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Pick 3 Paradox? Question of probability

Topic closed. 12 replies. Last post 11 years ago by Thoth.

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Findlay, Ohio
United States
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May 28, 2004
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 Posted: November 10, 2005, 1:12 pm - IP Logged

The standard (mechanical) way that a Pick 3 digit is selected is to draw one ball from a machine that contains a pool of ten balls.  This is done 3 separate times (there are 3 machines that each contain ten balls) to give us our winning Pick 3 combination.  Since there are ten balls in each machine we can calculate the total possible combinations by multiplying each of the machines ten balls by the other machines ten balls and so on.  When this is done we get 10x10x10, which equals 1000 possible combinations.  Additionally, we can easily see that the odds of getting a single particular digit out of a particular machine is 1 in 10.  For example, the odds of a 5 being drawn out of the first machine is 1 in 10 or the odds off getting a 5 out of the second machine is 1 in 10 and so on.

How about the odds of correctly guessing a digit from the first machine AND correctly guessing a digit from the second machine?  Say we were betting on 4 and 7 as the first two Pick 3 numbers (front pair).  The odds for the 4 to be drawn from the first machine would be 1 in 10 and the odds for the 7 being drawn from the second machine would also be 1 in 10.  The overall odds of correctly guessing those first two digits would be 1 in 100 because 1/10 times 1/10 equals 1/100.  Most everyone who plays Pick 3 knows the above situation and knows that any digit has a 1 in 10 chance of appearing in a selected position.

Now suppose that televised Pick 3 drawings were held out of the viewers watching eyes.  By this I mean that each machine is covered with a black cloth and viewers cannot see the action inside of each machine.  Now the viewer can hear the machines running and knows that the person performing the drawing is actually operating the machines and that balls are in fact being drawn, but no one can actually see what numbers were selected until the host reveals it to the audience by removing the black cloths one machine at a time.

For the first drawing the host reveals the first digit and it’s a 5, he then reveals the second digit and it is also a 5. What are the odds that the third machine will reveal a 5 once the cloth is removed?  Since we know that there are ten balls in that last machine and only one of those ten balls is a 5 we assume the odds to be 1 in 10.

In another drawing a similar situation is played out, but only this time the host reveals the third machine first and it’s a 2. He next reveals the first machine and it also drew a 2.  Now its time to reveal the second or middle machine and mathematically you know that there is supposed to be a 10% chance that it will also show a 2.

In yet another drawing the host reveals the third machine first.  It produced a 7.  He then reveals the second machine and it too produced a 7.  Lastly, the host grabs the cloth off of the first machine.  What are the odds that the first (position one) machine will also have produced a digit 7?  The odds should be 1 in 10 or 10% - or should they????

Imagine that every single time that your state draws a combination that contains AT LEAST 2 repeated digits that they are revealed in the above manner.  The machines and ball pools are separate.  If any two particular machines produced the same digit (i.e. 00, 11, 22, 33, etc…) the last machine revealed should in fact still have a 1 in ten or 10% chance of also producing the necessary digit for a Triple.

In thinking the above way you would statistically expect to have one Triple Digit combination (all three balls the same digit) drawn for every ten times a Double Digit Pair is drawn. Mathematically this would seem to make sense, however it just can’t be possible.  In 1000 combinations there are 270 doubles and only 10 triples.  In a statistically perfect run of 10,000 games you should have something like 2800 combinations that contain at least 2 repeated digits and 280 of those would be triples if each Double Digit Pair had at a 10% chance of becoming a Triple Digit combination.

In the real world, a statistically perfect run of 10,000 Pick 3 games would in fact have 2800 3 digit combinations that contain AT LEAST 2 repeated digits, but 2700 will be Double Digit combos and only 100 will be Triples.

Stop for one second at look at the following combinations.  The X represents an unknown digit or separate machine that contains 10 balls with the digits 0-9 in it.

11X,  1X1,  22X,  X22,  7X7,  99X,  4X4,  55X,  X00

Since each “X” has a 10% chance of being the digit needed to make all 3 numbers the same, you would think that we would see Triples more often.

The question I have for anyone who knows the answer is this:  What is the overall chance that Double will become a Triple and how is it calculated?

I know that each Double Digit pair can appear within 28 different straight combinations and that only one of those 28 is a Triple (leaving 27 doubles).  Also, the 27 straight Double Digit combos also represent 9 Boxed Double Digit combos (1 Triple is 10% when adding the 9 boxed for a total of 10).

Are the odds of a Double becoming a Triple 1 in 28 or a 10% chance of a 27% chance? (there are 270 doubles in the Pick 3 = 27%.)  Some of my data seems to support the later – i.e. .027 or 2.7%.

