Centerflder
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October 4th, 2019 at 8:17:09 AM permalink
This question may have been asked and answered already, but I could not find it! In CA. we have a Keno lottery game called Hotspot. So, as in Keno, there are 80 numbers to choose from an 20 balls drawn each game and 1 of the 20 is the Bull's Eye number. This number allows you to substantially increase your winnings should any of your numbers match the Bull's Eye number. I have just started looking for patterns regarding the Bull's Eye numbers. There are 300 draws each day. The draws are every 4 minutes during certain hours each day. I have recorded only part of 5 days worth of draws and have already found 30 occurrences where the same Bull's Eye number has come up twice at the same time of day. Even more noticeable was when in day 5, the number 61 came up for the third time at 9:32 am. My questions are, #1(what are the odds that the same number would come up as the Bull's Eye number at the same time of day in 3 out of 5 days?). And #2( can someone provide the formula needed to figure the odds of occurrences as I move forward with my search for patterns?). I would genuinely appreciate help with this matter. Thank you!
CrystalMath
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October 4th, 2019 at 8:46:11 AM permalink
It is 1 in 656 of getting 3 equal numbers out of 5 drawings. So, with 300 drawing a day, you should see this about every other day, on average.

For 1 specific number, the probability is (1/80)^3*(79/80)^2*combin(5,3). This is the binomial distribution.

For any number, multiply those results by 80.
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CrystalMath
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October 4th, 2019 at 8:58:41 AM permalink
There is an 11.809% chance of having exactly 2 matching numbers. With 300 drawings a day, if you looked at a whole week, you should see this about 35 times. Seems like you're right on track.
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Centerflder
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October 8th, 2019 at 8:40:38 PM permalink
Thank you for responding to my question. Based on your answer, I don't think I did a very good job of how I worded the information. The 300 draws I am referring to occur on a repetitive scheduled every 4 minutes. An example would be (9:00 am), (9:04 am), (9:08 am), (9:12 am), through (9:56 am). Then you would start the (10:00am) thru (10:56 am) hour with there being 15 draws every hour at :04, :08, :12, :16 and so on throughout the day. So, when I referred to the number 61 coming up as the Bull's Eye or Red ball number in 3 out of 5 days, I should clarify that on Sept. 2 at the 9:32 am draw, the number 61 came up as the one red ball out of the 20 that made up that draw. The next day, Sept. 3 at the 9:32 am draw, the number 61 came up as the red ball number again. Then on Sept 6th, the number 61 came up for the third time in 5 days as the red ball number on the 9:32 am draw. So, with 80 numbers that could come up, and only 20 drawn, and only one comes up red every draw, the chances that the same number came up red at the same time of draw out of the 300 possible draws during the day, in 3 out of 5 days has got to be something closer to astronomical than anything else. Right ?
KevinAA
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October 9th, 2019 at 1:38:48 AM permalink
If you look at past results and see something like this, it's probably just a coincidence. See if the same thing happens again in the future (i.e. the next 5 days). If it happens again, then I would say there is a good chance that the game is not drawing numbers randomly.

If you post this question on lotterypost.com you will get a bunch of replies from people who think that lottery drawings have patterns. Don't bother asking them what the probability of that happening is because all of them (except me) have absolutely no concept of probability. I visit that forum occasionally just for kicks to see what idiotic things people are saying. Some of them spend hours on spreadsheets trying to predict lottery numbers LOL
Wizard
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October 9th, 2019 at 6:15:54 AM permalink
First, I had some trouble understanding the rules based on the OP, so found them on the California Lottery web site: https://www.calottery.com/play/draw-games/hot-spot/how-to-play.

However, for the question at hand, as I understand it, the only pertinent rules are:

1. There are 80 balls.
2. The game will draw one "hot spot" ball per drawing.
3. There are 300 drawings per day.

The question is what is the probability that same number is drawn at the same time in 3 out of 5 consecutive days. I shall work on the answer...
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Wizard
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October 9th, 2019 at 1:25:42 PM permalink
I get an answer of 36.71%. The is the probability of at least one of the 300 drawing times over 5 days having exactly three of the days matching Hot Spot balls.

For any given drawing time, the probability of exactly 3 of the days having the same number is combin(5,3)*(1/80)^2*(79/80)^2 = 0.001523682.

