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Neb. lottery numbers drawn twice in row

Topic closed. 35 replies. Last post 8 years ago by savagegoose.

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Chief Bottle Washer
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May 31, 2000
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Posted: January 31, 2009, 9:19 am - IP Logged

I'll go back to what I said earlier.  The odds are 1 in 1 million for two draws in a row to hit one number that you pick in advance.  For example, if I say that the number "821" will come out in NJ today and tommorrow, the odds are 1 in 1 million that it will come true.

 

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    JADELottery's avatar - MeAtWork 03.PNG
    The Quantum Master
    West Concord, MN
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    Posted: January 31, 2009, 10:37 am - IP Logged

    I'll go back to what I said earlier.  The odds are 1 in 1 million for two draws in a row to hit one number that you pick in advance.  For example, if I say that the number "821" will come out in NJ today and tommorrow, the odds are 1 in 1 million that it will come true.

    This not what I am getting Experimentally.

    This comes from actual data taken from the Wisconsin Daily Pick 3 draws.

    Below is a short list of the calculation, however, you can download the Excel file with the following link.

    http://home.mchsi.com/~jadelottery/Pick3NextDrawMatchProbability.xls

    It reaffirms the calculation of (1 current known draw / 1000 next unknown possible draws).

     

    Wisconsin Daily Pick 3 - Taken from the Wisconsin Website http://www.wilottery.com

             

    if current draw matches the next draw, then 1, else 0

    This probability is just by analysis to support the actual calculation value.

     
             

    ¯

    Total Draws - 1 =

     5978
    IndexYearMonthDayDraw 

    Total Matches =

     6
    1199292180     
    219929225840

    Probability Current Draw Matches Next Draw =

     0.00100368
    319929234060

    (Total Matches) / (Total Draws - 1)

     
    41992924500

    or

     or

    519929256360

    Probability Current Draw Matches Next Draw =

     1 / 996.333333333333
    61992926150

    (Total Matches) / (Total Draws - 1)

      
    719929276930   
    819929288920   
    919929298670   
    1019929308600   
    1119921013420   
    1219921022700   
    1319921031200   
    1419921046950   
    151992105500   
    1619921064680   
    1719921072620   
    1819921085130   
    1919921094740   
    20199210102240   
    ......   
    ......   
    ......   
    59552009169370   
    59562009177880   
    59572009184560   
    59582009199490   
    595920091108540   
    596020091118430   
    596120091125760   
    59622009113870   
    596320091141260   
    596420091158950   
    596520091165970   
    596620091179410   
    596720091183810   
    596820091194870   
    596920091206150   
    597020091213250   
    597120091221000   
    597220091235670   
    597320091244560   
    597420091259150   
    597520091268550   
    597620091274900   
    597720091283940   
    597820091299570   
    597920091302480   

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    Jehocifer

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      Posted: February 1, 2009, 4:36 am - IP Logged

      Fair question. It doesn't have to be three days in a row. It's easier to visualize & solve it that way, though. Smile

      Here's my work out, for a 6 drawing set. Now, remember I was answering RJOh's question of the odds of a "threepeat". He didn't specify a certain number.

      Let's call our target number 000. That comes out on the first day.

      On the second day, 000 has 1 in 1000 chance of falling again. Overall odds so far: 1/1000

      On the third day, 000 has a 1 in 1000 chance of falling again. Overall odds so far: 1/1,000,000

      At this point, I'll stop calculating overall odds, for a minute

      On day four (this is where it gets interesting) we want 000 to NOT fall. The odds of that are 999/1000 : slightly less that one

      On day five, we want, again for 000 to NOT fall. The odds, again are 999/1000 : slightly less than one

      (Day 5 could be a repeat of day 4, though)

      On day six, we want to avoid 000, and we may want to avoid whatever fell on days 4 and 5, if 5 is a repeat of day 4. We're looking for a single threepeat, not two of them. Using the worst-case scenario (don't I always? Wink) of a repeat - that means we drop 2 numbers from consideration: 000 and whatever fell on days 4 & 5. Leaving us 998/1000 : slightly less than one. BTW, if day 4 is different than day 5, the odds go back to 999/1000

      So, putting them all together: 1st three days [1/1,000,000] * 2nd three days [999/1000 * 999/1000 * 998/1000]

      Because the second three days calculate to slightly less than one, the total product is reduced from 1 in 1,000,000 (10 in 10 million) to j_u_s_t under that. We can say our target 000 is a symbol (instead of a number), we can calculate the odds of that symbol showing-vs-not showing a certain number of times, given a certain number of trials. Once you look at it that way, you can see it doesn't matter the days the target number falls on.

