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# Pick 5 Filter Program

Topic closed. 6 replies. Last post 9 years ago by RJOh.

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United States
Member #31636
February 1, 2006
256 Posts
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 Posted: December 23, 2007, 11:48 pm - IP Logged

I wish I had a program for the pick 5 like Paurth's freewheelpick4. In a 5/39 game, I would like to see how many combinations there are for 1 or 2 numbers.

Example:
There are 575,757 combinations in a 5/39 game.
1-2-3-4-5 = 1 combination
1-2-3-4-x = 35 combinations
1-2-3-x-x = ?
1-2-x-x-x = ?
1-x-x-x-x = ?

So if I played one number, there won't be 575,757 combinations.
Also a way to filter out consecutive numbers. Also all odd, even.

So in probability, some combinations are not likey to ever be drawn.

mid-Ohio
United States
Member #9
March 24, 2001
19828 Posts
Online
 Posted: December 24, 2007, 12:24 am - IP Logged

I wish I had a program for the pick 5 like Paurth's freewheelpick4. In a 5/39 game, I would like to see how many combinations there are for 1 or 2 numbers.

Example:
There are 575,757 combinations in a 5/39 game.
1-2-3-4-5 = 1 combination
1-2-3-4-x = 35 combinations
1-2-3-x-x = ?
1-2-x-x-x = ?
1-x-x-x-x = ?

So if I played one number, there won't be 575,757 combinations.
Also a way to filter out consecutive numbers. Also all odd, even.

So in probability, some combinations are not likey to ever be drawn.

An odds calculator may be all you need.

1-2-3-4-5 = 5/5 = 1 possible combination
1-2-3-4-X = 1/35(39-4) = 35 possible combinations
1-2-3-X-X = 2/36(39-3) = 630 possible combinations
1-2-X-X-X = 3/37(39-2) = 7,770 possible combinations
1-X-X-X-X = 4/38(39-1) = 73,815 possible combinations
X-X-X-X-X = 5/39(39-0) = 575,757 possible combinations
All even numbers = 5/19 = 11,628 possible combinations
All odd numbers = 5/20 = 15,504 possible combinations

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

United States
Member #31636
February 1, 2006
256 Posts
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 Posted: December 24, 2007, 12:51 am - IP Logged

Let's say we took this for example.
1-X-X-X-X = 4/38(39-1) = 73,815 possible combinations.

Is there any way to filter it down. Filter out all odd and even.
Filter out 4 odd, 1 even. 4 even, 1 odd. Filter out consecutive 3 and 4 numbers.

Thanks.

mid-Ohio
United States
Member #9
March 24, 2001
19828 Posts
Online
 Posted: December 24, 2007, 2:05 am - IP Logged

1-X-X-X-X = 4/38(39-1) = 73,815 possible combinations

Remove all even/odd numbers would remove 19 numbers = 4/19 = 3,876 possible combinations

Remove 1 odd + 4 even would remove 5 numbers = 4/33 = 40,920 possible combinations

Removing other groups and figuring the possible combinations requires a little more thought.

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

United States
Member #31636
February 1, 2006
256 Posts
Offline
 Posted: December 24, 2007, 3:43 am - IP Logged

This looks really good or maybe I'm missing the boat.

"Remove all even/odd numbers would remove 19 numbers = 4/19 = 3,876 possible combinations"

So if I played one number, removing all even/odd numbers, that leaves
3,876 possible combinations. Therefore it looks like I should win the
jackpot about every time playing 3,876 possible combinations in a 5/39 game.
I would invest that amount to win \$100,000.

I think I must have missed the boat somewhere.

United States
Member #13130
March 30, 2005
2171 Posts
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 Posted: December 24, 2007, 3:55 am - IP Logged

3,876 is the number of combinations removed from the 575,757, holding one key number steady and dropping all-odds or all-evens, in the other four. With these lotto style games you're going to need to be either lucky with your number selection, or good at math.

In neo-conned Amerika, bank robs you.
Alcohol, Tobacco, and Firearms should be the name of a convenience store, not a govnoment agency.

mid-Ohio
United States
Member #9
March 24, 2001
19828 Posts
Online
 Posted: December 24, 2007, 4:39 am - IP Logged

This looks really good or maybe I'm missing the boat.

"Remove all even/odd numbers would remove 19 numbers = 4/19 = 3,876 possible combinations"

So if I played one number, removing all even/odd numbers, that leaves
3,876 possible combinations. Therefore it looks like I should win the
jackpot about every time playing 3,876 possible combinations in a 5/39 game.
I would invest that amount to win \$100,000.

I think I must have missed the boat somewhere.

You are missing the boat if you think covering 3,876 of the 575,757 possible combinations would guaranteed you a win every time.  You would only win if the winning combination had a number 01 and four even or four odd numbers depending on which ones you covered.

Ohio Rolling Cash5 is a 5/39 game that has had 1037 drawings so far.  Only 131 of those drawings had a combination with the number one in it. Only in eleven of those combinations were the other four numbers all even and in seven the other four numbers were all odd.

If you had spent \$4,019,412 to play the 3,876 combinations with the number one and four even numbers for every drawing, you would have won or shared eleven jackpots of \$100K-\$350K and won smaller prizes worth about \$55K.

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

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