What number is too high to play for a first ball? (mm/pb)

Reply #23

sandnan — Fri, 04 Apr 2014 12:39:39 GMT

Jade where you're going? won the lottery

Reply #22

JADELottery — Tue, 25 Mar 2014 15:56:23 GMT

- Correction -

We said, The percent probability is just the e -(x / m ) portion of the formula.

The correct probability is (1 / (2.45550799441)2 ) e -(x / 2.45550799441 ).

The (1 / (2.45550799441)2 ) propotion is required.

The proportion can also be expressed as (1917710 / 780983)2.

To get percentge, multiply by 100%.

Reply #21

JADELottery — Mon, 24 Mar 2014 20:08:32 GMT

Next, if we sum the frequencies in the first column for the range 01 to 12 we get:

1150626 + 1088430 + 1028790 + 971635 + 916895 + 864501 + 814385 + 766480 + 720720 + 677040 + 635376 + 595665

or

10230543

Now, this tells the frequency when that range hits, however, it's when it does not hit when the consecutive run ends.

So, we need the frequency of when that range does not hit; we simply subtract it from the total possible outcomes of 17259390 to get the correct frequency we need... [ More ]

Reply #20

JADELottery — Mon, 24 Mar 2014 19:29:15 GMT

Ok, to start we need to determine the total distribution of numbers by column.

We've also posted another topic many draws ago,

Combinatorial Distribution

In it there is a formula for determining any Pick r of n column distribution:

Combinatorial Distribution

Factorial - n! = n * (n -1) * (n - 2) * ... * 3 * 2 * 1 , and 0! = 1

Permutation - P(n,r) = n! / (n - r)!

Combination - C(n,r) = P(n,r) / r!

Combinatorial Distribution - D(n,r,c,z) = C(z - 1, c - 1) * C(n - z... [ More ]

Reply #19

JADELottery — Sun, 23 Mar 2014 19:16:15 GMT

It follows this thread we posted many draws ago.

Discharging Reoccurrence Distribution

To find the distribution you have to determine what it is by what it is not.

Reply #18

GoogilyMoogily — Sat, 22 Mar 2014 19:48:05 GMT

Formula please:)

I will state most emphatically that pseudo-random number generators are absolute junk.

Reply #17

JADELottery — Sat, 22 Mar 2014 15:07:41 GMT

2 last night (2014-03-21).

We ran a 10,000 draw simulation to see what kind of continuation this run of selecting in the range of 01 to 12 might look like.

A count of 1 in the left column means it selected in the range of 01 to 12 and the next draw did not; essentially the selection cycle starts again.

This last cycle makes it a count of 8 and the next cycle is 9 for 2014-03-25 draw.

Consecutive Run or # of Selections Drawn in Range 01 to 12 Percent of 10,000 Draw Sample Observ... [ More ]

Reply #16

LottoBoner — Fri, 21 Mar 2014 16:02:38 GMT

I think 74 might be too high too.

Reply #15

SergeM — Fri, 21 Mar 2014 15:01:57 GMT

What number is too high to play for a first ball? (mm/pb)

76 is too high too.

Reply #14

SkyLine69 — Thu, 20 Mar 2014 12:09:31 GMT

I'm surprised Jo one eon the prize. Ball 75 would be too high. Lol. Sorry guys had to say it

Reply #13

GoogilyMoogily — Wed, 19 Mar 2014 23:51:23 GMT

Statements like 1 to 5 cannot win are nonsense. I agree. The odds of 1-2-3-4-5 being drawn are the exact same odds as any other combination. My chart ignores the order they are drawn, just the numerical order they are reported.

Reply #12

JADELottery — Wed, 19 Mar 2014 14:26:25 GMT

11 last night.

13 or greater is more likely now and will continue to be more and more each time it's not selected.

just a guess.

Reply #11

SergeM — Wed, 19 Mar 2014 10:32:24 GMT

What number is too high to play for a first ball? (mm/pb)

MM 1 2 3 4 5

1 6.667 no no no no

2 6.306 0.360 no no no

3 5.961 0.691 0.015 no no

4 5.630 0.993 0.043 0.000 no

5 5.312 1.269 0.084 0.002 0.000

...

