New Gambling Odds (Or why the goblins are stupid)

[Phoxtrot]

Ok, they reduced the numbers range to 1-36, so here are the new odds:

 

0 Match     : 30.48%

1 Match     : 43.90%

2 Matches : 21.10%  (win 10)

3 Matches :   4.169% (win 50)

4 Matches :  1/299 chance  (win 250)

5 Matches :  1/10.821 chance (win 2500)

6 Matches :  1/ 1.947.972 chance  (Jackpot)

 

Since the tickets are sold for 10 silver, the average return (before jackpot) is about: 5.26 silver !

 

In other words, each silver you bet will give you back somewhat more than 52 copper on average.

 

Message Edited by Phoxtrot on 10-10-2005 03:11 PM

[Phoxtrot]

Oh I read that the actual bid is 10s and not 1s as indicated in the patch notes...

 

This means of course that the goblins are making money after all.

If working correctly with a 10s bid, you need a jackpot of 922 plat 59 gold and 20s to get an even return :p

[JFuller]

Is there any way to know what the jackpot is at any given time?

[legaleagle]

yes, click on the book next to the goblin.  Jackpots started at 10p on all servers. 

As for the odds of winning, they have to be less than your investment, otherwise the jackpot wouldn't grow.  If people won more smaller pots than they paid to play, the jackpot would actually decrease. 

[FreaklyCreak]

What are the odds of making a profit from spending 1plat playing?

[Eirgo]

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Phoxtrot wrote:

Ok, they reduced the numbers range to 1-36, so here are the new odds:

 

0 Match     : 30.48%

1 Match     : 43.90%

2 Matches : 21.10%  (win 10)

3 Matches :   4.169% (win 50)

4 Matches :  1/299 chance  (win 250)

5 Matches :  1/10.821 chance (win 2500)

6 Matches :  1/ 1.947.972 chance  (Jackpot)

 

Since the tickets are sold for 1 silver, the average return (before jackpot) is about: 5.26 silver !

 

In other words, each silver you vet will gives you back more than 5 silver on average.

 

Uh!

---

Ok its been a LONG time since I took statistics or any higher math class - could you explain how you got some of your numbers?

How do you have an almost 44% chance to match one?

If you only got one chance to match 1 number and the range was 1-36 that would be a 1/36 chance of getting one match ~= 2.8% chance

Given 6 tries to get one number wouldnt that be 6 * 2.8% = 16.8% chance to get one match??

So how I figure (which I freely admit could be wrong, and I assume the winning numbers are not dependent on each other ie jackpot could be 15 15 15 15 15 15)

0 Match     :   about 82.5%

1 Match     :   16.8%

2 Matches :  .462%  (win 10)

3 Matches :  .01284% (win 50)

4 Matches :  .000357  (win 250)

5 Matches :  .0000099% (win 2500)

6 Matches :  .000000276%  (Jackpot)

 

Sorry if my thinking is off, just got me curious

Message Edited by Eirgorn on 10-06-2005 11:32 AM

[TheRealDude]

the way i was thinking of it was there are 36 possible numbers for each of the 6 slots so the chances that you would get one of those right is 6:36 which is 16.6% chance

 

to match two numbers thats a 6 : (36*36) or 6:1296 ... a 0.462% chance

 

to match three would be (6*2) : (36*36*36) or 12:46656 .. a 0.02572% chance

 

and so on.

 

the chance you would miss completely is (35*35*35*35*35*35) : (36*36*36*36*36*36) or a 1838265625:2176782336 which is a 84.448% chance

Message Edited by TheRealDude on 10-06-2005 11:44 AM

[Eirgo]

Yep thats the way I thought TRD, thanks for the clear explanation.

 

So this is a COMPLETE rip lol.

 

If you played 1pp worth (10,000 games) you would on average:

 

Match 2 - 115.7 times x 10s  = 1157s

Match 3 - 4.29 times x 50s = 214.5s

Match 4 - 0.119 times x 250s = 29.75s

Match 5 - 0.00248 times x 2500s = 6.2s

match 6 - 0.0000276 times x Jackpot = depends on jackpot

 

For you to break even (on average) the jackpot needs to be:

 

Starting money: 100gp - 14.07g (winnings from other matches) = 85.93g needs to be made up on jackpot

 

so the jackpot needs to be 85.93g/0.0000276 = 3113405g = 31134.06pp

 

.......