~Probability=Odds in Motion~

New Jersey
United States
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June 28, 2005
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 Posted: November 10, 2005, 6:36 pm - IP Logged
Cash 3 Facts
============
Cash 3 uses number from 0 to 9 (10 digits) and has a possible of 1,000
combinations.

These numbers can be combined as follows
Combo Type       Sample #s   Occurrences
-----------------------------------------------
3 single digits   123        71%
Double digits       112        27%
Triple digits       111         1.37%

The above numbers are facts not conjecture, guessing, Dream books,
tic-tac-toe or anything else that is not based upon Factual draws.

So from looking at the combination percentages you should readily see
that there are certain combinations that you should totally avoid.

That means NO Triples! Eliminating these combos represent a total of
1.37%!

So concentrate on the more frequent combos... 98.63%: Single or double
digit combinations!

This also means that if your lottery has not had a double digit number
in 5 or more draws you should lean heavily in that direction.

Cash 4 Facts
============
Cash 4 uses number from 0 to 9 (10 digits) and has a possible of 10,000
combinations.

These numbers can be combined as follows
Combo Type           Sample #s     Occurrences
-----------------------------------------------
4 single digits       1234       50%
Double digits        1123       43%
Triple digits        1112       3%
Double Double digits 1122       2%

The above numbers are numerical facts just check with your local
lottery office to verify these results, they are freely available.

So from looking at the combination of percentages you should readily
see that there are certain combinations that you should totally avoid.

That means NO QUADS or Double double combinations... and probably no
triples. Eliminating these combos represent a total of 5.12%! Therefore,
using these number combinations you could win 5 days out of 100
providing you had the correct combination.

So concentrate on the more frequent combos... 94.88%: Single or double
digit combinations!

Findlay, Ohio
United States
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May 28, 2004
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 Posted: November 10, 2005, 8:30 pm - IP Logged

Thanks Raven,

I know that the 720 NoMatch #'s = 72% and that the 270 Doubles = 27% and the ten Triples = 1% of the total 1000 combos.  Any of the three number types will statistically perform closely to their odds: in 10,000 games there will be AROUND 7200 NoMatches, 2700 Doubles, and 100 triples.  Frequency can never change the odds, but only verify it.

Im not trying to guess the probabilty based on frequency or statistics, I am trying to verify the odds (or should I say find the odds/probability) of a somewhat hypothetical situation.  The situation is as follows;

If you knew that AT LEAST two of the 3 digits to be drawn were going to be the same (i.e. two 5's or two 4's etc...) what is the probabilty that the third digit will be the same?

After playing with the idea at work for a few hours I think I have found my answer.  There is a 2.71% chance that a double will be a Triple.

~Probability=Odds in Motion~

Manitoba
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July 11, 2004
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 Posted: November 12, 2005, 2:14 am - IP Logged

I'd say if you live in Western Canada,the odds are pretty good. The last three draws were

593, 953, 953. And here's me blaming it on our greedy government rather than mathematical odds.

Stone Mountain*Georgia
United States
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November 2, 2002
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 Posted: November 24, 2005, 10:11 am - IP Logged

Thanks Raven,

I know that the 720 NoMatch #'s = 72% and that the 270 Doubles = 27% and the ten Triples = 1% of the total 1000 combos.  Any of the three number types will statistically perform closely to their odds: in 10,000 games there will be AROUND 7200 NoMatches, 2700 Doubles, and 100 triples.  Frequency can never change the odds, but only verify it.

Im not trying to guess the probabilty based on frequency or statistics, I am trying to verify the odds (or should I say find the odds/probability) of a somewhat hypothetical situation.  The situation is as follows;

If you knew that AT LEAST two of the 3 digits to be drawn were going to be the same (i.e. two 5's or two 4's etc...) what is the probabilty that the third digit will be the same?

After playing with the idea at work for a few hours I think I have found my answer.  There is a 2.71% chance that a double will be a Triple.

Would like to know how the 2.71 chance works.  Thanks for this post this is very interesting. I have always wondered about this and it has perplexed me for years.

The only real failure .....is the failure to try.

Luck is a very rare thing....... Odds not so much.

Odds never change .....but probability does.

Win d

United States
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June 28, 2005
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 Posted: November 24, 2005, 10:32 am - IP Logged

GOOD SHOW !!

Good luck to everyone!!!

Stone Mountain*Georgia
United States
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November 2, 2002
10491 Posts
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 Posted: November 24, 2005, 10:43 am - IP Logged

Although... strange as it might seem... when getting the actual results back for this over the period of a year the expected ratio between these doubles and trips is almost right on the money.

Looking at doubles for the U.S. the ratio of doubles and their particular trips was right at the expected rate 10%.

11 doubles  213 ..... trips 111 ......26

22 double  185.....trips  222 .......13

33 double  191 ....trips 333.......25

44              180......trips 444.......17

55             172                          14

etc............ etc..............

