Centerflder Joined: Oct 4, 2019
• Posts: 5
October 6th, 2019 at 6:54:12 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!
dailysattaking Joined: Nov 1, 2019
• Posts: 1
November 1st, 2019 at 6:31:07 AM permalink
IF YOU ARE not get money on this game you can get lot of money on other game like satta matak game this type of game are so populer in India many people's in India have lot on money win on daily game.This type of game is now online you can visit and book your numbers and pay online like get your money online
Purdygirl1 Joined: Dec 13, 2019
• Posts: 1
December 13th, 2019 at 12:10:58 PM permalink
So, I started playing in Sept...I bet like 15dollars at most a week....if i win 4 or 2dollars ill keep going till i lose..or if i win 16 or 18 ill keep playing w that money too..I bet 5 I never do bullseye, just cuz thats 5dollars more and im more of a conservative gambler..well Ive won \$450 7 times , 5 out of 5numbers...I use same numbers everytime , they mean something to me..I would've won 3 other times but I wasn't wearing my glasses and i marked one wrong number..other 2times the bartender was screwing around and not paying attn to me..I was so pissed because i can FEEEEL when im gona win. Last night I got 4 numbers 4 times ...For me..I know when Im hot and when Im not...Now...5 out of those 7times i would've won. 900 cause i had bullseye number.but i didnt play bullseye...ohhhh..i also play 5games at a time for \$1 a game.... Joined: Feb 28, 2014
• Posts: 70
December 13th, 2019 at 12:53:51 PM permalink
Hi Nathan
Son you ain�t paying attention I�m cutting you but you ain�t bleeding - Foghorn Leghorn
Mission146 Joined: May 15, 2012
• Posts: 12989
December 13th, 2019 at 2:15:36 PM permalink
Quote: Centerflder

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!

The first one is pretty simple, really. Imagine that there are only two draws in two days: On day one, we know that a number MUST be the bullseye number, right? Some number has to be. So, on the second day, the probability that the number is the same as the first day is 1/80 because that's how many numbers there are.

I can also save you some time, if you go here:

https://www.calottery.com/draw-games/hot-spot/past-winning-numbers?query=2500000#search

You can simply look up all draws that have taken place within the last 180 days.

For example, here is Draw 2510000:

https://www.calottery.com/draw-games/hot-spot/past-winning-numbers?query=2510000#search

And, here is the most recent drawing as of this time:

https://www.calottery.com/draw-games/hot-spot/past-winning-numbers

So, you likely aren't going to find any patterns, but that will give you at least 51,464 results you can look at, probably a few more.

Same time of day

Not specific enough. First, I would need you to define a specific, "Time range," as in hours and minutes. Secondly, I would have to know how many draws take place during the time range.

Patterns

The thing about patterns, in general, is that the human brain is hard-wired to notice, "Patterns," in things, but patterns are also meaningless. It's for that reason that, if a human were asked to put together a series of coin flip results (without actually flipping a coin) that the human would likely avoid patterns thinking that patterns should not occur randomly. Unfortunately, the opposite is true, patterns can and do occur randomly...but they are meaningless for predicting the future.

Here's a pattern, starting from draw 2561457 and going forward nine additional draws, 7/10 draws had a bullseye number that contained either a, '4' or a, '7' somewhere in it. Wild, right? Okay, so we look at the total 32/80 numbers that have either a four or a seven, or both. You can confirm that if you wish. 40's ten across is ten, 70's ten across is another ten, six more fours going down (44 and 74 have already been counted) and six more sevens going down (47 and 77 already counted).

Okay, so what is the probability of any number with either a four or a seven showing?

32/80 = .4

So, we have a 40% shot happening seven out of ten times, what is that probability?