 

save your money hehe

 

(of course this assumes my calculations are correct - so please double check hehe)

Message Edited by Eirgorn on 10-06-2005 02:26 PM

[TheRealDude]

ok well ive read up on it a little bit and its:

 

in order to win the jackpot there are 6 numbers that have to be drawn and the order doesnt matter and they can be repeats. so there are 6^6 or 46656 different orders that will work in total there are 36^6 or 2176782336 different combonations that can possibly come out.

 

now when you divide 2176782336 by you 46656 chances of winning you get a 1 in 46656 chance or 0.0000214% chance of winning the jackpot

 

this is given that they dont need to be in the right order ... if they need to be in the  right order you then have a 1:2176782336 chance or 0.00000000045939% chance

 

for one number you have:

 

in order to get one number right you have a 1 in 36 chance of picking each number correctly. given that there are 6 numbers you have 6 chances so you have a 6:36 chance or 16.6% chance.

[Eirgo]

---

TheRealDude wrote:

 

 

to match two numbers thats a 6 : (36*36) or 6:1296 ... a 0.462% chance

 

---

Ok I think we are both off on the match 2, 3, 4 and 5

 

On match 2 - it can be any of the 2 so the probablity would be:

 

1/36 x 1/36 x ( 6! / (4! x 2!)) = 1/36 x 1/36 x 15 = 1.157%

 

We were multiplying by 6 instead of 15, but 15 should be right because there are 15 combinations of choosing 2 right ones out of the 6.

 

So Match3 would be:

 

1/36 x 1/36 x 1/36 x ( 6! / (3! x 3!) ) = 0.0429%

 

and Match4 would be:

 

1/36 x 1/36 x 1/36 x 1/36 x ( 6! / (3! x 3!) ) = 0.00119%

 

and Match5 would be:

 

1/36 x 1/36 x 1/36 x 1/36 x 1/36 x ( 6! / (2! x 4!) ) = 0.0000248%

 

so please disregard my previous statements, the odds are slightly better than I first thought (still a rip though hehe)

[Dyse]

I have no interest in this game atm.  But would be rather nice to blow all of the odds and win the jackpot on first spin    Ya, I live in a Lottery state and a casino city, and I know that'd never happen... but alas, I can still dream.

[Eirgo]

Bah just thought of another twist (not that anyone cares - just kinda fun to finally use my brain while at work)

 

I think I double counted some.  For instance if you match 3, I also counted that as matching 2 (because technically you did both) so the % for matching 3,4,5 or 6 should all be subtracted from the probability of winning due to matching 2 numbers.

[AbsentmindedMage]

I am not sure where you guys took statistics and probability theory.  But if I remember correctly, you consider each set random and independent from the last. So, if you are to match 1 number out of 36.  Your odds are 1/36.  You assume that all likelihoods are equal hence the 1/36 if there are 36 possibilites. The odds of matching 3 numbers correctly would be (1/36)(1/36)(1/36) = 1/46656 The odds of matching n numbers correctly would be (1/36)^n  = 1/36^n Your odds do not change or improve by buying more.  The odds are static.  You have the same odds of winning if you buy one ticket as you would if you bought 100 since it is all random.

[lisasdarr]

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AbsentmindedMage wrote:I am not sure where you guys took statistics and probability theory.  But if I remember correctly, you consider each set random and independent from the last. So, if you are to match 1 number out of 36.  Your odds are 1/36.  You assume that all likelihoods are equal hence the 1/36 if there are 36 possibilites. The odds of matching 3 numbers correctly would be (1/36)(1/36)(1/36) = 1/46656 The odds of matching n numbers correctly would be (1/36)^n  = 1/36^n Your odds do not change or improve by buying more.  The odds are static.  You have the same odds of winning if you buy one ticket as you would if you bought 100 since it is all random.