Total...941 doubles.....  guess what? ......95 trips for the yr.

That's as close to that expected 10% as they come

Now is it just a perception thing that we don't get that doubles to trips ratio?

After further research ....I do think that it is a perception thing and not a paradox problem at all. The perception problem comes out of the total number of double hits .....and using this figure against the Trips result totals. For example:

Only one third of the doubles hits should be used in the ratio. Using only the 1/3rd. of the doubles that begin in first and second position will always give us the correct expected ratio of around 10%

So ..... after all these years its not a paradox problem at all ....it was a perception puzzle  ....at least until now.

The only real failure .....is the failure to try.

Luck is a very rare thing....... Odds not so much.

Odds never change .....but probability does.

Win d

Stone Mountain*Georgia
United States
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November 2, 2002
10491 Posts
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 Posted: November 24, 2005, 11:27 am - IP Logged

Some really good FYI that might be of further interest .....

There were 5,410 doubles last year.

A little over 1/3 of those were the expected Front loaded Doubles ......or doubles starting out in Positions 1 and 2.  1,952 Front loaded doubles to be exact.

One would expect of the 1,952 doubles that we would get approx 195 Trips.

We actually got a few more than that .....we got an additional 21 trips. For a grand total of216 trips for the year. Very gratifying really.

It's actually.. 11 %  ........very pretty.

We can all thank Todd for the services that are available to us with the Premium Memberships that make this sort of research possible. Remarkable.

The only real failure .....is the failure to try.

Luck is a very rare thing....... Odds not so much.

Odds never change .....but probability does.

Win d

Findlay, Ohio
United States
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May 28, 2004
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 Posted: November 27, 2005, 7:13 pm - IP Logged

Good to see someone caught on - Indeed it is all about perception!

A front loaded double does in fact have a 10% chance of creating a triple.  In just the the same way a back loaded double and a split loaded double also have the same 10% chance, but when and only when they are looked at individually.  The problem arises in my so called paradox because of how the numbers were perceived or observed.  All the double types (front - back - split) do have the individual 10% chance of creating a triple when looked at, tracked, or counted seperate from one another.

In my hypothetical "unviewed drawings"  I imagined that the cloths were first removed from the double digit pairs, leaving only the single non-repeating digit within the combination to be guessed at.  In my three sample drawings the host revealed the front pair in the first game, the split pair in the second game, and the back pair in the third game.

Now suppose the combination was a in fact a triple......how would the host have known which two cloths to remove first?  Which pair (front back or split) would have actually had the 10% chance of creating the triple?  There would probably be no way to correctly answer this.

The perception problem comes with the fact that for all three of the drawings we were seeing two repeated digits and and asking what the chance was that the third unknow digit would be the same.

The ideal way to look at the problem should have been to first ask how many pick 3 combinations have AT LEAST 2 repeated digits somwhere within the combination?  Then next step should have been to ask how many of those total combinations had 3 repeated digits?

There are 270 doubles and 10 triples that meet that criteria to answer the first question, giving us a total of 280 combinations with AT LEAST two repeated digits.  Knowing that only 10 of those 280 are triples answers the second.  So with that right there we know that there is only a 10/280 chance that any double could be a triple. 10/280 = 1/28 or .03571 or 3.571%.

If we was goin by a specific doble digit (i.e. 55 in any 2 positions) then we would ask how many combos have AT LEAST two 5's in them.  The answer would be 28 (27 doubles and 1 triple).  Once again that again gives us our 1/28 or .03571 or 3.571%.

I had stated in an earlier post that "There is a 2.71% chance that a double will be a Triple.".....THAT WAS INCORRECT

I now know that there is a 1/28 or 3.57% chance that a double will be a Triple.

Two help verify that, we once again should imagine a statistically perfect run of 10,000 games.  We would have 2700 doubles and 100 triples that when combined will total 2800 combinations that had AT LEAST 2 repeated digits in them.  So if 3.571% is correct than the 2800 combinations with AT LEAST 2 repeated digits should yield 2800 X 3.571% = 99.988 (or 100 rounded up) combinations with three repeated digits (trips).  And its easy to see that it does.

Wow I made a long spiel out of that! lol   The reason i posed the question was because i needed to know the odds of a double becoming a triple so I could calculate the 50% probabilty median for "doubles between triples".

~Probability=Odds in Motion~

Tennessee
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 Posted: December 1, 2005, 11:13 pm - IP Logged

Thoth,

You posted this info about 6-way numbers:

"50% of the time a NoMatch (6 way boxed number) will hit within 116 draws of its last draw.
75% of the time it will hit within 231 draws of its last draw.
And....99.00% of the time a nomatch will hit within 765 games of its last draw."

Do you have the same type of data for triples?

