quiz pick-4 daily game odds

Reply #48

RL-RANDOMLOGIC — Sat, 30 Jan 2016 10:45:32 GMT

Jade

Thanks, I think I can use this to help extrapolate my estimate. I also think I will add a SC range filter

to my wheel.

RL

Reply #47

JADELottery — Wed, 27 Jan 2016 21:19:27 GMT

Reply #46

JADELottery — Wed, 27 Jan 2016 20:37:10 GMT

Reply #45

RL-RANDOMLOGIC — Wed, 27 Jan 2016 03:43:24 GMT

P.S. My analysis treats the step values as 12 independent variables. The codes are just

a means to break the digits into a series of smaller choices.

RL

Reply #44

RL-RANDOMLOGIC — Wed, 27 Jan 2016 03:24:20 GMT

Jade

I can calculate the odds for each S/C value. I feel that I am not adequately explaining what I am trying to do.

I am trying to offset the odds using my average hit rate as a starting point to calculate an estimate for being

able to hit all the values. This is kind of quasi-math guess at best but since my baseline hits are far above the

expected, I would feel comfortable as long as I know I am not missing something in my calculations. I am not

a mathematician so I can stray o... [ More ]

Reply #43

JADELottery — Tue, 26 Jan 2016 18:18:38 GMT

Reply #42

RL-RANDOMLOGIC — Tue, 26 Jan 2016 04:02:34 GMT

Jade

Based on my test play where the average hit rate is 9 of 12 step values, is it possible that

I can calculate my odds for hitting all 12 by calculating the odds for hitting 3 of 3 steps. In

my test play I managed 3 attempts with 11 out of 12. The other games ranged 8 to 10

the overall average was 9 of 12.

Most of my misses are step-1 values which are always 1 in 3. It seems logical that I could

get a good estimate by calculating the odds for hitting 3 or 4 step-1 values.

Reply #41

RL-RANDOMLOGIC — Tue, 26 Jan 2016 03:01:49 GMT

SergeM

The way the digits are distributed in the groups has no bearing other than that's the way the codes

are developed.

Step-1 has 3 choices.

If set to 1 then we are limited to the digits in group 1 or digits (0-1-2)

If set to 2 then we are limited to the digits in group 2 or digits (3-4-5-6)

If set to 3 then we are limited to the digits in group 3 or digits (7-8-9)

Step-2 has 2 choices 1 = play odd and 2=play even

Step-3 has 2 choices 1=lowest remaining digit... [ More ]

Reply #40

SergeM — Mon, 25 Jan 2016 18:15:59 GMT

Step 1: Group 1 {0, 1, 2}, Group 2 {3, 4, 5, 6}, Group 3 {7, 8, 9}

How about Group 1 {0, 9, 2}, Group 2 {3, 4, 5, 6}, Group 3 {7, 8, 1

Reply #39

SergeM — Mon, 25 Jan 2016 18:12:08 GMT

Why do you guys stick to groups

Reply #38

JADELottery — Mon, 25 Jan 2016 15:17:08 GMT

The percent probability of one group remaining the same is as follows:

Step 1: Group 1 30%, Group 2 40% Group 3 30%

Step 2: Group 1 50%, Group 2 50%

Step 3: Group 1 60%, Group 2 40%

The percent probability of one group changing to any other is as follows:

Step 1: Group 1 70%, Group 2 60%, Group 3 70%

Step 2: Group 1 50%, Group 2 50%

Step 3: Group 1 40%, Group 2 60

Reply #37

JADELottery — Mon, 25 Jan 2016 14:41:58 GMT

Well, we deciphered the steps.

We just wanted you to confirm.

