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Lotto Architect v2.2

Topic closed. 86 replies. Last post 13 years ago by lottoarchitect.

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Greece
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November 18, 2003
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 Posted: April 19, 2004, 11:54 am - IP Logged

Hi everyone, my questi

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mid-Ohio
United States
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March 24, 2001
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 Posted: April 19, 2004, 2:16 pm - IP Logged

I recently added a feature to my program that allows me to compare a drawing with the previous drawings (any amount) to see if the hit frequency(hot or cold) can be an indicator of which ones to wheel.  For example when looking at the previous 20 draws for a 5/37 game(Buckeye5), numbers that have hit 2-3 times repeat most. It also shows that numbers that  have hit 2-3 times is usually a list of 18-25 numbers and contain 4 of the winning numbers about 5% of the time.  With PowerBall (5/53) the choice is even more confusing.

RJOh

* you don't need to buy more tickets, just buy a winning ticket *

Dump Water Florida
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June 5, 2002
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 Posted: April 19, 2004, 2:21 pm - IP Logged

Just apply probability you'll get the same answer or the programs did even worse.  BobP

Greece
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November 18, 2003
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 Posted: April 19, 2004, 4:46 pm - IP Logged

So, RJOh you say that you pick 18-25 numbers and in 20 draws, your chance to have 4 correct numbers is 5% in a 5 of 37 game?

BobP, all those programs do worse than probability?

For example, if i pick X numbers in a game 5 of 37, then the probability is:

Picked>      10            15            20

0 match : 18.5%        6%          1.4%

1 match : 40.3%      25.2%      10.9%

2 match : 30.2%      37.1%      29.6%

3 match :  9.7%      24.1%      35.6%

4 match :  1.3%        6.9%      18.9%

5 match : 0.05%      0.7%        3.5%

So, if we pick 20 numbers for a game 5 of 37, our chances are 18.9% to have 4 numbers in our selection. RJOh, do you perform worse than luck? It looks strange you get only 5% for 4 match even if you use 18 numbers (chance 13.3% to match 4).

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mid-Ohio
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March 24, 2001
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 Posted: April 19, 2004, 6:19 pm - IP Logged

Apagogeas,

I was surprised at those results also because I was looking for a way to pick a few numbers to wheel and willing to accept losing if there was a good chance of a match5 in the near future.  I was figuring that any strategy that would work in the future, had probably worked in the past.

According to your figures, if I pick 20 numbers the group is more likely to have 3 numbers(35.6%) than 2 numbers (29.6%) than 1 number (10.9%).  I would have thought it would be just the opposite.  I know that figures don't lie, but I have to question those figures. The figures I posted was what I observed, I would have thought that calculated figures would have been closer to them..

RJOh

* you don't need to buy more tickets, just buy a winning ticket *

Greece
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November 18, 2003
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 Posted: April 19, 2004, 7:02 pm - IP Logged

There is a simple equation to determine the chance to match x numbers. I'll describe it to you. It works only for fields that don't allow same numbers to be drawn (no Pick-3/4 games).

Suppose you have a game a of b (eg 5 of 45, a=5 b=45). You want to pick n numbers (eg 10 numbers, n=10) and you want to find the chance to have m matches (eg 3 matches, m=3) in your selection.

Then, chance %= [ nCm * (b-m)C(a-m) / bCa ] * 100%

where nCk =n!/ (k!*(n-k)!)

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mid-Ohio
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March 24, 2001
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 Posted: April 19, 2004, 8:06 pm - IP Logged

I was going to use the equation to calculate the chance% or a match4 in a 5/37 game with 20 numbers,
a=5
b=37
n=20
m=4
but you didn't define n,C or k.  nCk looks like the combination formula.

RJOh

* you don't need to buy more tickets, just buy a winning ticket *

Greece
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 Posted: April 19, 2004, 8:26 pm - IP Logged

nCk is the combination formula

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Greece
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November 18, 2003
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 Posted: April 19, 2004, 8:35 pm - IP Logged

There is a mistake in the formula

game 5 of 45 -> a=5, b=45

picked numbers -> n

match -> m

Then, chance %= [ nCm * (b-n)C(a-m) / bCa ] * 100%

If you have something to do, at least do it well...

mid-Ohio
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March 24, 2001
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 Posted: April 20, 2004, 2:13 am - IP Logged

5/20 = 15504 possible combinations of 5's - 4845 combinations of 4's
5/37 = 435897 possible combinations of 5's - 66045 combinations of 4's
15504/435897 = 3.6%  and  4845/66045 = 7.3%
My guess is the correct solution is between those two figures.

* you don't need to buy more tickets, just buy a winning ticket *

Greece
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 Posted: April 20, 2004, 10:55 am - IP Logged

The division you apply is not valid because you apply it in different sets of numbers. This is why you get wrong results. I'll explain the above equation just to make it clear why it works.

The parameters are as above a,b,n,m

We have the equation %=[ nCm * (b-n)C(a-m) / bCa ] * 100%

Now, bCa = the total tickets available for our game.

nCm = In how many different ways we can have m numbers in a set of n numbers.

Now, for each of the above nCm ways, we can arange the remaining a-m numbers (to form a complete ticket) in (b-n)C(a-m) possible ways. Thus, all the tickets that have m matches in n numbers, is the multiplication of the above 2 sets, nCm * (b-n)C(a-m). This result shows all tickets that can have m matches in a set of n picketd numbers.

Finally we divide by bCa to get a rate from 0 to 1 and multiply with 100% to get the percentage.

If you have something to do, at least do it well...

Greece
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 Posted: April 22, 2004, 2:20 pm - IP Logged

Ouaou! Ok, no more ads! I boug

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Greece
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November 18, 2003
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 Posted: April 26, 2004, 7:40 pm - IP Logged

It seems there is a problem

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Dump Water Florida
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 Posted: April 27, 2004, 3:06 am - IP Logged

The odds of any set of twenty numbers containing the prize winning numbers in a 5/37 game are . . .

5 in 20: 1 in 28.12
4 in 20: 1 in 5.29
3 in 20: 1 in 2.81
2 in 20: 1 in 3.37 (positive)
1 in 20: 1 in 9.16 (positive)

I don't do math, but I have good software and I know how to use it. ;-) BobP

Greece
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November 18, 2003
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 Posted: April 27, 2004, 4:51 am - IP Logged

These rates are correct BobP,

Actually, you say it in other words, because

1 in 20 : 1 in 9.16=1/9.16=10.9%

2 in 20 : 1 in 3.37=1/3.37=29.7%

etc

If you have something to do, at least do it well...

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