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# Probability Question

Topic closed. 18 replies. Last post 12 years ago by johnph77.

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New Member
Ontario
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 Posted: March 28, 2005, 10:32 am - IP Logged

In a pick 3 lottery, each digit (0-9) has a probability of 1/10 of showing up in a particular position. I would like to know what the probability is for a number to show up? Is it 3/10?

mid-Ohio
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 Posted: March 28, 2005, 11:06 am - IP Logged

Of the 1000 possible combinations in a pick3 lottery there are three digits.  A number can appear 100 times in the first digit,10 times in second digit and 10 times in the third digit, that would be 120/1000.

RJOh

* you don't need to buy more tickets, just buy a winning ticket *

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 Posted: March 28, 2005, 11:17 am - IP Logged

RJOh, Thank you very much for the response! It got me to start thinking about the probabilites in a new light, with a new understanding. Thanks...

mid-Ohio
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 Posted: March 28, 2005, 11:39 am - IP Logged

Correction.

Quote: Originally posted by RJOh on March 28, 2005

Of the 1000 possible combinations in a pick3 lottery there are three digits.  A number can appear 100 times in the first digit,100 times in second digit and 100 times in the third digit, that would be 300/1000.

RJOh

You were right the first time.

* you don't need to buy more tickets, just buy a winning ticket *

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 Posted: March 28, 2005, 1:05 pm - IP Logged

I wrote up this little macro for excel to actually count how many occurances of a particular digit are within the range 000 to 999. The macro tells me that there are 271 numbers that can contain a particular digit. So that would make the probability 271/1000 = 27.1% slightly less then 30%.

[code]

Public Sub countNumbers()

Dim counts(9) As Long

Dim numberCounter As Long

Dim number As String

Dim numberToTest As Long

Const upperLimit As Long = 999

For numberCounter = 0 To upperLimit

number = Format(numberCounter, "00#")

For numberToTest = 0 To 9 'loop through the digits and see if any of them are in the draw number

If InStr(1, number, numberToTest) > 0 Then

counts(numberToTest) = counts(numberToTest) + 1

End If

Next

Next

MsgBox "0 = " & counts(0) & vbNewLine & _

"1 = " & counts(1) & vbNewLine & _

"2 = " & counts(2) & vbNewLine & _

"3 = " & counts(3) & vbNewLine & _

"4 = " & counts(4) & vbNewLine & _

"5 = " & counts(5) & vbNewLine & _

"6 = " & counts(6) & vbNewLine & _

"7 = " & counts(7) & vbNewLine & _

"8 = " & counts(8) & vbNewLine & _

"9 = " & counts(9) & vbNewLine

End Sub

[/code]

mid-Ohio
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 Posted: March 28, 2005, 6:01 pm - IP Logged

10 D=0:CLS:INPUT "NUMBER TO COUNT AND PRINT (0-9) ",N
20 IF N>9 THEN 10
30 FOR A=0 TO 9
40 FOR B=0 TO 9
50 FOR C=0 TO 9
60 C\$= CHR\$(A+48)+CHR\$(B+48)+CHR\$(C+48)
70 IF A=N OR B=N OR C=N THEN D=D+1:PRINT D;". ";C\$
80 NEXT C
90 NEXT B
100 NEXT A

I wrote the above basic program that counts and prints out every combinations that include the number N and it counts 271 also.

RJOh

* you don't need to buy more tickets, just buy a winning ticket *

CA
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 Posted: March 28, 2005, 6:11 pm - IP Logged

Fenris and RJOh are correct - 271. A 9 will appear 300 times in the 3000 digits present in the 1000 possibilities in a Pick 3 draw but due to doubles and the triple will appear only 271 times in all 3-digit possibilities

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.

=^.^=

mid-Ohio
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 Posted: March 28, 2005, 7:53 pm - IP Logged

That makes sense. The number 9 appears 300 times in 271 combinations.

009    019    029    039    049    059    069    079    089    090
091    092    093    094    095    096    097    098    099    109
119    129    139    149    159    169    179    189    190    191
192    193    194    195    196    197    198    199    209    219
229    239    249    259    269    279    289    290    291    292
293    294    295    296    297    298    299    309    319    329
339    349    359    369    379    389    390    391    392    393
394    395    396    397    398    399    409    419    429    439
449    459    469    479    489    490    491    492    493    494
495    496    497    498    499    509    519    529    539    549
559    569    579    589    590    591    592    593    594    595
596    597    598    599    609    619    629    639    649    659
669    679    689    690    691    692    693    694    695    696
697    698    699    709    719    729    739    749    759    769
779    789    790    791    792    793    794    795    796    797
798    799    809    819    829    839    849    859    869    879
889    890    891    892    893    894    895    896    897    898
899    900    901    902    903    904    905    906    907    908
909    910    911    912    913    914    915    916    917    918
919    920    921    922    923    924    925    926    927    928
929    930    931    932    933    934    935    936    937    938
939    940    941    942    943    944    945    946    947    948
949    950    951    952    953    954    955    956    957    958
959    960    961    962    963    964    965    966    967    968
969    970    971    972    973    974    975    976    977    978
979    980    981    982    983    984    985    986    987    988
989    990    991    992    993    994    995    996    997    998
999

RJOh

* you don't need to buy more tickets, just buy a winning ticket *

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Ontario
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 Posted: March 29, 2005, 11:46 am - IP Logged

Thanks guys I really appreciate the help. RJoh, thanks for the confirmation from a different language!

