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Clercx posted an excellent analogy concerning raindrops and probability.
Mostly concerning sections of the wheel that have failed to show.
If you compare this to rain, you can assume that both are random - but the largest dry area
will probably be receiving raindrops in the very near future since we know the eventual outcome
is complete coverage of the area (wheel).
I'm looking at it another way, although simillar.
Let's look at the 3 dozen sections in front of us at the table as 3 squares side by side (a sidewalk - for this analogy)
All three will receive raindrops randomly, but we can be almost 100%
assured that all 3 will never have the same identical pattern between
them.
This applies to the dozen sections when looking at the table layout in graphical layout.
Let's assume that after 37 spins we have sleepers in dozen 1 of #3, #8, #9 and #12
What are the odds that dozen 2 and dozen 3 both have those same identical sleepers in their layout ?
for dozen 2 it would be #s 15,20,21,24 and for dozen 3 #s 27,32,33,36
The odds of all three sections having the same pattern is nill in my opinion.
(I'm sure a possible "rare" event though of course)
Now let's assume we have sleepers in the dozen 1 section again of 3,8,9 and 12.
From the above we can "assume" that 15,20,21,24,27,32,33 and 36 "have" hits on them,
probably many but no doubt about average. Will these locations show a higher hit rate than
expected since their dozen 1 neighbor has these same locations as sleepers ?
Let's run off a quick test from actuals and see where the merit in this is (if any)
I chose a random s/h day, in this case Jan 1 2005
We can record the first 37 spins, and have 14 sleepers (as normal).
The sleepers on the image below are left on the layout, the numbers that hit
have been removed.
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What we see is that (as expected) the pattern between all 3 dozens is different.
If we look at the corresponding numbers to the sleepers, we can get the following info :
dozen 1 sleepers :
1,2,3,6,8,11,12
the same locations in dozens 2 and 3 :
13,14,15,18,20,23,24,25,26,27,30,32,35,36
(there are 14 numbers here, during the last 37 spins they had 18 hits on them :) )
dozen 2 sleepers :
13,19,22,24
the same locations in dozens 1 and 3 :
1,7,10,12,25,31,34,36
(there are 8 numbers here, during the last 37 spins they had 8 hits on them :) )
dozen 3 sleepers :
28,34,35
the same locations in dozens 1 and 2 :
4,10,11,16,22,23
(there are 6 numbers here, during the last 37 spins they had 8 hits on them :) )
in summary to this point -
we have 14 sleepers total, 28 numbers that are corresponding to those sleepers,
and a total of 34 hits on them.
===
The advantage here ?
Some people will say "so what ? of course the sleepers had 0 shows and the others had
more than average, that's common sense"
also, "what good is this info ? It's from the past and doesn't effect the future spins where
we are going to be playing"
This isn't the case in my opinion.
We can assume alot with this game,
above we assumed that all 3 dozens wouldn't have the same pattern of sleepers.. and they
don't (and probably almost never will)
Now, we can assume the the next 37 spins will not be identical to the 37 that just happened...
(leaving the exact same sleepers in the exact same locations)
What does this tell us if anything ?
Let's look at the dozen with the largest number of sleepers.
This will be our base dozen.
I'll assume that during the next 37 spins there will be hits on these sleepers (somewhere)
and I don't need to know which ones or when actually.
Now, since all 3 dozens won't follow the same pattern, I can assume the the "opposite"
if these sleepers in the other 2 dozens will also show during the next 37.
This leaves us with a bet selection that is very simple.
We are going to bet on the sleepers in the dozen that had the most of them,
in this case Dozen 1
We are also going to bet in the opposite of these numbers in the other two dozens.
From this day - Jan 1 2005 it means the following :
Bets :
Dozen 1 : 1,2,3,6,8,11,12
Dozen 2 : 16,17,19,21,22
Dozen 3 : 28,29,31,33,34
17 number total flat betting $5.00 each
On any win, we simply add to the winner $5.00 (there's no sense in
removing the bet after it wins if we believe that it will show again)
let's look at the next 37 spins in order :
19 win #1
24
3 win #1
26
15
21 win #1
31 win #1
2 win #1
24
32
25
30
35
12 wn #1
20
9
2 win #2
35
19 win #2
21 win #2
16 win #1
22 win #1
17 win #1
17 win #2
3 win #2
31 win #2
32
18
23
12 win #2
5
10
13
22 win #2
2 win #3
12 win #3
22 win #3
= end of second 37 spin cycle =
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+ $1,315.00 for only 37 spins.
So far live I've had one +5K day and one +4K day of play using this approach.
Would I call it the holy grail ?
Of course not.
But a better form of bet selection that most systems use, yes.
The probability guys should be able to back up this method also,
although in the end it may fair no better than any other method of play.
Hope the info was found to be usefull.
Feel free to email me or post at GG, VIP, or this site's forum
which is linked on the main page.
Cheers
Ed
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