CA. Hotspot Draw Lottery

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!

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.

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.

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 ?

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

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...

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?

Agree. My initial response was for just one time of day.

Quote: CrystalMath

Agree.

Thanks!

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