We can use a binomial distribution:

http://vassarstats.net/binomialX.html

N = 10 Chances

K = 7 Successes

P = .4 (40% likely)

Exactly Seven Times: 0.042467328

Seven or More Times: 0.0547618816

Does this matter? Absolutely not. Let's add another ninety draws and see what happens:

You can confirm this if you want, but the previous ninety draws and the ones already included ended up being 44/100. Even starting with a biased (7/10) sample of numbers containing either a four or a seven, in this particular instance, it retreated to the mean pretty quickly.

N = 100 Attempts

K = 44 Success

p = 40% probability

The probability of seeing 44, or more, out of 100 is 0.236531180169.

Anyway, that's just an illustration of the fact that, the larger a sample size becomes, the more likely it is to approach the mean. Worrying about particular times of day and going out of your way (or even detecting unintentionally) patterns is a bit of a dangerous thing because you're going to see stuff that simply isn't even really there, or, if it is there, is just a function of randomness.

What You're Up Against

Okay, so I am going to ignore 8-10 spots because, apparently, those prizes are all shared if they are hit (8/8, 9/9, 10/10) during the same drawing, so that just throws a wrench in the whole thing.

Apparently, you can either play Hot Spot Only or Hot Spot with the Bullseye. It says your Bullseye Wager TOTAL (including \$1 Hotspot) must be at least \$2 and must also equal your Hot Spot Wager.

Hot Spot apparently pays:

7:7 = \$2000
6:7 = \$150
5:7 = \$10
4:7 = \$3
3:7 = \$1

So, we go here:

https://wizardofodds.com/games/keno/calculator/

And, we find that the return on the Hot Spot Game is:

0.576567804415906 if betting a single dollar.

Okay, but then we have HotSpot with the Bullseye game, so let's figure that out for the seven-spot.

The first thing that we have to acknowledge is that we have a (7/80) probability of hitting our Bullseye Bonus ball.

7/80 = 0.0875

The first thing that we have to do is figure out our probabilities assuming that we hit the Bullseye Bonus Ball. We must bet \$1 on the regular Hot Spot card (return above) and \$1 on the Bullseye Bonus. The Bullseye for a seven-spot pays as follows:

7:7 + Bullseye = \$8,000
6:7 + BS = \$500
5:7 + BS = \$60
4:7 + BS = \$18
3:7 + BS = \$3
2:7 + BS = \$2
1:7 + BS = \$4

Okay, so now we have to get into a little combinatorics to see how our other draws go. The first thing that we know is that there is a 73/80 probability that we just lose our \$1 Bullseye Bonus bet because we miss the Bullseye ball completely:

73/80 * -1 = -.9125

Okay, so now we win the rest of the time, so let's figure that out:

The first thing we need to know is that there are nCr(79,19) = 883829035553043580 ways to select 19 balls out of 79. The way the rest of the formula will work is we want to see how many of our balls get selected vs. how many other balls get selected and then divide that by the total number of ways, here's the example for zero:

nCr(6,0)*nCr(73,19)/nCr(79/19) = 0.180110002894813

-Okay, so what that means is that there is a probability of 18.01100002894813% that, if we know our Bullseye Ball has already hit, that we will get zero other balls.

We can keep going:

nCr(6,1)*nCr(73,18)/nCr(79,19) = 0.373318915091067 Bullseye Ball +1

nCr(6,2)*nCr(73,17)/nCr(79,19) = 0.2999884139124645 BS + 2

nCr(6,3)*nCr(73,16)/nCr(79,19) = 0.1192936382809801 BS + 3

nCr(6,4)*nCr(73,15)/nCr(79,19) = 0.0246814424029614 BS + 4

nCr(6,5)*nCr(73,14)/nCr(79,19) = 0.0025099771935215 BS + 5

nCr(6,6)*nCr(73,13)/nCr(79,19) = 0.0000976102241925 BS + 6

PROOF: 0.180110002894813 + 0.373318915091067 + 0.2999884139124645 + 0.1192936382809801 + 0.0246814424029614 + 0.0025099771935215 + 0.0000976102241925 = 1

