MISSOURI IS THE WORST STATE FOR PICK 3 & PICK 4 LOTTERY PLAYERS AND STATS BACK IT UP.Prev Topic Next Topic

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Newkids06

Charlotte,Nc
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January 17, 2014
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Oct 18, 2026, 11:04 pm

MISSOURI IS TERRIBLE, IN THE PAST 7 YEARS AFTER TRACKING DATA IT SHOWS FAVORABLE PUBLIC NUMBERS SUCH AS TRIPS AND QUADS ARE PLAYED AT A MUCH LOWER RATE THAN OTHER STATES. AIN THE PAST 7 YEARS MISSOURI HAS DROPPED TRIPS 39X WHICH IS 12X LESS THAN THE EXPECTED AVERAGE. YEAR AFTER YEAR THEY ALWAYS GO BELOW THE AVERAGE OF EXPECTED PLAYS. AN QUADS FORGET ABOUT IT. THEY HAVENT PLAYED A QUAD IN 3.5YRS AND PROBABLY HAVE THE BALLS PROGRAMMED TO NOT DROP THEM. HERES A COMPARISON TO A CUSTOMER FRIENDLY STATE SUCH AS VIRGINIA. MISSOURI HAS ALSO MASTERED THE 1 AWAY SO INSTEAD OF GIVING YOU 111 THEY GIVE YOU 110 OR 112. MR LESTER ELDER DIRECTOR YOU HAVE SOME EXPLAINING TO DO.
FrostyGambling

Southwest USA
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September 9, 2025
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Yesterday, 7:43 pm

Why would trips or quads be favorable over any other combination? The odds are 1 in 10,000.

The more seeds of opportunity you plant the luckier you will become!
MzDuffleBaglady

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F-I-L-T-E-R-S
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October 16, 2009
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Yesterday, 8:12 pm

It's the repeats that take me out.
I rarely play quads, because, I called the 4444's and didn't play them, but, my oldest son did.
At one time, I was calling triples, but the game changed, no matter what the statistics show!
I will still take a chance on 50 cent straight plays for a $300.00 win.

389, Midday specific: .
parlayman

50

bliss
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July 31, 2016
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Yesterday, 9:18 pm

Quote: Originally posted by FrostyGambling on Oct 19, 2026
Why would trips or quads be favorable over any other combination? The odds are 1 in 10,000.

Because "Quads" are not a number, they are a Set of Ten numbers.
So you odds calculation is off by an order of magnitude.

-p
parlayman

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bliss
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July 31, 2016
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Yesterday, 9:33 pm
Quote: Originally posted by Newkids06 on Oct 18, 2026
MISSOURI IS TERRIBLE, IN THE PAST 7 YEARS AFTER TRACKING DATA IT SHOWS FAVORABLE PUBLIC NUMBERS SUCH AS TRIPS AND QUADS ARE PLAYED AT A MUCH LOWER RATE THAN OTHER STATES. AIN THE PAST 7 YEARS MISSOURI HAS DROPPED TRIPS 39X WHICH IS 12X LESS THAN THE EXPECTED AVERAGE. YEAR AFTER YEAR THEY ALWAYS GO BELOW THE AVERAGE OF EXPECTED PLAYS. AN QUADS FORGET ABOUT IT. THEY HAVENT PLAYED A QUAD IN 3.5YRS AND PROBABLY HAVE THE BALLS PROGRAMMED TO NOT DROP THEM. HERES A COMPARISON TO A CUSTOMER FRIENDLY STATE SUCH AS VIRGINIA. MISSOURI HAS ALSO MASTERED THE 1 AWAY SO INSTEAD OF GIVING YOU 111 THEY GIVE YOU 110 OR 112. MR LESTER ELDER DIRECTOR YOU HAVE SOME EXPLAINING TO DO.
I can't speak to quads, but I just ran the trips in Missouri's Pick 3 draw.
Over the entire history of the Missouri Pick 3, there have been 18,638 draws.
The odds of a trip being drawn are 1 in 100, so the expected number is 18,638 / 100 = 186 (rounded to whole number)
By my count, there have been 174 trips drawn in the entire history of the Missouri Pick 3.
You are right in thinking that is low, but is it out of bounds?
I asked Google: In 18,638 pick 3 draws, the expectation for trips is 186.38. Is an actual draw result of 174 trips out of bounds statistically?
Google answered:
No, an actual draw result of 174 trips is not statistically out of bounds. It falls well within normal statistical variations, sitting less than 1 standard deviation away from the expected 186.38 trips.
To determine if the variance is statistically significant, we can model the draws using a binomial distribution (or a normal approximation):

1. Identify the Variables
- Total Draws ($n$): $18,638$
- Probability of a trip ($p$): $0.01$ (Since there are 10 triples—000 through 999—out of 1,000 possible combinations in Pick 3)
- Expected Value ($E$): $186.38$
- Actual Result ($X$): $174$
2. Calculate the Variance and Standard Deviation
The standard deviation ($\sigma $) measures the typical variation you can expect from the average by chance:
$\text{Variance\ }(\sigma ^{2})=n\times p\times (1-p)$
$\sigma ^{2}=18638\times 0.01\times 0.99=184.5162$
$\text{Standard\ Deviation\ }(\sigma )=\sqrt{184.5162}\approx 13.58$
3. Calculate the Z-Score
The Z-score reveals how many standard deviations the actual result ($174$) is from the expected mean ($186.38$):
$Z=\frac{X-E}{\sigma }$
$Z=\frac{174-186.38}{13.58}=\frac{-12.38}{13.58}\approx -0.91$
Conclusion
In statistics, a result is generally considered "out of bounds" (statistically significant) if the Z-score is less than $-1.96$ or greater than $+1.96$ (which corresponds to a 95% confidence level).
Because your Z-score is only $-0.91$ (with a two-tailed p-value of approximately $0.362$), this deviation is very common. There is about a 36.2% chance of seeing a result this far or farther from the expectation purely due to random luck.
I don't have the numbers for all the other lotteries to compare, but the Missouri Pick 3 trips rate is not statistically out of bounds.
-p
FrostyGambling

Southwest USA
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September 9, 2025
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Today, 2:16 am

Whether you pick 7777 or 7091 the odds are the same.

The more seeds of opportunity you plant the luckier you will become!