Number 4 is too low as it will never be drawn as fifth ball. I posted that one longer ago? Any other number can be drawn in a winning combination as fifth ball. Statements like 1 to 5 cannot win are nonsense. Look for developer in the members list, she

Reply #10

GoogilyMoogily — Tue, 18 Mar 2014 23:20:40 GMT

I believe that's what my chart illustrates:) (hopefully) If you ask me, How many times does the number 23 fall as the second digit in MM? , I would reply, It falls 2.817% of the time. Each of my chart's columns add up to 100%.

Reply #9

JADELottery — Tue, 18 Mar 2014 23:17:23 GMT

no.

...for every possible combination of 5 out of 75.

Reply #8

GoogilyMoogily — Tue, 18 Mar 2014 23:11:24 GMT

Are you saying =average(b2:b76) ? Because that would equal 1.408. That falls between 24 and 25.

Reply #7

JADELottery — Tue, 18 Mar 2014 23:06:26 GMT

looks correct for that approach.

However, the split point we are giving you is derived a different way.

It's the average of all the numbers in the first column for every possible combination of 5 out of 75.

Reply #6

GoogilyMoogily — Tue, 18 Mar 2014 22:49:31 GMT

Explain like I'm 5. My spreadsheet has the game title MM in cell A1. Cells B1 thru F1 are 1 thru 5. Cells A2 - A76 are 1 thru 75. The formula in Cell B2 = =(COMBIN($A2-1,B$1-1)*COMBIN(LARGE($A$2:$A$76,1)-$A2,LARGE($B$1:$F$1,1)-B$1))/COMBIN(75,5)*100

Is that formula correct?

Edit: Please note the formula is set up so i can copy/paste it over and over and it doesn't matter the range of numbers. If the lottery system was a 1-80, for example, the large() portion of the formula would pick... [ More ]

Reply #5

GoogilyMoogily — Tue, 18 Mar 2014 22:44:12 GMT

@jade, here is my powerball percent chart. The other table format fell apart on copy/pasta, so this may also look equally bad. I figure 1 thru 7 = 48.087% of the draws for the first number.

PB 1 2 3 4 5

1 8.475 no no no no

2 7.890 0.584 no no no

3 7.336 1.107 0.031 no no

4 6.812 1.572 0.089 0.001 no

5 6.317 1.982 0.172 0.004 0.000

6 5.849 2.340 0.275 0.011 0.000

7 5.408 2.649 0.397 0.021 0.000

8 4.992 2.912 0.535 0.036 0.001

9 4.600 3.132 0.685 0.... [ More ]

Reply #4

JADELottery — Tue, 18 Mar 2014 22:43:26 GMT

No, it's like a coin flip, Heads (01 to 12) or Tails (13 to 71).

The Mega Millions first column split point is actually 76/6 or 12.666... so your percentage sums are not going add up to 50% exactly.

Reply #3

GoogilyMoogily — Tue, 18 Mar 2014 22:36:45 GMT

The sum of the MM numbers 1 thru 9 in the first column = 48.220% (6.667+6.306+5.961+5.630+5.312+5.009+4.719+4.441+4.176)

Is 50% the cut off? Anything outside of this 50% range, I should not consider?

Edits for accuracy and removing unnecessary copy/pasta of table.

Reply #2

JADELottery — Tue, 18 Mar 2014 22:36:42 GMT

Power Ball's first column half split is about (01 to 10) and (11 to 55).

We say about, because the split point is 10, exactly.

This gives the range 01 to 10 a slight advantage.

Reply #1

JADELottery — Tue, 18 Mar 2014 22:23:43 GMT

This is an indirect answer to your question for the Mega millions.

On average, Half of numbers selected in the first column are 01 to 12.

The other Half are 13 to 71, on Average.

If we look at the total draws to date (2014-03-18), we can see numbers in the first column were selected 6 times in the range of 01 to 12.

Our best guess is the next selection in the first column is going to be a number 13 or greater.

Just a guess.

*first column numbers 01 to 12

MM DD YYYY 1... [ More ]

What number is too high to play for a first ball? (mm/pb)

GoogilyMoogily — Tue, 18 Mar 2014 05:02:11 GMT

Likewise, what number is too low to play for a last ball? And the 2/3/4 balls in between? I am assuming numbers are in numerical order, as reported after the draw, not in the order they are drawn. I'm trying to determine the maximum range of allowable balls in the system I am working on.

Here are the percentages of how often a ball will land in each position, for the megamillions:

MM 1 2 3 4 5

1 6.667 no no no no

2 6.306 0.360 no no no

3 5.961 0.691 0.015 no no

4 5.630... [ More ]