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That would be true if SOE hadn't broken the standard lotto rules of only allowing a number to be chosen once or drawn once, since they allow both  the odds get a lot more complex and messy.

[Moorgard]

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lisasdarren wrote:

That would be true if SOE hadn't broken the standard lotto rules of only allowing a number to be chosen once or drawn once, since they allow both  the odds get a lot more complex and messy.

---

Thankfully we have upheld some lottery traditions. If your character isn't elderly and from the midwest, forget it! You have no chance. :smileyvery-happy:

 

You can also watch an episode of My Name is Earl to gauge the effects of karma on your character's chances to win. Have fun! :smileywink:

[Furyslay]

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AbsentmindedMage wrote:
I am not sure where you guys took statistics and probability theory.  But if I remember correctly, you consider each set random and independent from the last.

So, if you are to match 1 number out of 36.  Your odds are 1/36.  You assume that all likelihoods are equal hence the 1/36 if there are 36 possibilites.

The odds of matching 3 numbers correctly would be (1/36)(1/36)(1/36) = 1/46656

The odds of matching n numbers correctly would be (1/36)^n  = 1/36^n


Your odds do not change or improve by buying more.  The odds are static.  You have the same odds of winning if you buy one ticket as you would if you bought 100 since it is all random.

---

Wow you can not be more far from right.  I have only played like 200 times and i have already matched 3 like 2-3 times already.  Even the person below seems to agree with your theory.  So next time pay more attention in your statistics class, and dont correct people if u dont have a clue yourself.  There are 36 numbers, but there are 6 numbers u can match.  So its not 1/36 its 6/36.  Now it would be (6/36)(6/36)(6/36) which is 1/216.  Now if they remove that number once picked like it should, it would be (6/36)(5/35)(4/34) = 1/357.  However, these numbers are stats for if only 3 numbers are picked, but we get 6 chances dont we.  So the numbers we need are way different. 

[Furyslay]

I dont know what you all are fussing about anyway, the first post is 100% correct.  Well may be slightly off if SoE is letting you actually get the same # twice.  But the odds wont differ from what is listed very much.  These are the odds if u can not get same # twice.

Odds:

match 1= 1/2.28

match 2= 1/4.74

match 3= 1/23.99

match 4= 1/298.50

match 5= 1/10,821.07

match 6= 1/1,947,792.00 

[Eirgo]

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AbsentmindedMage wrote:
I am not sure where you guys took statistics and probability theory.  But if I remember correctly, you consider each set random and independent from the last.

So, if you are to match 1 number out of 36.  Your odds are 1/36.  You assume that all likelihoods are equal hence the 1/36 if there are 36 possibilites.

The odds of matching 3 numbers correctly would be (1/36)(1/36)(1/36) = 1/46656

The odds of matching n numbers correctly would be (1/36)^n  = 1/36^n


Your odds do not change or improve by buying more.  The odds are static.  You have the same odds of winning if you buy one ticket as you would if you bought 100 since it is all random.

---

Your reasoning is abit off but I see what you are talking about. 

Let me try to clarify (and of course like I stated before I could be off so please dont flame me or question the college I went too, its been several years since I took any of those classes)

What you are saying holds true if you take events one at a time.  Say I flip a coin 100 times and it lands on heads everytime - the probability it lands on heads the 101st time is still only 50% because prior events have no effect one the outcome of event n + 1.

BUT

We are talking about a game that gives you 6 opportunities to match numbers.  So to match one of the numbers the probability isnt 1/36 it is 6/36 because you get 6 attempts.  Thus 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 6/36 or 1/6

To take this example to an extreme end, would you say the probability of getting one number right on a range of 1-36 would still be 1/36 even if in one game you got 36 picks?  That way if you picked one of each number you are sure to match at least one.  So 1/36 + ...... + 1/36 = 36/36 gives you the probability of 1. 