The probability of anything else is 1 - 0.001523682 = 0.998476318.

The probability of not having 3 out 5 of matches over 300 drawings is 0.998476318^300 = 63.29%.

Thus, the probability of at least one drawing time having 3 out of 5 Hot Spot matches is 1 - 63.29% = 36.71%.

Anyone agree or disagree?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
CrystalMath
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October 9th, 2019 at 3:18:20 PM permalink
Agree. My initial response was for just one time of day.
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Wizard
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October 9th, 2019 at 4:54:45 PM permalink
Quote: CrystalMath

Agree.



Thanks!
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ThatDonGuy
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October 9th, 2019 at 7:10:48 PM permalink
Quote: Wizard

I get an answer of 36.71%. The is the probability of at least one of the 300 drawing times over 5 days having exactly three of the days matching Hot Spot balls.

For any given drawing time, the probability of exactly 3 of the days having the same number is combin(5,3)*(1/80)^2*(79/80)^2 = 0.001523682.

Anyone agree or disagree?


Disagree. It's (5)C(3) x (1/80)3 x (79/80)2, or about 1 / 52,504.
The probability of it not happening at all in 300 chances is (1 - 1 / 52,504)300 = about 1 / 175.5135.

Some quick simulation gets the same result.

Note that if you are looking for three or more Hot Spot matches at the same time, the first number is:
( (5)C(3) x (1/80)3 x (79/80)2 + (5)C(4) x (1/80)4 x (79/80) + (5)C(5) x (1/80)5 )
= (10 x 79 x 79 + 5 x 79 + 1) / 805 = about 1 / 52,173
and the probability of this not happening in 300 plays per day is 1 - (1 - (1 / 52,173))300 = about 1 / 174.41
Wizard
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October 9th, 2019 at 7:54:54 PM permalink
Quote: ThatDonGuy

Disagree. It's (5)C(3) x (1/80)3 x (79/80)2



But it can be any Hot Spot number, not a specific one.
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ThatDonGuy
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October 9th, 2019 at 8:01:53 PM permalink
Quote: Wizard

Quote: ThatDonGuy

Disagree. It's (5)C(3) x (1/80)3 x (79/80)2



But it can be any Hot Spot number, not a specific one.


Something tells me I am misunderstanding the rules of the game...
Wizard
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October 9th, 2019 at 8:51:49 PM permalink
Quote: ThatDonGuy

Something tells me I am misunderstanding the rules of the game...



A ball is drawn from 1 to 80 that is the Hot Spot. What is the probability that in five drawings, exactly three have the same Hot Spot number?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Centerflder
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October 10th, 2019 at 11:01:35 PM permalink
This info is for clarification regarding my earlier posts. The reason for my looking into the possibilities of finding a pattern within the CA. Hot Spot draws was based on an article I read a few weeks ago The article talked about a man who frequented a local casino in his area and started charting the draws from the Keno game. The one difference between this casino's Keno game and most other casinos is that this Keno game did not run 24/7 as most do. The article went on to explain that the 6 hour gap of down time had some effect on the algorithms that the RNG operated on and allowed for patterns to occur. This man spotted the patterns and was able to take advantage them to make a lot of money. The article stressed that the only reason that these patterns were made possible, was because of the starting and stopping of the RGN. So, this made me think of the Hot Spot Keno game here in CA. that has a 4 hour gap each day! My thinking was that if I could find a pattern with regard to the (Bull's Eye or Red ball) draws that I might be able to take advantage of this information. When the number 61 came up as the red ball for the 9:32 am draw on 9/2, the 9:32 am draw on 9/3, and the 9:32 am draw on 9/6, I hoped that I might be on to something. To a non-actuary, when you start with 80 numbers, and eliminate 75% of those numbers because only 20 are drawn for each game, and then further eliminate 95% of those numbers to leave just one number as the Red Ball number. Then you add to that the fact that it occurred on the exact same time of draw each day at (9:32 am), so that eliminates 299 of the 300 draw times each day from consideration, I assumed that the odds of all of that happening on three of the first five days of charting was significant. I hope this helps clear up any confusion about how to figure the math. I truly appreciate everyone's help and contributions with this matter!
onenickelmiracle
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October 11th, 2019 at 4:20:38 PM permalink
Quote: Centerflder

