Welcome Guest
You last visited January 17, 2017, 9:54 am
All times shown are
Eastern Time (GMT-5:00)

# quiz pick-4 daily game odds

Topic closed. 48 replies. Last post 12 months ago by RL-RANDOMLOGIC.

 Page 2 of 4
NY
United States
Member #23835
October 16, 2005
3502 Posts
Offline
 Posted: January 22, 2016, 2:41 pm - IP Logged

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%

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, but there 6 ways to get it right, not 3: AB, AC, AD, BC, BD, CD. So getting at least 2 right is .1*.1 *6 = .06. Once again you have to get the others (2 this time)wrong, which is .9*.9 = .81. Thus .81*.06 = .0486, or 4.86%. 1 in 20.58

To get at least 3 numbers is .1^3 = .001 (that's .1%, not .01%), and there are 4 ways to have 3 right (ABC, ABD, ACD, BCD). That gives us 4*.001 = .004. You've only got to get one number wrong, so .9*.004 = .0036, or .36%.  1 in 277.78

That last one is really easy to check against the actual numbers. If the winning number is 1234 there are 10 ways you could have 123?, but one of the 10 would give you all 4. That leaves 9 ways to have the 123 correct and the 4 wrong. The same is true for each of the other numbers. There are 10 ways to have 12?4, and one of them is 1234. Thus 36 of the 10,000 possibilities would give you 3 of the 4, so 36/10,000 = .0036

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: January 22, 2016, 4:17 pm - IP Logged

(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

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: January 22, 2016, 4:41 pm - IP Logged

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.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: January 22, 2016, 4:46 pm - IP Logged
 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

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

United States
Member #59354
March 13, 2008
4054 Posts
Offline
 Posted: January 22, 2016, 11:42 pm - IP Logged

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 tickets with 3 positional digits.  Average 74/3=24.67.  This is over 10x better than the

odds would suggest but close to a straight, not at all.  I guess if my attempts were consistant

then I could expect a hit on around 1000 lines but that's just a guess-ta-ment.

RL

P.S.  I had hoped you would hold off a little longer before posting.  LOL

Working on my Ph.D.  "University of hard Knocks"

I will consider the opinion that my winnings are a product of chance if you are willing to consider

they are not.  Many great discoveries come while searching for something else

Trump / 2016 & 2020

United States
Member #59354
March 13, 2008
4054 Posts
Offline
 Posted: January 22, 2016, 11:51 pm - IP Logged

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

Working on my Ph.D.  "University of hard Knocks"

I will consider the opinion that my winnings are a product of chance if you are willing to consider

they are not.  Many great discoveries come while searching for something else

Trump / 2016 & 2020

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: January 23, 2016, 1:20 am - IP Logged

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

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

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: January 23, 2016, 1:28 am - IP Logged

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

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

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: January 23, 2016, 1:34 am - IP Logged

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.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

United States
Member #59354
March 13, 2008
4054 Posts
Offline
 Posted: January 23, 2016, 5:34 am - IP Logged

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 = 36

4 = 1

OK, it just hit me, thanks for the correction.

RL

Working on my Ph.D.  "University of hard Knocks"

I will consider the opinion that my winnings are a product of chance if you are willing to consider

they are not.  Many great discoveries come while searching for something else

Trump / 2016 & 2020

United States
Member #116344
September 8, 2011
3938 Posts
Offline
 Posted: January 23, 2016, 8:56 am - IP Logged

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?

What are the odds of matching 2 digits to the correct position with 4 selections?

What are the odds of matching 3 digits to the correct position with 4 selections?

RL

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(???)

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 times, large picks is 'put off' for  some folks(scared of cost), but if 504 picks can earn me 5k within 5 draws, then the consistency should be tested. Note that the number of picks  can be reduced by the size of  N.

Good Luck

The Quantum Master
West Concord, MN
United States
Member #21
December 7, 2001
3680 Posts
Offline
 Posted: January 23, 2016, 10:09 am - IP Logged

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 = 36

4 = 1

OK, it just hit me, thanks for the correction.

RL

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.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

United States
Member #59354
March 13, 2008
4054 Posts
Offline
 Posted: January 23, 2016, 4:38 pm - IP Logged

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 select out digits.  I am still trying to come up with the

probabilities / odds for the steps.

We use a Same/change chart and my steps program to select digits.  There are 3 steps for each of the 4 digit

positions for a total of 12 values that have to be set.  We are averaging 9 to 11 of the 12 steps game to game

but I am having a hard time figuring out the odds.  The steps values selected are converted to digits so IMHO

I should get a better sense of how I am doing by calculating the odds for the steps.  Hitting 11 out of 12 steps

seems to be doing better than the odds for a 3 positional hit.  The steps program converts the digits into a

3 digit code.

The three digit code is a follows.

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.

Step codes

121=digit 0

111=digit 1

122=digit 2

211=digit 3

221=digit 4

212=digit 5

222=digit 6

311=digit 7

321=digit 8

312=digit 9

codes 112 and 321 are not used.

Step-1 sets the group to use. Step-2 says it will be odd or even and step-3 takes the lower or highest remaining value.

There are still 10 total choices for each digit selected so the overall odds are are the same but I find it's much easier to

select the correct digit by analyzing the codes 1 step at a time.

Here is another problem I am working to solve.  What are the odds for hitting 9, 10 11 etc of the 12 step values needed

in a pick-4 game?  This is straights only because I am working on the digits by position.  If I hit 11 of the 12 I will only

match 3 positional digits but steps wise I had 91.6% correct.  I want to calculate the expected cost and number of plays

to hit a straight estimated on my test plays.

I also want to figure out is how much advantage if any I am gaining from using the steps vs picking the digits from a pool

of 0 to 9.  Pick-4 is a hit-miss game, a 3 of 4 don't count.

Test play data.

I have played 18 games and 74 lines total.  My average for 2 positional digit hits is better than 1 to 1 overall and in the

18 games played I have had 3ea lines with 3 positional hits hitting 11 of the 12 steps.   My average for the 3 positional

hits is 1 in 24.7.

From this data I should be able to come up with a good estimate of how long I can expect to play and the cost of winning

a p-4 straight.

Here is a scan of one of the 11 of 12 step plays.  You can see that I missed the 4th digit by one step value.  Based on

the steps this play came closer than a 3 of 4 hit.  In the 4 position I managed to hit step-1 with odds of 1 in 3 and step-2

with odds of 1 in 2.  If looking only at the digits, I hit 3 of 4 for 75% but steps wise I hit 91.6%   What are the odds for

hitting 9, 10 11 etc.. of the 12 steps?  Remember that step-1 has odds of 1 in 3 and step-2 and step-3 are 1 in 2.

Working on my Ph.D.  "University of hard Knocks"

I will consider the opinion that my winnings are a product of chance if you are willing to consider

they are not.  Many great discoveries come while searching for something else

Trump / 2016 & 2020

United States
Member #59354
March 13, 2008
4054 Posts
Offline
 Posted: January 23, 2016, 5:03 pm - IP Logged

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

often manage more than one line with 2 positional digit hits and in a few games every line had

two positional digits.

RL

Working on my Ph.D.  "University of hard Knocks"

I will consider the opinion that my winnings are a product of chance if you are willing to consider

they are not.  Many great discoveries come while searching for something else

Trump / 2016 & 2020

Economy class
Belgium
Member #123700
February 27, 2012
4035 Posts
Offline
 Posted: January 23, 2016, 6:37 pm - IP Logged

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

 Page 2 of 4