CA
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December 10, 2003
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 Posted: December 2, 2005, 4:52 am - IP Logged

The information is misleading in a way. There are 270 combinations with two numbers the same and only ten triples. But the premise at the beginning was that you already had the same two numbers and were drawing a third.

For instance, if you were picking the numbers in consecutive order (Position 1, Position 2 and Position 3), if the same two numbers were drawn for Positions 1 and 2, then there is a 1 in 10 possibility for a triple.

The 270 double combinations include those combinations with the same numbers in the first and second positions, the first and third positions and the second and third positions. So you might not have the same two numbers for the first and second positions but duplicate one of those numbers in the third position.

gl

j

Blessed Saint Leibowitz, keep 'em dreamin' down there.....

Next week's convention for Psychics and Prognosticators has been cancelled due to unforeseen circumstances.

=^.^=

Findlay, Ohio
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 Posted: December 5, 2005, 9:14 pm - IP Logged

Thoth,

You posted this info about 6-way numbers:

"50% of the time a NoMatch (6 way boxed number) will hit within 116 draws of its last draw.
75% of the time it will hit within 231 draws of its last draw.
And....99.00% of the time a nomatch will hit within 765 games of its last draw."

Do you have the same type of data for triples?

The Triples, as a whole, make up 10/1000 or 1% of the Total combinations.  This reduces to 1/100.

When a ANY Triple is drawn, there is a 50% chance that you will see another triple drawn WITHIN the next 69 games.

75% of the time, a triple will be drawn WITHIN 138 games of the last triple hit.

99% of the time a triple will hit WITHIN 459 games of the last triple hit.

The above info is NOT for any specific Triple, it treats all 10 Triples as a group.

If your tracking a specific Triple, such as 999, then there is a 50% chance that it could be drawn again sometime WITHIN the next 693 drawings of its last hit.

75% is within 1386 games of its last hit and 99.00% is within 4603 games of its last hit.

Hope this helps,

Scott

~Probability=Odds in Motion~

Findlay, Ohio
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 Posted: December 5, 2005, 11:14 pm - IP Logged

The information is misleading in a way. There are 270 combinations with two numbers the same and only ten triples. But the premise at the beginning was that you already had the same two numbers and were drawing a third.

For instance, if you were picking the numbers in consecutive order (Position 1, Position 2 and Position 3), if the same two numbers were drawn for Positions 1 and 2, then there is a 1 in 10 possibility for a triple.

The 270 double combinations include those combinations with the same numbers in the first and second positions, the first and third positions and the second and third positions. So you might not have the same two numbers for the first and second positions but duplicate one of those numbers in the third position.

gl

j

It's not that I was trying to make the info misleading.....It was about observation.

In the hypothetical situation, the balls were drawn under a cover and then two repeated digits were always revealed first (regardless of position) -  leaving the last unknown digit to be guessed at and what were the odds that it would be the same.

The whole point was, that when it is observed in this way, that it is impossible to give the last digit revealed a true 10% chance of being a duplicate of the first two digits revealed - which is what most would assume the odds to be.  When looked at this way the odds would have to be 1 in 28 or 3.57%.

In the same sense, I think it is wrong to give a front loaded double a 10% chance of being a triple without applying the same chances to the back and split loaded doubles.  Everything should be equal (odds) in every angle as far as math is concerened.  It is probably not mathematically correct to only apply the odd to JUST the front loaded doubles and not the others.

Again, in a statistically perfect run of 10,000 games there would be 2700 doubles, 900 of those would be front loaded doubles, 900 would be back loaded doubles and 900 would be split loaded doubles.  Giving each type its own 10% chance to create a double would lead us to a total of 270 Triples having been drawn when we know that there should be (and is) only be 100 Triples for the 10,000 games.

And so another very intriguing thought arrises; If you did give ONLY the front loaded doubles the 10% chance to create a Triple, then in 10,000 games there would be 900 front loaded doubles with 10% or 90 of those being Triples. !!! ONLY 90 Triples for 10,000 games???  We need 100 to be statistacally and mathematically correct.

The only way to make the math work out is to say that in 10,000 games, there would be 2800 numbers that contained AT LEAST 2 repeated digits.  1 in every 28 would be a triple, giving you a total of 100 triples and leaving the rest to be the 2700 doubles that you should have for 10,000 games.

....which bye the way, is now showing me that i am also wrong.  The only way my odds of 1 in 28 or 3.57% for a double to become a triple works out, is if you take 1/3 of the triples (33) and add them to each of the 900 double groups.....

933 x 1/28 (.03571428571) = 33.3214 triples produced for each doubles group

33.3214 X 3 = 99.96 Total triples.

The better odd for a true double to be a triple would be 1/27 or .037037...3.7%

900 x 3.7% =  33.33 Triples Per Double Type.....X 3 Types = 99.99 Total Triples

~Probability=Odds in Motion~

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