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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
New Jersey
United States
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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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The Quantum Master
West Concord, MN
United States
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December 7, 2001
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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 Index Year Month Day Draw Total Matches = 6 1 1992 9 21 80 2 1992 9 22 584 0 Probability Current Draw Matches Next Draw = 0.00100368 3 1992 9 23 406 0 (Total Matches) / (Total Draws - 1) 4 1992 9 24 50 0 or or 5 1992 9 25 636 0 Probability Current Draw Matches Next Draw = 1 / 996.333333333333 6 1992 9 26 15 0 (Total Matches) / (Total Draws - 1) 7 1992 9 27 693 0 8 1992 9 28 892 0 9 1992 9 29 867 0 10 1992 9 30 860 0 11 1992 10 1 342 0 12 1992 10 2 270 0 13 1992 10 3 120 0 14 1992 10 4 695 0 15 1992 10 5 50 0 16 1992 10 6 468 0 17 1992 10 7 262 0 18 1992 10 8 513 0 19 1992 10 9 474 0 20 1992 10 10 224 0 . . . . . . . . . . . . . . . . . . 5955 2009 1 6 937 0 5956 2009 1 7 788 0 5957 2009 1 8 456 0 5958 2009 1 9 949 0 5959 2009 1 10 854 0 5960 2009 1 11 843 0 5961 2009 1 12 576 0 5962 2009 1 13 87 0 5963 2009 1 14 126 0 5964 2009 1 15 895 0 5965 2009 1 16 597 0 5966 2009 1 17 941 0 5967 2009 1 18 381 0 5968 2009 1 19 487 0 5969 2009 1 20 615 0 5970 2009 1 21 325 0 5971 2009 1 22 100 0 5972 2009 1 23 567 0 5973 2009 1 24 456 0 5974 2009 1 25 915 0 5975 2009 1 26 855 0 5976 2009 1 27 490 0 5977 2009 1 28 394 0 5978 2009 1 29 957 0 5979 2009 1 30 248 0

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Jehocifer

NY
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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.

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? ) 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.

Charlotte NC
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June 18, 2005
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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

New Member

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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.

adelaide sa
Australia
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April 11, 2006
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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

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