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.
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?
florida did the same thing with number 354 a couple of days ago
Happens all the time in KY.
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.
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.
What kind of RNG does LP use on the qp's, Todd?
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.
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.
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.
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.
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.
Happens in Ga all the time!
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.
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 =
Total Matches =
Probability Current Draw Matches Next Draw =
(Total Matches) / (Total Draws - 1)
or
or
Probability Current Draw Matches Next Draw =
(Total Matches) / (Total Draws - 1)
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.