Step 1: Group 1 {0, 1, 2}, Group 2 {3, 4, 5, 6}, Group 3 {7, 8, 9}

Step 2: Group 1 {1, 3, 5, 7, 9}, Group 2 {0, 2, 4, 6, 8}

Step 3: Group 1 {0, 1, 3, 4, 7, 8}, Group 2 {2, 5, 6, 9}

Here's the step chart

Reply #36

RL-RANDOMLOGIC — Mon, 25 Jan 2016 06:52:15 GMT

I need to correct a error

Lets say that I then set step-2 = 2 the pool is reduce to odd digits (3,5)

should read

Lets say that I then set step-2 = 1 the pool is reduce to odd digits (3,5)

Step code = 2-1-1 = digit 3

RL

Reply #35

RL-RANDOMLOGIC — Mon, 25 Jan 2016 02:29:14 GMT

Jade

In step-1 we have 3 choices. Group-1 = (0,1,2) Group-2 = (3,4,5,6) and Group-3 = (7,8,9)

If step-1 is thought to be 2 the the pool is reduce to digits (3,4,5,6)

Lets say that I then set step-2 = 2 the pool is reduce to odd digits (3,5)

Lets say that I then set step-3 to 1 the pool is reduced to the lowest of the two remaining

digits. Step code = 2-2-1 = digit 3

Note! Codes 1-1-2 and 3-2-2 are not used so there are 10 possible outputs, 0 to 9.

The overall odds... [ More ]

Reply #34

JADELottery — Sun, 24 Jan 2016 16:37:05 GMT

Step-1 breaks the digits into 3 groups, 1=(012), 2=(3456) and step-3=(789)

Step-2 is odd or even 1=odd 2=even

Step-3 is 1 = low and 2=high.

What numbers are associated with each group: even, odd, low and high

Reply #33

RL-RANDOMLOGIC — Sun, 24 Jan 2016 13:11:41 GMT

Guys

Thanks for the info. What I am doing now is this. Using my steps program and my S/C chart for analysis

I pick one digit for each of the four positions. I never know which digits are correct until after the drawing

but I have a very high hit rate for at least two of my selections showing in the correct spot. There are 6

positional pairs in a 4 digit line. AB, AC, AD, BC, BD and CD. It takes 486 lines to cover my 4 digit selections

to cover all six possible pairings. If any two... [ More ]

Reply #32

AllenB — Sun, 24 Jan 2016 03:56:42 GMT

The Chart Below show the Number of possible Straight Combinations that are Possible for 0,1,2,3 or 4 Numbers Repeating from the Previous Draw. In the Left Column the number left of the decimal point is the number of Repeat. The number Right of the decimal point is the number of digits that can repeat.

From the chart it is possible to see the number of combinations resulting from Your answer the First Question. How many numbers from the previous game Repeat?

Next you can see the number of

Reply #31

adobea78 — Sun, 24 Jan 2016 02:42:53 GMT

RL, you're on right track. You need to make a choice between the options ' front pair, front triads and any pair, any triad', each has different lines. Good Luck

Front pair 100lines

Any pair filter target pair contained in 10000picks for a straight hit

front triad 10 lines

any triad filter target triad contained in 10000 picks for a straight hit.

Reply #30

RL-RANDOMLOGIC — Sun, 24 Jan 2016 02:04:37 GMT

SergeM

I seem to doing that quite a bit since I started messing around with the daily games.

With the number games, I know what I am doing but the daily games require both

combinations and permutations. I need to think about things more before jumping

in. I am starting to get the feel but still got a ways to go.

RL

Reply #29

SergeM — Sat, 23 Jan 2016 23:37:20 GMT

For the odds it is often easier and faster to generate and count them.

Reply #28

RL-RANDOMLOGIC — Sat, 23 Jan 2016 22:03:38 GMT

P.S.

Playing large is a option as my 2 positional digit hits are better than 1 to 1 per play overall. To cover

a 6-way p-4 pair bet require 486 lines. MO pays $6K for a $1.00 P-4 str8, the only downside is filling

out 486 boards every day or twice a day if mid and evening games are played.

