Crunch question

Reply #11

johnph77 — Fri, 30 Apr 2004 08:59:09 GMT

It is probable that these conditions can be made with as little as 525 draws, and possibly 506. Not quite sure of the numbers or formula yet. I'm on the road for the next 3 weeks. I'll do more when I get back - intriguing.....

Reply #10

RJOh — Mon, 26 Apr 2004 19:59:32 GMT

waynr,Once you get the complete list(or file), with most any lottery program, you can check it for match 4 duplications, eliminate them and use the same program to check it with a lottery history file to eliminate additional duplications. RJOh

Reply #9

waynr — Mon, 26 Apr 2004 11:15:57 GMT

Quote: Originally posted by johnph77 on April 26, 2004Another way to attack the problem. What is being dealt with is a 5/37 lottery. If the number in the first position is 13, and 13 only, then there are 4 additional picks to make between 14 and 37, which is the same as a 4/24 draw, with 10,626 possibilities. Eliminating the 21 combinations with 4 consecutive numbers leaves 10,605 possibilities. Good work, hypersoniq.gljohnjohn, it's not consecutives out. It's 4 number repeats. Let me give you a... [ More ]

Reply #8

johnph77 — Mon, 26 Apr 2004 10:50:56 GMT

Another way to attack the problem. What is being dealt with is a 5/37 lottery. If the number in the first position is 13, and 13 only, then there are 4 additional picks to make between 14 and 37, which is the same as a 4/24 draw, with 10,626 possibilities. Eliminating the 21 combinations with 4 consecutive numbers leaves 10,605 possibilities. Good work, hypersoniq.gljohn

Reply #7

hypersoniq — Mon, 26 Apr 2004 02:30:24 GMT

i have it done in excel... it is 10,605 after removing the 4 consecutives...the macro...Option ExplicitDim B As Integer, C As IntegerDim D As Integer, E As IntegerDim N As LongSub List_Comb()Range( A1 ).SelectApplication.ScreenUpdating = FalseN = 0For B = 13 + 1 To 34For C = B + 1 To 35For D = C + 1 To 36For E = D + 1 To 37ActiveCell.Value = 13 - B - C - D - EActiveCell.Offset(1, 0).SelectN = N + 1If N = 65001 ThenN = 0ActiveCell.Offset(-65001, 1).SelectApplication.Screen... [ More ]

Reply #6

waynr — Mon, 26 Apr 2004 00:38:59 GMT

Quote: Originally posted by keystonechas on April 25, 2004The high and low answer is the same,correct?ChasChas, the high, the low and everything in between is what the no-repeats are to be chosen from.hypersoniq, this is going to be very hard in excel.golotto, look at the numbers as mere ball labels. Just think of what lottery programs would look like if number one was white, the highest black and everything in between grayscale.

Reply #5

waynr — Sun, 25 Apr 2004 22:42:29 GMT

Quote: Originally posted by RJOh on April 25, 2004If you are considering Buckeye5, a 5/37 game which has had 2576 drawings since 04/06/93, there have been 331 drawings that included the number 13 and only 13 of those had 13 as the lowest number. If you are only considering combinations starting with 13 then you are only talking about 10,626 combinations in total including the 13 that have already occurred rather than all the 376,992 combinations that contain the number 13.RJOhI'm talking about... [ More ]

Reply #4

golotto — Sun, 25 Apr 2004 22:34:34 GMT

Many may consider number 13 unlucky but I'm certinly not one of them. I've included 13 in virtually every 'my pick' lottery ticket that I've ever purchased. Although this thread is about Pick 5, when speaking of Powerball, I've also included 13 either as one of the white or otherwise as the +red in almost every non quick pick ticket I've ever purchased. Since day one, I've always had the feeling that if I ever do hit one of the nicer PB prizes, 13 will be along for the ride. Thirteen unlucky? No... [ More ]

Reply #3

RJOh — Sun, 25 Apr 2004 21:18:48 GMT

If you are considering Buckeye5, a 5/37 game which has had 2576 drawings since 04/06/93, there have been 331 drawings that included the number 13 and only 13 of those had 13 as the lowest number. If you are only considering combinations starting with 13 then you are only talking about 10,626 combinations in total including the 13 that have already occurred rather than all the 376,992 combinations that contain the number 13.RJOh

Reply #2

keystonechas — Sun, 25 Apr 2004 20:46:18 GMT

The high and low answer is the same,correct?Chas

Reply #1

hypersoniq — Sun, 25 Apr 2004 19:21:16 GMT

159,979 combos with no 4 repeats? I need to check it out in excel quick...

Crunch question

waynr — Sun, 25 Apr 2004 15:40:14 GMT

The number 13 is generally considered unlucky. Suppose some unlucky person wanted to play all their tickets with 13 in the first position in a 5/37 game. The only limits the person wants in their lines are no 4 repeats and max number of lines for more luckiness. Anyone here wants to, give it a shot. Max lines with no 4 number repeats for the range below. How many?Low.......13-14-15-16-17High......13-34-35-36-37