The probability of 2 matches if you only had 3 attempts would be 1/36 x 1/36 x 1/36 = 1/46656 like you said - but since you get 6 attempts that changes things.  You may be familiar with the term 6 choose 3 to describe this.  This is written ( 6! / ( 3! x 3!) or in the long form ( 6 x 5 x 4 x 3 x 2 x 1) / (3 x 2 x 1 x 3 x 2 x 1) = 20

You have to mulitpy the 1/46656 by 20 because with 6 tries you have 20 combinations of what 3 numbers could have matched.  Lets label the 6 attempts A, B, C, D, E, and F.  The following subsets are possible:

ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF        BCD, BCE, BCF, BDE, BDF, BEF    CDE, CDF, CEF    DEF

[AbsentmindedMage]

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Furyslayer wrote:

---

AbsentmindedMage wrote:I am not sure where you guys took statistics and probability theory.  But if I remember correctly, you consider each set random and independent from the last.So, if you are to match 1 number out of 36.  Your odds are 1/36.  You assume that all likelihoods are equal hence the 1/36 if there are 36 possibilites.The odds of matching 3 numbers correctly would be (1/36)(1/36)(1/36) = 1/46656The odds of matching n numbers correctly would be (1/36)^n  = 1/36^nYour odds do not change or improve by buying more.  The odds are static.  You have the same odds of winning if you buy one ticket as you would if you bought 100 since it is all random.

---

Wow you can not be more far from right.  I have only played like 200 times and i have already matched 3 like 2-3 times already.  Even the person below seems to agree with your theory.  So next time pay more attention in your statistics class, and dont correct people if u dont have a clue yourself.  There are 36 numbers, but there are 6 numbers u can match.  So its not 1/36 its 6/36.  Now it would be (6/36)(6/36)(6/36) which is 1/216.  Now if they remove that number once picked like it should, it would be (6/36)(5/35)(4/34) = 1/357.  However, these numbers are stats for if only 3 numbers are picked, but we get 6 chances dont we.  So the numbers we need are way different. 

---

That is really funny because what you are saying is absolutely INCORRECT.  Each selection is random and independent of the other.  You have 6 slots and each slot could have 1 of 36 numbers.  If every number has an equal chance of being picked then the chance of getting the correct number in that slot is 1/36.  Each slot is also independent of the other slot.  You might roll a 1 in slot 1. The odds of rolling that was 1/36.  Now you might also roll a 1 in slot 2.  The odds of that is also 1/36.  Now, the odds of rolling a 1 in slot 1 and also in slot 2  is (1/36)(1/36)=1/36^2 You need to go back and review things.  If you actually think your chances are increasing with more tickets you are really lost in terms of probability theory.

[AbsentmindedMage]

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Eirgorn wrote:

---

AbsentmindedMage wrote:I am not sure where you guys took statistics and probability theory.  But if I remember correctly, you consider each set random and independent from the last.So, if you are to match 1 number out of 36.  Your odds are 1/36.  You assume that all likelihoods are equal hence the 1/36 if there are 36 possibilites.The odds of matching 3 numbers correctly would be (1/36)(1/36)(1/36) = 1/46656The odds of matching n numbers correctly would be (1/36)^n  = 1/36^nYour odds do not change or improve by buying more.  The odds are static.  You have the same odds of winning if you buy one ticket as you would if you bought 100 since it is all random.

---

Your reasoning is abit off but I see what you are talking about. 

Let me try to clarify (and of course like I stated before I could be off so please dont flame me or question the college I went too, its been several years since I took any of those classes)

What you are saying holds true if you take events one at a time.  Say I flip a coin 100 times and it lands on heads everytime - the probability it lands on heads the 101st time is still only 50% because prior events have no effect one the outcome of event n + 1.

BUT

We are talking about a game that gives you 6 opportunities to match numbers.  So to match one of the numbers the probability isnt 1/36 it is 6/36 because you get 6 attempts.  Thus 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 6/36 or 1/6

To take this example to an extreme end, would you say the probability of getting one number right on a range of 1-36 would still be 1/36 even if in one game you got 36 picks?  That way if you picked one of each number you are sure to match at least one.  So 1/36 + ...... + 1/36 = 36/36 gives you the probability of 1. 