Here are a couple pics of my positional wheel and the prize matches. winning line = 7664

P.S. What I mean by the above hit rate is that I play anywhere from 1 to 5 lines per game and I

o... [ More ]

Reply #27

RL-RANDOMLOGIC — Sat, 23 Jan 2016 21:38:05 GMT

adobea

I did mention straights only just above the odds questions. I am new to the daily games and as such a bit

behind in my math for these games. When I first considered the odds I made estimates without working

them out, turns out, my estimates were way off. I built a program and was surprised to see that the odds

for matching 3 positional digits in a pick-4 game was only 1 in 276.87. Before this I thought I was really doing

well. The wife then brought up the method we use to s... [ More ]

Reply #26

JADELottery — Sat, 23 Jan 2016 15:09:46 GMT

ok, good.

the lottery industry odds are the following:

1 : (10n / (C(n, m) 9(n - m)))

it's just off by 1 and the reciprocal of the probability, p, or 1 / p.

this works for any Daily Game Pick Size lottery: Pick 0, Pick 1, Pick 2, Pick 3, Pick 4, ...

where n 0, m 0 and m n.

Reply #25

adobea78 — Sat, 23 Jan 2016 13:56:14 GMT

Maybe you should rephrase your question:

Math odds quiz. Straight only

How many picks for each position one digit(504 or 1000)

How many picks for a Front Pair(100) or any pair(???)

How many picks for Front Triad(10) or any triad (260??)

Probability, hence odds , is a derivative of the binomial format NPr or NCr. So why not concentrate on this filter for your strategy and cost analysis. If you're very good with posed scenario, then start testing even with the large picks.Often tim... [ More ]

Reply #24

RL-RANDOMLOGIC — Sat, 23 Jan 2016 10:34:15 GMT

Jade

I built a calculator to get my values and the way it works is to first calculate the number of lines

needed to cover the number of matches then it divides, 10,000/36=277.778. In your post above

your odds for a matching-3 positional digits is listed as 1 in 276.778. I figured it was just a typo.

I don't understand the calculation for matching 0 digits as I get 10,000/6,561 = 1 in 1.524.

My calculations to cover each match from 0 to 4

0 = 6,561

1 = 2916

2 = 486

3... [ More ]

Reply #23

JADELottery — Sat, 23 Jan 2016 06:34:33 GMT

The match 0 was written 1 : 0.52 just to be consistent with 1 : least number of failures.

It could have been written 2 : 1.04, or 2 successes for approximately 1.04 failures.

Reply #22

JADELottery — Sat, 23 Jan 2016 06:28:34 GMT

In the odds for matching 4 it is 1 : 9999.

It means for every 1 possible success there are 9999 possible failures.

Reply #21

JADELottery — Sat, 23 Jan 2016 06:20:51 GMT

Those are true odds, the ratio of success to failure, typically reduced to 1 success to failure or 1 : least number of failures.

If p is the probability of success, then the odds become 1 : (1 / p) - 1

Reply #20

RL-RANDOMLOGIC — Sat, 23 Jan 2016 04:51:02 GMT

KY

I missed your post so it looks like you beat jade to the punch. Jade looks to have a typo as it's 277.78

for a 3 positional match not 276.78

RL

Reply #19

RL-RANDOMLOGIC — Sat, 23 Jan 2016 04:42:02 GMT

Jade

Bingo. In another post I started playing and posting a few pick-4 games and my hit rates for

2 positional digits were very high and along the way I also had a few lines with 3 positional

digits. I was sure that I was close to hitting a straight which led me to work out the odds.

Turns out that a 3 digit positional hit is way, way and I do mean way far away from a 4 of 4.

