Probability Question

Reply #18

johnph77 — Sat, 02 Apr 2005 06:05:37 GMT

Fenris -Not offhand, and I'm on the road right now. I'll do some research and get back to you when I return next week.glj

Reply #17

fenris — Fri, 01 Apr 2005 13:25:13 GMT

Quote: Originally posted by johnph77 on April 1, 2005 Fenris -The chances of a 9 appearing in any particular draw, whether it has appeared in the immediate previous draw or not, is still 271::1,000.Your math for a 9 not appearing in consecutive draws is correct. And the way you would determine the odds of a 9 not appearing in x number of consecutive draws is to take the chances of a 9 not appearing - 729::1,000, or 0.729 - and determining its result by using your consecutive draw number as an ex... [ More ]

Reply #16

johnph77 — Fri, 01 Apr 2005 05:52:28 GMT

Fenris -The chances of a 9 appearing in any particular draw, whether it has appeared in the immediate previous draw or not, is still 271::1,000.Your math for a 9 not appearing in consecutive draws is correct. And the way you would determine the odds of a 9 not appearing in x number of consecutive draws is to take the chances of a 9 not appearing - 729::1,000, or 0.729 - and determining its result by using your consecutive draw number as an exponent in the same manner you did before - say (0.729

Reply #15

biglooooser — Fri, 01 Apr 2005 03:51:13 GMT

here is my advicedo what the lottery is doing, dont try to outsmart it, dont try to out think yourselfplay your gutPlay a number that wont leave you alone it keeps naggin at ya night and day.you hear it knockin but something is telling you , no way and so you pass over it again and again and again and againthat's the numberwell it is if your me.I always think like this..I try to picture myself after the drawing, I have now seen the winning numbers and I hear myself sayingMan i knew that number... [ More ]

Reply #14

LANTERN — Thu, 31 Mar 2005 20:33:24 GMT

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.

Reply #13

fenris — Thu, 31 Mar 2005 20:17:38 GMT

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 1/185.4So 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 wi... [ More ]

Reply #12

RJOh — Thu, 31 Mar 2005 18:34:26 GMT

Quote: Originally posted by LANTERN on March 31, 2005This 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

Reply #11

LANTERN — Thu, 31 Mar 2005 17:38:01 GMT

Quote: Originally posted by RJOh on March 28, 2005That 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

Reply #10

fenris — Thu, 31 Mar 2005 17:09:58 GMT

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 = .271Ahh, 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 = 0.073441Also are there any good books or references out ther... [ More ]

Reply #9

Bingo Long — Thu, 31 Mar 2005 16:21:11 GMT

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

Reply #8

fenris — Tue, 29 Mar 2005 16:46:06 GMT

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

Reply #7

RJOh — Tue, 29 Mar 2005 00:53:06 GMT

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

Reply #6

johnph77 — Mon, 28 Mar 2005 23:11:44 GMT

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 possibilitiesglj

Reply #5

RJOh — Mon, 28 Mar 2005 23:01:41 GMT

10 D=0:CLS:INPUT NUMBER TO COUNT AND PRINT (0-9) ,N20 IF N>9 THEN 1030 FOR A=0 TO 940 FOR B=0 TO 950 FOR C=0 TO 960 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 C90 NEXT B100 NEXT AI wrote the above basic program that counts and prints out every combinations that include the number N and it counts 271 also. RJOh

Reply #4

fenris — Mon, 28 Mar 2005 18:05:10 GMT

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 LongDim numberCounter As LongDim number As StringDim numberToTest As LongConst upperLimit As Long = 999For numberCounter = 0 To upperLimit number = Format(numb... [ More ]

Reply #3

RJOh — Mon, 28 Mar 2005 16:39:51 GMT

Correction.Quote: Originally posted by RJOh on March 28, 2005Of 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.

Reply #2

fenris — Mon, 28 Mar 2005 16:17:39 GMT

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...

Reply #1

RJOh — Mon, 28 Mar 2005 16:06:21 GMT

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

Probability Question

fenris — Mon, 28 Mar 2005 15:32:25 GMT

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