      My calculation was for any one and only one pick 3 number showing up straight, exactly three times (no more) in a six drawing period.

      Now, if you want a particular number to show up or want certain days, then that is a more complex operation.

      Making it easier to visualize and solve is all well and good, but sometimes it's too easy to simplify part of the answer out of the calculations, and we both did that.

      By calculating the chances of getting the same number on the first 3 days without getting it on any of the other 3, you've excluded the chance that it could happen on other days, and still only happen 3 times. Even if we limit ourselves to 3 consecutive days instead of any 3 days, there are still 4 possible outcomes.  The 3 consecutive days can start on day 1, day 2, day 3 or day 4. That means 3 in a row is about 4 times as likely as you thought. I'd also say you've added an unnecessary restriction to the question. The chances of getting the same number 3 times in a given period also includes the possibility of getting it more than 3 times, so unless the question is for 3, and only 3 times, I wouldn't reduce the chances based on the  (very slim) possibility of more than 3 occurrences. That makes the chances of 3 (or more) occurrences on (any) 3 consecutive days in a 6 day period about 4 in 1 million.

      It also sounds like you think your result would also apply to non-consecutive days, but that's incorrect. Here's another way to simplify it. For any 3 days in 6, it's the same as calculating any other combination of 3 in 6. There are 20 ways for a result to on 3 of 6 days (4 of which are 3 consecutive days):
      123 124 125 126 134 135 136 145 146 156
      234 235 236 245 246 256
      345 346 356
      456
      For all 20 of these, whatever the first day result is, there's a 1 in 1 million chance that the other 2 days will be the same. That gives us a total of 20 chances in 1 million, or 1 in 50,000.

      I made an error similar to yours in my previous post, which was based on 3 repeats in 7 drawings. I simplified by calculating the chances of a number repeating on 2 of the first 6 days, and then happening again on the 7th day. That ignored the chances that the 3rd repeat would also happen during the first 6 days.

        Tenaj's avatar - michellea
        Charlotte NC
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        Posted: February 8, 2009, 6:08 pm - IP Logged

        north carolina has also happan at 24 january 2009 asme no for day and evnining 923

        and they use balls  NC repeats a lot - especially in the last few months

        takeemtothebank

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          Posted: February 13, 2009, 11:39 am - IP Logged

          It happens at least once a year in Maine on the Pick 4 which is even higher odds of happening.

            savagegoose's avatar - ProfilePho
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            Posted: March 8, 2009, 11:33 am - IP Logged

            it depends on what you are calculating,  if they are sayinng any 3 numbers repeating, then the odds are 1 in 1000.

             

            the fact is  you dont care what 3 numbers are drawn makes the 1st draw irellevant, say 111 is drawn , who care is could have 222 or 345.

            that sets  the number you are testing for the second draw, which is 1 in 1000

             

            infact you are stating unity for the test of the 1st draw, the odds of any number being drawn is 1000 in 1000 or 1

             

            so the odds are in fact 1x1000 = 1 in 1000

             

            now if they stated  the odds of number 138 being drawn twice in a row it would be 1 in 1 mill. as you have odds of 1 in 1000 for the 1st drawm then 1 in 1000 for the second 1kx1k  = 1 mill

             

            infact that reminds me many years ago i  started playing the results  form the previous draw on a $1 game. just to try and stick it  lady luck. i might actually start doing that again

            its not that ucommon for numbers to repeat,  for 138 to repeat twice gets a little harder.

             

            but im  just being pedantic.

            i hope i made sense.

            2014 = -1016; 2015= -1409; 2016 JAN = -106; FEB= -81; MAR= -131; APR= - 87: MAY= -91; JUN= -39; JUL=-134; AUG= -124; SEP = -123; OCT= -84  NOV=- 73 TOT= -3498

            keno historic = -2291 ; 2015= -603; 2016= JAN=-32, FEB= +12 , MAR= -86, APR = -77. MAY= -48, JUN= -29, JUL=-71; AUG = -52; SEPT= -43; OCT = +56 NOV = -33 TOT= -3297