A 3 positional hit is not really worth mention. In 74 lines spread out over around 18 plays I

had 3 ti... [ More ]

Reply #18

JADELottery — Fri, 22 Jan 2016 21:46:10 GMT

Match Probability Odds

0 65.61% 1 : 0.52

1 29.16% 1 : 2.43

2 4.86% 1 : 19.58

3 0.36% 1 : 276.78

4 0.01% 1 : 9999.00

Reply #17

JADELottery — Fri, 22 Jan 2016 21:41:18 GMT

the odds are 1 : (10n / (C(n, m) 9(n - m))) - 1

9(n - m) is 9 raised to (n - m)

superscript not high enough visually.

Reply #16

JADELottery — Fri, 22 Jan 2016 21:17:21 GMT

(C(n, m) 9(n - m)) / 10n

n - pick size

m - match size

C(n, m) = n! / (m! (n - m)!)

n! is the factorial; where n! = n (n - 1) (n - 2) 3 2 1 and 0! = 1

Reply #15

KY Floyd — Fri, 22 Jan 2016 19:41:02 GMT

Not quite.

The first is almost as straightforward as what you did, but you missed one detail. For getting at least 1 of the 4 numbers right you've got a 1 in 10 chance for each of the 4 positions, so 4 * 10% = 40%. To get only 1 of 4 correct you have to get the other 3 wrong. That means you need to multiply the probability of getting at least 1 right by the probability of getting 3 of them wrong. That makes it .4 * .9^3 = .2916, or 29.16%. 1 in 3.43

For getting 2 right it's the same thing,... [ More ]

Reply #14

phileight — Fri, 22 Jan 2016 12:48:25 GMT

What are the odds of matching 1 digit to the correct position with 4 selections? 10% * 4 40%

What are the odds of matching 2 digits to the correct position with 4 selections? 1% * 3 3%

What are the odds of matching 3 digits to the correct position with 4 selections? .01% *2 .02

Reply #13

RL-RANDOMLOGIC — Fri, 22 Jan 2016 12:38:02 GMT

The solution is simple.

RL

Reply #12

RL-RANDOMLOGIC — Fri, 22 Jan 2016 12:24:36 GMT

In around 18 attempts I hit 3ea 3 of 4 digit straights playing less than 5 lines or less with a minimum

of 2 lines.

To evaluate my system I needed to know the odds for hitting 3 correct positional digits out of 4. The

odds for a pick-4 straight are 1 in 10K. My first thought was that I could expect to hit 3 positional out

of 4 around 1 in 1000 plays.

RL

Reply #11

RL-RANDOMLOGIC — Fri, 22 Jan 2016 12:17:05 GMT

Teaser.

The odds for hitting a pick-3 straight are 1 in 1000, so the odds for hitting 3 of 4 in a pick-4 should be less than 1 in 1000.

True or false.

RL

Reply #10

RL-RANDOMLOGIC — Fri, 22 Jan 2016 12:02:37 GMT

Let's say that we start with 4 positions. Each position has 1 in 10 odds for selecting the correct digit.

If we hit 3 digits in the correct positions and miss one then what are the odds of doing so. Remember

that we don't know which 3 are correct until after the drawing. We have 4 possible outcomes.

A+B+C

A+B+D

A+C+D

B+C+D

A______ B______ C______ D_______

Hint. We get 4 choices to hit 3 correct, the problem is further complicated by the fact that any 1 of the 4... [ More ]

Reply #9

MonEl — Fri, 22 Jan 2016 11:23:03 GMT

Even if it was a pick 3 game, getting 1 straight pick 3 number right out of 1000 pick 3 numbers with only 1 straight pick 3 number is very hard to do, if it wasn't hard, Would there still be pick 3 games?

It is not impossible, people have done it even with only 1 straight pick 3 number, but not very often.

Reply #8

MonEl — Fri, 22 Jan 2016 11:03:35 GMT

At my age right now and with as little formal education as I have, doing that took a lot out of me and was very hard, I am not used to thinking too much and mostly now-days, I can't even predict the pick 3 boxed any more as I used to.

I just no longer have the energy and the free time that I need.