MA
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 Posted: March 31, 2005, 11:21 am - IP Logged

Or, in probability terms,

P(a 9 somewhere) = 1- P(no nines) = 1 - (P(first not 9)*P(second not 9)*P(third not 9)) = 1 - .9*.9*.9 = 1 - .729 = .271

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 Posted: March 31, 2005, 12:09 pm - IP Logged
Quote: Originally posted by Bingo Long on March 31, 2005

Or, in probability terms,

P(a 9 somewhere) = 1- P(no nines) = 1 - (P(first not 9)*P(second not 9)*P(third not 9)) = 1 - .9*.9*.9 = 1 - .729 = .271

Ahh, that makes sense. I was trying to directly calculate the probability.

So, then what would the probability of at least 2 digits (say 9) appearing in a particular drawing be?

p(2 nines) = P(a 9 somewhere)*P(a 9 somewhere) =0.271*0.271 = 0.073441

Also are there any good books or references out there on this type of probability work?
Tx
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 Posted: March 31, 2005, 12:38 pm - IP Logged
Quote: Originally posted by RJOh on March 28, 2005

That makes sense. The number 9 appears 300 times in 271 combinations.

009    019    029    039    049    059    069    079    089    090
091    092    093    094    095    096    097    098    099    109
119    129    139    149    159    169    179    189    190    191
192    193    194    195    196    197    198    199    209    219
229    239    249    259    269    279    289    290    291    292
293    294    295    296    297    298    299    309    319    329
339    349    359    369    379    389    390    391    392    393
394    395    396    397    398    399    409    419    429    439
449    459    469    479    489    490    491    492    493    494
495    496    497    498    499    509    519    529    539    549
559    569    579    589    590    591    592    593    594    595
596    597    598    599    609    619    629    639    649    659
669    679    689    690    691    692    693    694    695    696
697    698    699    709    719    729    739    749    759    769
779    789    790    791    792    793    794    795    796    797
798    799    809    819    829    839    849    859    869    879
889    890    891    892    893    894    895    896    897    898
899    900    901    902    903    904    905    906    907    908
909    910    911    912    913    914    915    916    917    918
919    920    921    922    923    924    925    926    927    928
929    930    931    932    933    934    935    936    937    938
939    940    941    942    943    944    945    946    947    948
949    950    951    952    953    954    955    956    957    958
959    960    961    962    963    964    965    966    967    968
969    970    971    972    973    974    975    976    977    978
979    980    981    982    983    984    985    986    987    988
989    990    991    992    993    994    995    996    997    998
999

RJOh

"Ten measures of beauty descended to the world, nine were taken by Jerusalem."

mid-Ohio
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 Posted: March 31, 2005, 1:34 pm - IP Logged

Quote: Originally posted by LANTERN on March 31, 2005

This means that if you now for sure that the number 9 will be coming out next on any one of the 3 pick 3 positions and you can to play it straight including doubles and triples then you need to buy 271 straight pick 3 combos to win straight........

For pick3 games, Maybells Quikwheel (a wheel program from the old Maybell website) wheels 271 lines for a straight hit if key number hits and only 54 lines for a box hit if the key number hits.

RJOh

* you don't need to buy more tickets, just buy a winning ticket *

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Ontario
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 Posted: March 31, 2005, 3:17 pm - IP Logged

I was thinking more along the lines of this:

If nine hit in the last three draws, what are the odds that it will hit in the next draw. In other words, what is the probability that 9 will hit 4 times in a row?

That would be, .271^4 =0.00539 or 0.539% or 1/185.4

So it is a good chance that 9 won't show up in the next draw. Now if you kept track of all of these digits and did a similar thing, then you may have a good way of reducing the number of potentials. Keep in mind a similar thing can be done with consecutive misses.

Tx
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 Posted: March 31, 2005, 3:33 pm - IP Logged

because of all the undue manipulations by the state lotteries such as predraws and whatever else, fact(s) and expected mathematical probabilities might not corelate too well, you would be surprised as to what is possible when a lot of money is at stake and they have the ability to manipulate the numbers and results in almost anyway that they want to.

"Ten measures of beauty descended to the world, nine were taken by Jerusalem."

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