Neb. lottery numbers drawn twice in rowPrevious TopicNext Topic

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).

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.

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

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

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.