The probability of 2 matches if you only had 3 attempts would be 1/36 x 1/36 x 1/36 = 1/46656 like you said - but since you get 6 attempts that changes things.  You may be familiar with the term 6 choose 3 to describe this.  This is written ( 6! / ( 3! x 3!) or in the long form ( 6 x 5 x 4 x 3 x 2 x 1) / (3 x 2 x 1 x 3 x 2 x 1) = 20

You have to mulitpy the 1/46656 by 20 because with 6 tries you have 20 combinations of what 3 numbers could have matched.  Lets label the 6 attempts A, B, C, D, E, and F.  The following subsets are possible:

ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF        BCD, BCE, BCF, BDE, BDF, BEF    CDE, CDF, CEF    DEF

---

I am not flaming you or questioning your college.  I am just saying that each of the slots is random and independent as far as I can see after playing it.  You could roll the same number the same time in each of the 6 slots.  The actual order isnt important either. The probability of getting 1 match is 1/36.  Doesnt matter how many slots or attempts.  Yes, even if there were 36 slots.  The reason is because it is random for each slot.  You are saying that if there were 36 slots then the probability of getting a match would be unity!  That should prove that the way you are calculating is not right. The probability of getting 2 matches is 1/36^2.  The thing that I am seeing is you guys seem to think that somehow more slots or playing more actually improves your odds.  It doesnt.  Here is a short lesson on Probability of Independent Events:  http://www.mathgoodies.com/lessons/vol6/independent_events.html

[Sol-the-Wi]

These aren't independent events. You have six chances to match one of six numbers from a set of thirty-six. An easy way to figure this out is to use the complement. A probability of 1 (100%) means a definite match, unconditional. If P(x) is the probability of something happening, and ~P(x) is the probability of something NOT happening, then 1 - ~P(x) = P(x). or 100% - (won't happen %) = (will happen %) The probability of roll1 NOT containing any of the six chosen values is 30/36; thirty are wrong out of thirty-six. The probability of roll 1 AND roll 2 NOT containing any of the six chosen is (30/36) * (30/36). The probability of NONE of the rolls containing any of the values is (30/36)^6, or approximately 0.335. Therefore: ~P(x) = 0.335 P(x) = 1 - ~P(x) = 1-0.335 The probability of rolling at least one match is 0.665, or 66.5%. This works on the same principle as flipping a coin, which may be easier to understand. If one were to flip a coin six times, what is the probability of at least one of them being heads? ~P(x) = (1/2)^6      --> probability of one flip being NOT heads is 1/2; 6 flips becomes (1/2)^6 P(x) = 1 - ~P(x) = 0.984 As you'd expect, the probability of getting at LEAST one heads is very high, 98.4%.

[Eirgo]

I completely see where you are coming from and I think that we are both wrong and have been oversimplifying the matter!

The way I see it is you have 4 ways of running this lotto.

1.  Each number can be used only once and order counts.

2. Each number can be used only once and order doesnt matter

3. Each number can be used as many times as there are draws and order counts.

4. Each number can be used as many times as there are draws and order doesnt  matter.

I am assuming this game follows rule set #4.

You are describing the behavior of a game following rule set #3 where order does count.  This leads to the total number of combinations to be 36^6 = 2,176,782,336

 

In scenario #4 where order doesnt matter, the total number of outcomes would actually be  (41!) / (35! x 6!) or (41*40*39*38*37*36) / (6*5*4*3*2*1) = 4,496,388

 

The formula for the above is (n + k - 1)! / ((n-1)! k!)  where we choose k objects from a set of n (in our case n=36 and k=6)

 

A smaller example to show what I mean is using 2 picks from the numbers 1, 2, 3 where they can be replaced and order doesnt matter.

 

If order did matter, the total number of outcomes would be 3^2 = 9  [1,1; 1,2; 1,3; 2,1; 2,2; 2,3; 3,1; 3,2; 3,3] <- see 9 outcomes in the subset

 

But since order doesnt matter there are only 6 outcomes [1,1; 1,2; 1,3; 2,2; 2,3; 3,3]

 

So n=3 and k=2 so (3+2-1)! / (3-1)! 2!) = 4! / (2! x 2!) = 4 x 3 x 2 / (2 x 2) = 3 x 2 = 6

 

So this is all more complicated than we thought because order does not matter and the items are being replaced.