Of course, those are not our probabilities, because this game would be totally awesome if they were. Nope, now we have to go back and factor in the probability of even hitting the bonus ball to begin with. We have our losing probability, so let's bring this table back down here:

73/80 * -1 = -.9125

That is our probability of losing our Bullseye bet outright and the expected monetary result associated therewith. Now, remember, the game does not give you the winning result AND return your bet, so we have to subtract the bet amount to get our actual profit:

7:7 + Bullseye = \$7,999
6:7 + BS = \$499
5:7 + BS = \$59
4:7 + BS = \$17
3:7 + BS = \$2
2:7 + BS = \$1
1:7 + BS = \$3

Okay, now we have to determine our overall probability of hitting the bonus ball, multiply that by our extra ball probabilities and then multiply that by the actual monetary result to get our expected values:

BS BALL + 0 = (7/80) * (0.180110002894813 ) * (3) = 0.04727887575
BS BALL + 1 = (7/80) * (0.373318915091067) * 1 = 0.03266540507
BS BALL + 2 = (7/80) * (0.2999884139124645) * 2 = 0.05249797243
BS BALL + 3 = (7/80) * (0.1192936382809801) * 17 = 0.17744928694
BS BALL + 4 = (7/80) * (0.0246814424029614) * 59 = 0.1274179464
BS BALL + 5 = (7/80) * (0.0025099771935215) * 499 = 0.10959187921
BS BALL + 6 = (7/80) * (0.0000976102241925) * 7999 = 0.06831861604

Okay, so let's put all of this together and get an expected return of your \$1 Bullseye bet:

(0.04727887575 + 0.03266540507 + 0.05249797243 + 0.17744928694 + 0.1274179464 + 0.10959187921 + 0.06831861604) -.9125 = -0.29728001816

Okay, so the good news is technically you're improving your overall percentage return with this, even though it still makes your expected loss total substantially worse.

FINAL PROOF: .9125 + ((7/80) * (0.180110002894813)) + ((7/80) * (0.373318915091067)) + ((7/80) * (0.2999884139124645)) + ((7/80) * (0.1192936382809801)) + ((7/80) * (0.0246814424029614)) + ((7/80) * (0.0025099771935215)) + ((7/80) * (0.0000976102241925)) = 1

CONCLUSION

Therefore, even if you could identify a repeatable and predictable pattern (you can't) it would have to be one hell of a predictable pattern to overcome the House Edge. This is especially true when you consider the fact that your pattern you're looking for has to do with the Bullseye Ball that causes you to incur a greater expected loss.

EXPECTED LOSS HOTSPOT GAME ALONE: (1-0.576567804415906) = 0.42343219558 or a 42.343219558% House Edge

EXPECTED LOSS BULLSEYE BALL ALONE: 0.29728001816 or a 29.728001816% House Edge

COMBINED HOUSE EDGE: (0.29728001816+0.42343219558)/2 = 0.36035610687 or 36.035610687%

COMBINED EXPECTED \$\$\$ LOSS: 0.36035610687*2 = \$0.72071221374 for every \$2 bet.

Yeah, lottery games are terrible. They are worse, in some cases, than what the casinos can even legally do...depending on the game in question. For instance, almost every video keno game in the entire country (not lottery, but casino) would be legally required to be better than this one. You'll only cause yourself a loss of time and a headache to try to identify patterns, there won't be any, any you find will be based on poor sample sizing and, even if there are patterns, they won't be enough to turn this into a positive game.

