Neb. lottery numbers drawn twice in row

Jan 23, 2009, 10:27 am (35 comments)

Nebraska Lottery

OMAHA, Neb. — The odds are against something this odd.

A Nebraska Lottery official says the winning numbers for the state's Pick 3 lottery on Tuesday were exactly the same as the winning combination from the night before.

Lottery spokesman Brian Rockey says one of two lottery computers that randomly generate combinations picked the numbers 1, 9 and 6 — in that order — for Monday night's drawing. He says the other computer picked the same three numbers Tuesday in the same sequence.

The odds of such an occurrence? One in a million.

Rockey says one person won the game's top prize of $600 Monday but didn't win again the next night. Three other people won Tuesday's prize.

Thanks for jackpotismine for the tip.

AP

Comments

diamondpalace's avatardiamondpalace

It's not odd at all when dealing with computer generated combinations. :P

I wonder how it is calculated to be 1 in a million?

jgdr

florida did the same thing with number 354 a couple of days ago

spy153's avatarspy153

Happens all the time in KY.

Todd's avatarTodd

I personally think when computerized drawings are involved, you can't use regular mathematical odds.

It really depends on what kind of RNG the computer is using.

With computers that use the decay of radioactive material as their basis for the random numbers, then I feel like it is sufficiently random, the odds are probably a good measure.  (Although nothing makes me feel that computers should be used for ANY lottery drawing — they are too susceptible to hacking or errors, thereby bypassing the RNG.)

But for other RNGs, such as using the RNG that comes packaged in the operating system/programming language, that is NOT sufficiently random.  In fact, it will produce patterns and someone who knows how the RNG works may be able to successfully predict it.

So one in a million?  If Nebraska uses a regular RNG, then the odds are not going to be mathematically precise, they are going to be whatever the RNG's pattern dictates.

JADELottery's avatarJADELottery

It's not 1 in a million. The probability of the previous night's draw mathing exactly the next night's draw is 1 in 1000. Exactly the same if someone plays one single straight play for that night's draw.

spy153's avatarspy153

What kind of RNG does LP use on the qp's, Todd?

Todd's avatarTodd

Quote: Originally posted by JADELottery on Jan 23, 2009

It's not 1 in a million. The probability of the previous night's draw mathing exactly the next night's draw is 1 in 1000. Exactly the same if someone plays one single straight play for that night's draw.

That makes sense, from the standpoint that in reality it has happened many times in the past. 

I think it would be 1 in 1 million if you chose one particular combination in advance, and that combination hit on two consecutive days.

Todd's avatarTodd

Quote: Originally posted by spy153 on Jan 23, 2009

What kind of RNG does LP use on the qp's, Todd?

The Quick Picks generator available by clicking the link on the left uses a regular old RNG.  I do make an effort to use constantly-changing seed values, but it is still a regular RNG.

I am in the process of changing that to a much better, cryptographically-advanced RNG, and I'm halfway finished.

In fact, if you use the iPhone/iPod Touch version of Lottery Post today, the quick picks generator in that site uses the new RNG.  The Quick Picks generator in the main LP site will eventually look and work just like the iPhone version, which is also easier to use.

JADELottery's avatarJADELottery

Quote: Originally posted by Todd on Jan 23, 2009

That makes sense, from the standpoint that in reality it has happened many times in the past. 

I think it would be 1 in 1 million if you chose one particular combination in advance, and that combination hit on two consecutive days.

That is correct. If the same exact pick 3 number hit 3 times in a row, that would be 1 in a 1,000,000.

Raven62's avatarRaven62

Same Combination 3 Times in a Row would be more like 1 in a 1,000,000,000.

JADELottery's avatarJADELottery

Quote: Originally posted by Raven62 on Jan 23, 2009

Same Combination 3 Times in a Row would be more like 1 in a 1,000,000,000.

no, that is not correct. it's 1 in a 1,000,000.

RJOh's avatarRJOh

Quote: Originally posted by JADELottery on Jan 23, 2009

no, that is not correct. it's 1 in a 1,000,000.

What would be the odds of the same number coming up three times in one week for example on Monday, Wednesday and Friday?

JADELottery's avatarJADELottery

Quote: Originally posted by RJOh on Jan 23, 2009

What would be the odds of the same number coming up three times in one week for example on Monday, Wednesday and Friday?

That's for any straight three draws in row.

Since all possible straight repeating drawn three times in a row is as follows:

 1st  -  2nd  - 3rd
0 0 0 - 0 0 0 - 0 0 0
0 0 1 - 0 0 1 - 0 0 1
0 0 2 - 0 0 2 - 0 0 2
.
.
.
9 9 7 - 9 9 7 - 9 9 7
9 9 8 - 9 9 8 - 9 9 8
9 9 9 - 9 9 9 - 9 9 9

There are only 1,000 possible ways this can happen.

The total possible ways for any straight is as follows:

  1st  -  2nd  -  3rd
0 0 0 - 0 0 0 - 0 0 0
0 0 0 - 0 0 0 - 0 0 1
0 0 0 - 0 0 0 - 0 0 2
.
.
.
9 9 9 - 9 9 9 - 9 9 7
9 9 9 - 9 9 9 - 9 9 8
9 9 9 - 9 9 9 - 9 9 9

This is a total of 1,000,000,000 possible ways this can happen.

This is 1,000 out of 1,000,000,000 or 1,000 / 1,000,000,000 total possible ways any straight can occur three draws in a row.

This reduces down to 1 / 1,000,000 or 1 in 1,000,000.

chouquounette

Happens in Ga all the time!

jr-va

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  Confused

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

JADELottery's avatarJADELottery

Quote: Originally posted by chouquounette on Jan 23, 2009

Happens in Ga all the time!

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.

BobP's avatarBobP

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


 

KY Floyd's avatarKY Floyd

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.

LckyLary

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.

time*treat's avatartime*treat

Quote: Originally posted by RJOh on Jan 23, 2009

What would be the odds of the same number coming up three times in one week for example on Monday, Wednesday and Friday?

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.

sonvan's avatarsonvan

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

KY Floyd's avatarKY Floyd

Quote: Originally posted by time*treat on Jan 26, 2009

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.

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.

time*treat's avatartime*treat

Quote: Originally posted by KY Floyd on Jan 27, 2009

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. 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.

Stew12's avatarStew12

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).

amininflorida

Quote: Originally posted by diamondpalace on Jan 23, 2009

It's not odd at all when dealing with computer generated combinations. :P

I wonder how it is calculated to be 1 in a million?

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

amininflorida

Quote: Originally posted by amininflorida on Jan 28, 2009

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??

ThatScaryChick's avatarThatScaryChick

Quote: Originally posted by Stew12 on Jan 27, 2009

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).

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

LotteryTechInc

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

JADELottery's avatarJADELottery

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.

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