United States Member #132100 August 26, 2012 1079 Posts Offline

Posted: January 22, 2016, 4:11 am - IP Logged

Quote: Originally posted by RL-RANDOMLOGIC on January 22, 2016

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

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 this turns out to be box or any order and not straight so: 0.3 X 3 = 0.9

With 4 positions maybe: 0.3 X 4 = 1.2 chance or 4 times the chance as with only 1 position box (any-order not straight).

1 position = 30 % chance, 4 positions = 120% Chance, Any-Order or Box and Not Straight?

I am probably wrong! (?).

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

1 Digit and 1 position = 1/10

1 Digits 2 positions = 2/10

1 Digit and 3 positions = 3/10

1 Digit and 4 positions = 4/10

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).

Make it 10 positions and the chance for 1 digit right on any one position of the 10 positions might be 100%

United States Member #132100 August 26, 2012 1079 Posts Offline

Posted: January 22, 2016, 4:35 am - IP Logged

Quote: Originally posted by MonEl on January 22, 2016

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 this turns out to be box or any order and not straight so: 0.3 X 3 = 0.9

With 4 positions maybe: 0.3 X 4 = 1.2 chance or 4 times the chance as with only 1 position box (any-order not straight).

1 position = 30 % chance, 4 positions = 120% Chance, Any-Order or Box and Not Straight?

I am probably wrong! (?).

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

1 Digit and 1 position = 1/10

1 Digits 2 positions = 2/10

1 Digit and 3 positions = 3/10

1 Digit and 4 positions = 4/10

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).

Make it 10 positions and the chance for 1 digit right on any one position of the 10 positions might be 100%

I am not sure about any of that!

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.

United States Member #132100 August 26, 2012 1079 Posts Offline

Posted: January 22, 2016, 4:48 am - IP Logged

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

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

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3 Digits right on 3 particular straight positions maybe (1/10)/3 = 0.0333 to infinity.

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Maybe you really mean box for all 4 positions and not straight as you said.

If so, see my other answer before this one: (Maybe) = 0.1333

United States Member #132100 August 26, 2012 1079 Posts Offline

Posted: January 22, 2016, 5:16 am - IP Logged

Quote: Originally posted by MonEl on January 22, 2016

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 this turns out to be box or any order and not straight so: 0.3 X 3 = 0.9

With 4 positions maybe: 0.3 X 4 = 1.2 chance or 4 times the chance as with only 1 position box (any-order not straight).

1 position = 30 % chance, 4 positions = 120% Chance, Any-Order or Box and Not Straight?

I am probably wrong! (?).

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

1 Digit and 1 position = 1/10

1 Digits 2 positions = 2/10

1 Digit and 3 positions = 3/10

1 Digit and 4 positions = 4/10

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).

Make it 10 positions and the chance for 1 digit right on any one position of the 10 positions might be 100%

I am not sure about any of that!

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!

bgonÃ§alves Brasil Member #92564 June 9, 2010 2126 Posts Offline

Posted: January 22, 2016, 5:27 am - IP Logged

We also havethe question ofrepetition of adrawthe lastdigitto the next Draw,but notin the same positionexample 478 823 He repeatedthe digit8but notsame position Inpick4we also havethe twinsdigits=00-99 Thatmay be carried forwardinpick4in 6positions 1,21,42,32,43,41,3 The other twodigitslackseenin the verticalpositionof eachfrequency andStatistics

United States Member #132100 August 26, 2012 1079 Posts Offline

Posted: January 22, 2016, 6:03 am - IP Logged

Quote: Originally posted by MonEl on January 22, 2016

"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 = 1/4000

Am I right or am I wrong again?

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 done that? Educated or not?

United States Member #132100 August 26, 2012 1079 Posts Offline

Posted: January 22, 2016, 6:23 am - IP Logged

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.

United States Member #59354 March 13, 2008 3985 Posts Offline

Posted: January 22, 2016, 7:24 am - IP Logged

Quote: Originally posted by MonEl on January 22, 2016

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.

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

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