This info is for clarification regarding my earlier posts. The reason for my looking into the possibilities of finding a pattern within the CA. Hot Spot draws was based on an article I read a few weeks ago The article talked about a man who frequented a local casino in his area and started charting the draws from the Keno game. The one difference between this casino's Keno game and most other casinos is that this Keno game did not run 24/7 as most do. The article went on to explain that the 6 hour gap of down time had some effect on the algorithms that the RNG operated on and allowed for patterns to occur. This man spotted the patterns and was able to take advantage them to make a lot of money. The article stressed that the only reason that these patterns were made possible, was because of the starting and stopping of the RGN. So, this made me think of the Hot Spot Keno game here in CA. that has a 4 hour gap each day! My thinking was that if I could find a pattern with regard to the (Bull's Eye or Red ball) draws that I might be able to take advantage of this information. When the number 61 came up as the red ball for the 9:32 am draw on 9/2, the 9:32 am draw on 9/3, and the 9:32 am draw on 9/6, I hoped that I might be on to something. To a non-actuary, when you start with 80 numbers, and eliminate 75% of those numbers because only 20 are drawn for each game, and then further eliminate 95% of those numbers to leave just one number as the Red Ball number. Then you add to that the fact that it occurred on the exact same time of draw each day at (9:32 am), so that eliminates 299 of the 300 draw times each day from consideration, I assumed that the odds of all of that happening on three of the first five days of charting was significant. I hope this helps clear up any confusion about how to figure the math. I truly appreciate everyone's help and contributions with this matter!

I'm just curious how you were able to justify being around to see so many draws? Could they be seen online live or can you look back at the history quickly and easily or did you sit inside some place like a bar all day? The lotteries are the most corrupt of legalized gambling, hope you figure something out.
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ThatDonGuy
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October 11th, 2019 at 5:33:14 PM permalink
Quote: onenickelmiracle

I'm just curious how you were able to justify being around to see so many draws? Could they be seen online live or can you look back at the history quickly and easily or did you sit inside some place like a bar all day? The lotteries are the most corrupt of legalized gambling, hope you figure something out.


Hot Spot draw results, with time stamps, are available here.
ThatDonGuy
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October 11th, 2019 at 5:37:07 PM permalink
Quote: Wizard

A ball is drawn from 1 to 80 that is the Hot Spot. What is the probability that in five drawings, exactly three have the same Hot Spot number?


In that case, note that at most one number can be the Hot Spot number at least three out of five times.
The probability that it is number N is (5)C(3) x (1/80)3 x (79/80)2
Since N is equally likely to be any integer from 1 to 80, the total probability is 80 times that, or (5)C(3) x 80 x (1/80)3 x (79/80)2,
which is (5)C(3) x (1/80)2 x (79/80)2
which, er, is what you said before I disagreed with that answer...
Centerflder
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October 12th, 2019 at 9:50:24 AM permalink
Maybe I am thinking incorrectly about the relevance of the point that it's the fact that the same number is showing up as the red ball number at the same time of day, (9:32 am) in this case. I understand that the same number could come up as Red in 3 out of any 5 draws during the 300 draws per day or the 1500 draws over 5 days. I am assuming that the same number as Red at the same time of draw each day changes the odds significantly!
Agree or disagree? With an explanation if you disagree, please and thank you!
ThatDonGuy
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October 12th, 2019 at 10:43:33 AM permalink
Quote: Centerflder

Maybe I am thinking incorrectly about the relevance of the point that it's the fact that the same number is showing up as the red ball number at the same time of day, (9:32 am) in this case. I understand that the same number could come up as Red in 3 out of any 5 draws during the 300 draws per day or the 1500 draws over 5 days. I am assuming that the same number as Red at the same time of draw each day changes the odds significantly!
Agree or disagree? With an explanation if you disagree, please and thank you!

We are already calculating the probability that, in a particular 5-day period, the same number will be the Hot Spot at least three times at the same time of day in at least one of the 300 times of day.

If you limit it to a specific time of day - for example, it has to happen at 9:32 AM - then the probability is 1 / 656.3
However, you want to know what the probability is that it happens at least once over all 300 possible times of day - that is, three of the five Hot Spot numbers at 8:00 AM match, or three at 8:04 AM match, or three at 8:08 AM match, and so on; the probability of this is about 3/8.
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