Neb. lottery numbers drawn twice in rowPrevious TopicNext Topic

interesting

on one hand I see that ANY 2 draws have 1,000,000 possible 2-sets because it's 1,000x1,000

However, asking for odds of ANY TWIN of previous draw does seem that you would divide 1,000,000 by 1,000 possible sets of twins ...and it equlas 1:1,000.  (So do the odds of having a SPECIFIC TWIN equal 1:1,000 because of having 1,000 possible twins, but it must be a day of twins first) .   I am still somewhat confused 

A lot has to do with precisely the way the question is asked.  The question then has to be transformed to math precisely.

You are right and this should be true for all lotteries.

The average rate of reoccurrence for a draw to repeat itself the next night is once every 1,000 draws.

That's once every 2.738 years.

It's an average only, meaning it's more likely to happen sooner than later.

You can find the distribution of these rates of reoccurrence using a topic we posted a while ago, Discharging Reoccurrence Distribution and a related topic, Random Number Transforms - Reoccurrence Distribution.

Pick-3 is like a slot machine with three ten number reels.

1:10 x 1:10 x 1:10 = 1:1,000

Think of two slot machines with three ten number reels sitting side by side having the same three number combination come up in the same order.

1:1,000 x 1:1,000 = 1:1,000,000

However, once the first machine has been pulled and we know that combination the odds of the second machine matching it are 1:1,000.

1 x 1:1,000 = 1:1,000

At least that's what I think.  BobP


 

jr-va said  "A lot has to do with precisely the way the question is asked," which is a good point. I'd say that most people get the chances of the same number coming up twice in a row wrong because they're answering the wrong question.

The chance that any specific 3 digit number will come up in one drawing is 1 in 1000, so the chance of a specific number coming up in two consecutive drawings is 1/1000 * 1/1000, or 1 in 1 million. Since having a drawing means that some number will be drawn, the chance of a number being drawn on the first day are 1 in 1. The chances that  the same number will repeat in the next drawing is therefore 1 in 1000. Even then we're making an assumption that we mean two specific consecutive drawings. The chances of any specific number coming up in the next 7 drawings is about 7 times more likely. That makes the chance of a specific number coming up consecutively in 2 of 7 drawings about 7 in 1,000,000, or 1 in 167,000.

I interpret RJ's question to mean what are the chances that some number would be drawn 3 times in 7 drawings (leaving out the question of whether there are  1 or 2 drawings per day). In order for that to happen, one of the numbers drawn in the first 6 drawings will have to be drawn twice. It could be any of the 5 numbers, and it could happen on any 2 of the 6 days. The chances of that are about 1 in 66.7.  Here's how that works:

In drawing 1 a number is drawn, and there are 5 chances for it to repeat by drawing 6, or 5 in 1000.
In drawing 2 a number is drawn, and there are 4 chances for it to repeat by drawing 6, or 4 in 1000.
In drawing 3 a number is drawn, and there are 3 chances for it to repeat by drawing 6, or 3 in 1000.
In drawing 4 a number is drawn, and there are 2 chances for it to repeat by drawing 6, or 2 in 1000.
In drawing 5 a number is drawn, and there's a 1 in 1000 chance it will repeat in drawing 6.

That means the chances that one of the numbers drawn in the first 5 days will be drawn on day 6 is about 15 in 1000, 1 in 66.67. There's then a 1 in 1000 chance of  the 6th number being drawn again in the 7th drawing,  or 1 in 66,667.

One final caveat. Those are the chances for one specific state. With 30 states holding drawings, it's 30 times as likely that any given result will occur in 1 of the states.

The odds of one particular drawing to the next matching is 1 in 1000. The odds of it matching one drawing to the next "at some point sometime" is almost certain! The odds of *the next two drawings* matching a given drawing i.e. the same Pick-3 3 times STR, is 1 in 1 million. In the slot machine analogy (sorta) maybe they were thinking the odds of 2 specific draws in a row matching a *certain* STR # chosen arbitrarily.

I'd love to have access to what kind of algorithm is used to run a computerized  drawing. People would be less leery of them if independent (hobbyist etc.) programmers could check the software for bugs, and there is no reason (if they are truly random and no way to predict) why this information can't be made public. Even to know what source is used (quartz crystal? radioactive decay?) to generate the random seeds. I imagine the QP generator in the terminals is different than the one HQ uses.

They draw 6 times per week. I got 9.96 in 10 million chance of drawing the same number exactly 3 times straight, out of 6 drawings. Just under 1 in 1 million.

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

When you say "the same number exactly 3 times straight" do oyu mean 3 days in a row?  If the chances of having the same number come up 3 times in a row during a 3 day period is 10.00 in 10 million, wouldn't it be substantially more likely to happen 3 days in a row during a 6 day period? After all, there's only one way it can happen in 3 days, vs 4 ways it could happen over 6 days. For any 3 days in a 6 day period it would be about 1 in 100,000.

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.

It depends how you look at the odds. [semantics] 

If you say 'What are the chances that 000 will hit tonight and tomorrow night?', That's 1:1,000,000 (1000*1000).

If you say 'Since 000 *already* hit last night, what are the chances that 000 hits tonight?', That's 1:1000.

If you say 'Since 000 *already* hit last night, what are the chances it will hit for the next 2 days [making 3 in a row]?', That's 1:1,000,000 (1000*1000 since the previous night has no bearing since it has already occurred.

If you say 'What are the chances 000 will hit 3 nights in a row (before it hits the first time)?', That's 1:1,000,000,000 (1000*1000*1000).

Yes it is not truly a ball drop so I am not that amazed I mean it happened here in Florida with 3-5-4 last week...to answer your question... they somehow figure that the formula of "independent events" applies, whereas it really does not because here is why: 

    If I ROLL A pair of dice, the odds of both the dice being rolled landing "6" is .... 36:1 because the formula is...the odds are calculated as 1/6 x 1/6 = 1/36 (SIX Faces on a  die)

The reason why this does NOT apply to the Nebraska lottery drawing is because there was no "set" amount of possible results .... think about it....  when you roll the dice you KNOW there are only 6 Faces...so had this been a BALL DROP then yes the odds of that occurrence would have been ...

1/1000 x 1/1000 = 1:1,000,000

Okay I was thinking about it (And I know I am responding to my own post :D) ... i GUESS STATISTICALLY IT IS 1 in A million... but the mysterious part is that this is a "computer" so we dont know what operations are at work ... you know what i mean??

I like this. Very clear and easy to understand for someone who is math challenged like me.

In FL 354 was drawn on 01/19 on the 20th 765 was drawn so there was no repeat of the same numbers!!

Alright, ONE MORE TIME... the probability that the same exact combination will hit the next draw is 1/1000...  PERIOD.

NOT 1/1000000...  IT'S 1/1000.

As soon as the draw is made, there are no probabilities to calculate...  it's just 1, because it's a KNOWN combination.

Probabilities that calculate numbers like 1/2, 1/10, 1/100 or 1/1000 are for the UNKNOWN.

Hence, because the current draw is KNOWN and next nights draw is UNKNOWN, the probability is 1 x 1/1000 or 1 / 1000.

Once you've established the current draw the probability is 1.

When you calculate the next draw is 1/1000 or (1 for the current draw / 1000 for all the possible next draws).

GD...  you people are way over thinking this one.

Stop racking your brain, you'll smash the pea.