works.   Quid pro quo.  BobP

Cheers

Im not sure what state you are playing in Bob but maybe I should pack up and go play there - especially if they are paying $75.00 on a .50 cent boxed bet and $450 on a .50 cent straight bet.

Michigan pays $41 for .50 cent boxed bet and $250 for a .50 cent straight bet.  Ohio also only pays $250 for a .50 cent straight bet and $41.50 on a boxed bet.  Most states payout about the same on their Pick 3.

So for your figures above we should have:

9,723 x $8.00 = $77,784. or 6,942.00 x $8.00 = $55,536

Figure 431 box wins at $41 each and 1/6th of the boxed wins to also be straight for an additional $250 each.

$41.00 x 431 = $17,671 + $250 x 72 = $18,000 = 35,761 total won.

$77,784 - $35,761 = A loss of $42,023

Or if you somehow only managed to play all (and only) the NoMatch games, then $55,536 - $35,761 = A loss of $19,775

truecritic's avatar - PirateTreasure
truecritic

Thoth

When Bob stated:playing at the no name place.
I do believe he means at online sites that pay $900
for a $1 win.

Amazing how much difference a decent payout makes.

Thoth's avatar - binary
Thoth

Ahh gotchya - Anyone ever have any problems gettin paid there?

WIN  D's avatar - q05Q0
WIN D

 Back when they use to count wins and loses by hand they had a problem a couple of times with mine. They were quick to fix them then and I haven't had a problem since then. Now.....I trust.... but verify.

 

 P.S. Never play online casino games where the casino controls the results. 

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