By the way, here is a page on the HotSpot Lottery General info:

https://www.lottery.net/california/hot-spot

And, on this page, they say the probability of 7:7 with Bullseye Ball is: 1 in 117,084

Let's take what I got for it above (showing you how to do combinatorial math):

1/((7/80) * (0.0000976102241925)) = 117083.753501

Therefore, my math agrees with the lottery and that is proven.
Last edited by: Mission146 on Dec 13, 2019
Vultures can't be choosers.
gordonm888 Joined: Feb 18, 2015
• Posts: 2733
Thanks for this post from: December 13th, 2019 at 4:13:17 PM permalink
Mission, I think what the OP is getting at is whether the RNG for the game might possibly use some simple version of the time of day for a seed when picking a bullseye number.

For the OP's observed occurrences of the same bullseye number at the same time of day, one might calculate what the confidence level is that the game is random, or, conversely, what the confidence level is that the time of day is indeed weakly/strongly correlated with the selected bullseye number.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
odiousgambit Joined: Nov 9, 2009
• Posts: 8409
Thanks for this post from: December 13th, 2019 at 4:57:23 PM permalink
I wish you had touched on a subject Mission had studied before and knew something about ... j.k.!
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: �Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!� She is, after all, stone deaf. ... Arnold Snyder
Mission146 Joined: May 15, 2012
• Posts: 12989
December 13th, 2019 at 7:10:00 PM permalink
Quote: gordonm888

Mission, I think what the OP is getting at is whether the RNG for the game might possibly use some simple version of the time of day for a seed when picking a bullseye number.

For the OP's observed occurrences of the same bullseye number at the same time of day, one might calculate what the confidence level is that the game is random, or, conversely, what the confidence level is that the time of day is indeed weakly/strongly correlated with the selected bullseye number.

You could definitely do that, but again, we'd need to be asking a more specific question. I guess I'd ask the OP what time range we want to be analyzing, especially since we have the data for 50-some thousand draws at our disposal.

ADDED: Another thing is it would have to be pretty meaningful to even offer an advantage, which is why I highlighted how bad the game is. For instance, let's take our probabilities here:

BS BALL + 0 = (7/80) * (0.180110002894813 ) * (3) = 0.04727887575
BS BALL + 1 = (7/80) * (0.373318915091067) * 1 = 0.03266540507
BS BALL + 2 = (7/80) * (0.2999884139124645) * 2 = 0.05249797243
BS BALL + 3 = (7/80) * (0.1192936382809801) * 17 = 0.17744928694
BS BALL + 4 = (7/80) * (0.0246814424029614) * 59 = 0.1274179464
BS BALL + 5 = (7/80) * (0.0025099771935215) * 499 = 0.10959187921
BS BALL + 6 = (7/80) * (0.0000976102241925) * 7999 = 0.06831861604

Except, what we are going to do is let's say that getting the bonus ball right is twice as likely because we have identified whatever:

BS BALL + 0 = (14/80) * (0.180110002894813 ) * (3) = 0.09455775151
BS BALL + 1 = (14/80) * (0.373318915091067) * 1 = 0.06533081014
BS BALL + 2 = (14/80) * (0.2999884139124645) * 2 = 0.10499594486
BS BALL + 3 = (14/80) * (0.1192936382809801) * 17 = 0.35489857388
BS BALL + 4 = (14/80) * (0.0246814424029614) * 59 = 0.25483589281
BS BALL + 5 = (14/80) * (0.0025099771935215) * 499 = 0.21918375842
BS BALL + 6 = (14/80) * (0.0000976102241925) * 7999 = 0.13663723208

(0.13663723208+0.21918375842+0.25483589281+0.35489857388+0.10499594486+0.06533081014+0.09455775151) - (66/80) = 0.4054399637

But, factoring in what we lose on the base game (that has to be played for an equal amount):

(0.4054399637-0.42343219558)/2 = -0.00899611594

So, you would still be eating a 0.9% House Edge even if we could assume that the bonus ball is twice as likely as it should be to be one of our numbers. I guess what I am saying is it would have to be one hell of a pattern.

Something with reseeding on particular numbers is theoretically possible, but the last time I ever heard of that was the illegal little stand-alone machines.
Vultures can't be choosers.