Singapore Pool I want you to come clean!

only_lonely

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If someone had knowledge that the first digit was 1 and bet 1000 numbers $1 Big each from 1000 to 1999 (1000, 1001, 1002… 1998, 1999) he would have won:

First prize: $2,000
Starter: $1,250 (5 x $250)
Consolation: $180 (3 x $60)

Total gross win: $3,430

Total net win: $2,430

That’s a $1 Big per bet for 1000 numbers.

If he’s rich enough to buy $10 Big per bet for the same 1000 numbers costing $10,000, he would have won $24,300!
The system draws number after sales close leh.
They will compute and see which grp of number noone bet or has less bet amount then let you tiok abit. In overall cannot lose must profit.
 

powerrr

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This one the most Tua Kong (dodgy):

Sat, 05 Dec 2020Draw No. 4631
1st Prize6152
2nd Prize6163
3rd Prize6158

52 to 63 only 12 sets of numbers to buy:

6152
6153
6154
6155
6156
6157
6158
6159
6160
6161
6162
6163
 

buyacat

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This one the most Tua Kong (dodgy):

Sat, 05 Dec 2020Draw No. 4631
1st Prize6152
2nd Prize6163
3rd Prize6158

52 to 63 only 12 sets of numbers to buy:

6152
6153
6154
6155
6156
6157
6158
6159
6160
6161
6162
6163
Oh yes I remember this.

And people can say this is randomness at work. I can say this isn't random.
 

Cryptology

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i believe sgpools has biased ball sets. not all balls are equal
They have multiple machines and multiple sets of balls, not just one. Unless you tell me they dont replace new balls often then difficult to have biased balls.

In the old days roulette tables have biases cuz owners dont regularly change balls and the spinners, now they regularly change to beat those who seek predictive patterns.
 

dexboi

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If it's digitally generated draws, then it is at your advantage.
Technology must always follow a logic..

Now that you realised the numbers are "frequently repeated",
don't you think you have an "unfair" advantage? 🤭
 

BHills

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9 (including 1st prize) out of 23 winning numbers starts with 1.

9/23 = about 40%

If arrange to have 40% for 1 to appear as 1st digit in winning numbers, then buying 1xxx will huat big.
 

cleffa3000

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while you are it ...

T9fWw5p.png
Wah no sot alr
 

BHills

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This is just randomness at work la.

Unless all 1st 2nd and third prize same number then come talk.
If you have knowledge of statistics and theory of randomness, then you would know that the probability for 9 out of 23 numbers start with 1 is extremely extremely extremely low as the expected number of winning numbers starting with 1 is only 2.3 out of 23.
 

thisisnotme1212

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This is just randomness at work la.

Unless all 1st 2nd and third prize same number then come talk.
unless its skeweds :(
ifs chius knows the algos maybe certains months/days some lumbers gots highers chances :(
theys shlds opens sources thes sources codes :(
 

BHills

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This is just randomness at work la.

Unless all 1st 2nd and third prize same number then come talk.
9 (including 1st prize) out of 23 winning numbers starts with 1.

9/23 = about 40%

If arrange to have 40% for 1 to appear as 1st digit in winning numbers, then buying 1xxx will huat big.

If you have knowledge of statistics and theory of randomness, then you would know that the probability for 9 out of 23 numbers start with 1 is extremely extremely extremely low as the expected number of winning numbers starting with 1 is only 2.3 out of 23.

Winning number(s) starting with :
"0" = 1 time
"1" = 9 times
"2" = 3
"3" = 2
"4" = 0
"5 = 0
"6" = 2
"7" = 2
"8" = 2
"9" = 2
======
Total 23

If each number has equal chance of appearing, then each number is expected to appear
Expected = 23/10 = 2.3 times

So 9 times number 1 is way off the expected 2.3 times.

If use statistics (we need some data to calculate standard deviation), it can be shown that the probability of any number appearing 9 times instead of the expected 2.3 times is extremely extremely extremely low.
 

fantasyland

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Winning number(s) starting with :
"0" = 1 time
"1" = 9 times
"2" = 3
"3" = 2
"4" = 0
"5 = 0
"6" = 2
"7" = 2
"8" = 2
"9" = 2
======
Total 23

If each number has equal chance of appearing, then each number is expected to appear
Expected = 23/10 = 2.3 times

So 9 times number 1 is way off the expected 2.3 times.

If use statistics (we need some data to calculate standard deviation), it can be shown that the probability of any number appearing 9 times instead of the expected 2.3 times is extremely extremely extremely low.

Its rare, but not impossible. Odds are close to 1 in 2500 4d draws.
 
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