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UK LOTTO HOT PICKS betting with Fibonacci series - round 6
From
=?UTF-8?B?546L5Lm+5oe/IFdhbmcgUWlhb@21:1/5 to
All on Thu Mar 3 04:08:47 2022
Leela is the winner in round 2.
Prize for Leela: 635 euro.
MJ is the winner in round 5 as the new record-maker.
Prize for MJ: 2020 euro
Playing times for each round.
1,1,2,3,5,8,13,21,34,55,89,144,233,377
7 ROUNDS TILL ROUND 20
PLAY TOP MATCH3 WITH TOP 11 NUMBERS
PLAY UK LOTTO HOT PICKS
Round 6 will start to play 8 times.
If we get a 3 in 11 in round 6, the overall winning would be: (8*800)-{165*1*(1+1+2+3+5+8)}=6400-3300=3100 euro
Good luck!
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From
=?UTF-8?B?546L5Lm+5oe/IFdhbmcgUWlhb@21:1/5 to
All on Thu Mar 3 04:28:18 2022
However, if a player chooses to play 2 sets after round 5, he would still have lost 200 euro.
If we get a 3 in 11 in round 6 playing 2 sets, the overall loss would be: (8*800)-{165*2*(1+1+2+3+5+8)}=6400-6600=-200 euro
Therefore, the best strategy to guarantee the maximum chance of winning is to play 2 sets of 11 numbers within 5 rounds and stop playing.
Nan
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From
=?UTF-8?B?546L5Lm+5oe/IFdhbmcgUWlhb@21:1/5 to
All on Thu Mar 3 04:24:10 2022
I would suggest each player could post two sets for the Fibonacci series for later new competitions.
It can be seen that even if at least one set out of two generates a 3 in 11 till round 5, the chance of making profits would be still 100%.
However, when a player chooses to play 3 sets, he would have lost, even at round 2 ( losing 190 euro) and would have lost more up to 1940 euro at round 5, unless he makes a 3 in 11 among 3 sets at Round 1, winning (1*800)-{165*3*1)}=800-495=305 euro.
For example, if MJ played two sets including his and Jerome's till round 5, he would have still won.
(5*800)-{165*2*(1+1+2+3+5)}=4000-3960=40 euro
if MJ played three sets till round 5, he would have lost 1940 euro. (5*800)-{165*3*(1+1+2+3+5)}=-4000-5940=- 1940
if Nan played his set and Leela's till round 2, he would have still won: (1*800)-{165*2*(1+1)}=800-660=140 euro.
if Nan played three sets including Leela's and Alphazero's, he would have lost 190 euro.
(1*800)-{165*3*(1+1)}=800-990= - 190 euro
The sooner you make a 3 in 11, the more prize you would won.
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From
nigel@21:1/5 to
All on Thu Mar 3 17:34:01 2022
王乾懿 Wang Qianyi (Nan) wrote:
Sorry, not only did my system fail to cough up my desired 9, it couldn't
even manage an 11 :-(
Evil Nigel
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From
=?UTF-8?B?546L5Lm+5oe/IFdhbmcgUWlhb@21:1/5 to
All on Thu Mar 3 17:35:09 2022
在 2022年3月4日星期五 UTC+8 01:34:09，<nigel> 写道：
王乾懿 Wang Qianyi (Nan) wrote:
Sorry, not only did my system fail to cough up my desired 9, it couldn't even manage an 11 :-(
Evil Nigel
Two rounds to go and you still have the chance to win thousands of euro as long as there is a 3 in 11.
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From
=?UTF-8?B?546L5Lm+5oe/IFdhbmcgUWlhb@21:1/5 to
All on Thu Mar 3 17:36:47 2022
Random
55 17 39 41 44 48 31 27 40 57 30
Jerome's Scratchings:
3 47 34 19 23 2 5 8 29 40 10
Nigel
26,47,53,20,50,1,58,46,36,31,8
NN
51, 29, 53, 1, 18, 36, 17, 32, 59, 13, 30
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From
Marmaduke Jinks@21:1/5 to
All on Fri Mar 4 12:06:23 2022
"??? Wang Qianyi (Nan)" <
[email protected]> wrote in message news:
[email protected]...
However, if a player chooses to play 2 sets after round 5, he would still have lost 200 euro.
If we get a 3 in 11 in round 6 playing 2 sets, the overall loss would be: (8*800)-{165*2*(1+1+2+3+5+8)}=6400-6600=-200 euro
Therefore, the best strategy to guarantee the maximum chance of winning is
to play 2 sets of 11 numbers within 5 rounds and stop playing.
Nan
I must confess that when I extract the Jinks selections that I use a random sort on an excel spreadsheet. It's just if something is in the top 11 then
it's down to an excel =RANDBETWEEN(1,100). [Did you notice "if" and "then"
btw ;-) ]
I've no idea what goes on with this Fibonacci thing. So if I don't give you any more selections after round 20 then please use my last set.
:-)
MJ
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From
=?UTF-8?B?546L5Lm+5oe/IFdhbmcgUWlhb@21:1/5 to
All on Sun Mar 6 19:59:34 2022
Random
55 17 39 41 44 48 31 27 40 57 30
0 IN 11
Jerome's Scratchings:
3 47 34 19 23 2 5 8 29* 40 10*
2 IN 11
Nigel
26,47,53,20,50,1,58,46,36,31,8,
0 IN 11
NN
51*, 29*, 53, 1, 18, 36, 17, 32, 59, 13, 30,
2 IN 11
Nan
21* 25 26 27 28 56 11 15 16 17 18
1 IN 11
Alphazero
21* 23 25 26 27 52 11 13 15 16 17
1 IN 11
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