[Eirgo]

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Sol-the-Wise wrote:
These aren't independent events. You have six chances to match one of six numbers from a set of thirty-six.

An easy way to figure this out is to use the complement.

A probability of 1 (100%) means a definite match, unconditional. If P(x) is the probability of something happening, and ~P(x) is the probability of something NOT happening, then

1 - ~P(x) = P(x).
or
100% - (won't happen %) = (will happen %)

The probability of roll1 NOT containing any of the six chosen values is 30/36; thirty are wrong out of thirty-six. The probability of roll 1 AND roll 2 NOT containing any of the six chosen is (30/36) * (30/36). The probability of NONE of the rolls containing any of the values is (30/36)^6, or approximately 0.335.

Therefore:
~P(x) = 0.335
P(x) = 1 - ~P(x) = 1-0.335

The probability of rolling at least one match is 0.665, or 66.5%.

---

This assumes the numbers arent replaced after each draw for the 6 draws that comprise one game.  I was under the impression you could get 1,1,1,1,1,1 as a possible outcome of a lotto drawing.

Message Edited by Eirgorn on 10-06-2005 09:12 PM

[Sol-the-Wi]

You can roll 1,1,1,1,1,1; that changes nothing in my proof. The lotto target numbers are unique, however, and do not change until someone hits the jackpot.

[Eirgo]

I havent been able to play since this lotto has gone live, could you explain how it works?  apparently Im missing something.

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Sol-the-Wise wrote:
You can roll 1,1,1,1,1,1; that changes nothing in my proof. The lotto target numbers are unique, however, and do not change until someone hits the jackpot.

---

So you dont actually get to pick 6 numbers?  you just roll randomly 6 numbers from 1-36 and hope they match the "winning" numbers that are in clear sight to everyone that plays?

[Sol-the-Wi]

There are twelve slots, six empty on top of six filled with numbers. You buy a ticket and random numbers fill the top. You win or lose based on those random numbers matching the bottom ones, which only change when someone hits the jackpot.

Message Edited by Sol-the-Wise on 10-06-2005 09:22 PM

[Eirgo]

Bah!

 

Thanks for the explanation Solnus!

 

Well in this case the probability of matching x numbers would depend entirely on the jackpot numbers then wouldnt it?

 

You odds are much worse if you have to match 1,1,1,1,1,1  (only one roll possible)

 

opposed to if the numbers are 1, 2, 3, 4, 5, 6 (120 possible rolls would win this one - 6!)

[Sol-the-Wi]

No, you still misunderstand how it works. Let's try a visual. It starts with this: |__|__|__|__|__|__| |08|12|22|32|04|06| You buy a ticket, and it fills with these. |01|22|25|29|05|09| |08|12|22|32|04|06| You match one (22). The bottom numbers are all unique and do not change until someone jackpots. The top numbers are random and can be repeated.

[Eirgo]

Right I got that :smileyhappy:

I was just saying the odds of matching numbers depends on the jackpot numbers.   Maybe an smaller scale example could address what Im trying to say.  I'll refer back to the example I used a few posts up before I realized how the lotto works in game - it should still hold true here.

The jackpot consists of 2 numbers from a range of 1-3

The jackpot could be either

1,1       1,2      1,3     2,1   2,2    2,3    3,1  3,2   3,3                9 possibilities

If the jack pot numbers are doubles - 1,1; 2,2; or 3,3 then you only have 1/9 chance of winning the jackpot

However if the jackpot at the time is any non-double number you have a 2/9 chance on winning.  If it is 1,2 you can win with either 1,2 or 2,1

So the more repeat numbers there are, the less chance of substitution you have and therefore a lower probability of winning.

I could be wrong though, it is getting late :smileyhappy:

[Eirgo]

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Sol-the-Wise wrote:

The bottom numbers are all unique and do not change until someone jackpots. The top numbers are random and can be repeated.

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Ack you mean the jackpot numbers NEVER have doubles?

Ok then I see your point and I'll go to bed now.  That simplifies things alot.