It is a good thing that when I want to, I just don't give up, if I did, I would have never broken the boxed pick 3 game as I did so many years ago in so very many ways and How many people have do... [ More ]

Reply #7

MonEl — Fri, 22 Jan 2016 10:52:16 GMT

Lets say that we match 3 of the 4 digits to the correct position, what are the odds for doing this.

Maybe my question would be:

How many Straight pick 3 numbers are there on a pick 4 game?

OXXX

XOXX

XXOX

XXXO

I wonder if that is right?

If it is then:

1000 X 4 = 4000 Straight Pick 3 numbers On A Straight Pick 4 Game.

So maybe to have 1 particular straight pick 3 right on a straight pick 4 = 1/4000

So maybe the odds of 3 straight digits right on 4 straight digits

Reply #6

MonEl — Fri, 22 Jan 2016 10:41:46 GMT

I was wrong, this is way beyond me

Reply #5

dr san — Fri, 22 Jan 2016 10:27:23 GMT

We also have the question of repetition of a draw the last digit to the next

Draw, but not in the same position example

478

823

He repeated the digit 8 but not same position

In pick 4 we also have the twins digits = 00-99

That may be carried forward in pick4 in 6 positions

1,2 1,4 2,3 2,4 3,4 1,3

The other two digits lack seen in the vertical position of each frequency and Statistics

Reply #4

MonEl — Fri, 22 Jan 2016 10:16:13 GMT

Sorry my mistake, this is not possible:

If you have 1 position and 3 digits then the odds of getting 3 right might be 3/10 or 0.3

If you have 1 position you can't get 3 digits right on it.

-------

Forget about this also:

if you have 2 positions and 3 digits to get right maybe 0.3 X 2 = 0.6 as you now have 2 positions and that might be you double the chances (?).:

-------------

Forget about that whole post it is probably all wrong

Reply #3

MonEl — Fri, 22 Jan 2016 09:48:00 GMT

The odds of 1 digit right for 1 position will always be 1/10 or 0.1 for each straight position, but if you have 4 positions and you want the digit right on all 4 positions then maybe 1/40 chance.

If you want the digit right on any-one of the 4 positions 1 time or more times as chance might have it, but at least 1 time then maybe 4/10 or 0.4

------------

2 Digits right on 2 particular straight positions maybe (1/10) / 2 = 0.05

--------------

3 Digits right on 3 particular straight

Reply #2

MonEl — Fri, 22 Jan 2016 09:35:07 GMT

So maybe, each one of the 3 digits might have a 4/10 % chance for all 4 positions, but I am not sure if straight or any-order, but probably Box (Any-Order).

----------------

But because you want all 3 digits right on ANY 4 positions (Boxed?) at the same time maybe:

0.4 /3 = 0.13333333333333333 to infinity for all 3 digits on the 4 positions maybe any-order or box, but I am not sure.

Reply #1

MonEl — Fri, 22 Jan 2016 09:11:15 GMT

I don't know anything whatsoever about Math, but:

If you have 1 position and 10 possible digits the odds of getting 1 particular digit right might be 1/10 or 0.1

If you have 1 position and 3 digits then the odds of getting 3 right might be 3/10 or 0.3

if you have 2 positions and 3 digits to get right maybe 0.3 X 2 = 0.6 as you now have 2 positions and that might be you double the chances (?).

With 3 positions and the same 3 digits now you might have triple the chances, by the way thi... [ More ]

quiz pick-4 daily game odds

RL-RANDOMLOGIC — Fri, 22 Jan 2016 05:00:02 GMT

I started playing pick-4 a while back and while thinking about the odds I came to the conclusion

that it's not as easy as one might think. Lets say that we match 3 of the 4 digits to the correct

position, what are the odds for doing this. One might think matching 3 digits would be 1 in 1000

but this is not true unless you only select 3 of the 4 digits places to play.

Math odds quiz. Straight only

What are the odds of matching 1 digit to the correct position with 4 selections... [ More ]