New Gambling Odds (Or why the goblins are stupid) [Archive]

Eirgo

10-06-2005, 10:15 PM

<BR> <BLOCKQUOTE> <HR> Phoxtrot wrote:<BR> <DIV>Ok, they reduced the numbers range to 1-36, so here are the new odds:</DIV> <DIV> </DIV> <DIV>0 Match : 30.48%</DIV> <DIV>1 Match : 43.90%</DIV> <DIV>2 Matches : 21.10% (win 10)</DIV> <DIV>3 Matches : 4.169% (win 50)</DIV> <DIV>4 Matches : 1/299 chance (win 250)</DIV> <DIV>5 Matches : 1/10.821 chance (win 2500)</DIV> <DIV>6 Matches : 1/ 1.947.972 chance (Jackpot)</DIV> <DIV> </DIV> <DIV>Since the tickets are sold for 1 silver, the average return (before jackpot) is about: 5.26 silver !</DIV> <DIV> </DIV> <DIV>In other words, each silver you vet will gives you back more than 5 silver on average.</DIV> <DIV> </DIV> <DIV>Uh!</DIV><BR> <HR> </BLOCKQUOTE> <P><BR>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?</P> <P>How do you have an almost 44% chance to match one? </P> <P>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</P> <P>Given 6 tries to get one number wouldnt that be 6 * 2.8% = 16.8% chance to get one match??</P> <P>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)</P> <DIV>0 Match : about 82.5%</DIV> <DIV>1 Match : 16.8%</DIV> <DIV>2 Matches : .462% (win 10)</DIV> <DIV>3 Matches : .01284% (win 50)</DIV> <DIV>4 Matches : .000357 (win 250)</DIV> <DIV>5 Matches : .0000099% (win 2500)</DIV> <DIV>6 Matches : .000000276% (Jackpot)</DIV> <DIV> </DIV> <DIV>Sorry if my thinking is off, just got me curious</DIV><p>Message Edited by Eirgorn on <span class=date_text>10-06-2005</span> <span class=time_text>11:32 AM</span>

TheRealDude

10-06-2005, 10:41 PM

<DIV>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.<U>6</U>% chance</DIV> <DIV> </DIV> <DIV>to match two numbers thats a 6 : (36*36) or 6:1296 ... a 0.462% chance</DIV> <DIV> </DIV> <DIV>to match three would be (6*2) : (36*36*36) or 12:46656 .. a 0.02572% chance</DIV> <DIV> </DIV> <DIV>and so on.</DIV> <DIV> </DIV> <DIV>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</DIV><p>Message Edited by TheRealDude on <span class=date_text>10-06-2005</span> <span class=time_text>11:44 AM</span>

Eirgo

10-06-2005, 10:58 PM

<DIV>Yep thats the way I thought TRD, thanks for the clear explanation.</DIV> <DIV> </DIV> <DIV>So this is a COMPLETE rip lol.</DIV> <DIV> </DIV> <DIV>If you played 1pp worth (10,000 games) you would on average: <EDITED to use my numbers a couple posts down></DIV> <DIV> </DIV> <DIV>Match 2 - 115.7 times x 10s = 1157s</DIV> <DIV>Match 3 - 4.29 times x 50s = 214.5s</DIV> <DIV>Match 4 - 0.119 times x 250s = 29.75s</DIV> <DIV>Match 5 - 0.00248 times x 2500s = 6.2s</DIV> <DIV>match 6 - 0.0000276 times x Jackpot = depends on jackpot</DIV> <DIV> </DIV> <DIV>For you to break even (on average) the jackpot needs to be:</DIV> <DIV> </DIV> <DIV>Starting money: 100gp - 14.07g (winnings from other matches) = 85.93g needs to be made up on jackpot</DIV> <DIV> </DIV> <DIV>so the jackpot needs to be 85.93g/0.0000276 = 3113405g = 31134.06pp</DIV> <DIV> </DIV> <DIV>.......</DIV> <DIV> </DIV> <DIV>save your money hehe</DIV> <DIV> </DIV> <DIV>(of course this assumes my calculations are correct - so please double check hehe)</DIV><p>Message Edited by Eirgorn on <span class=date_text>10-06-2005</span> <span class=time_text>02:26 PM</span>

TheRealDude

10-06-2005, 11:22 PM

<DIV>ok well ive read up on it a little bit and its:</DIV> <DIV> </DIV> <DIV>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.</DIV> <DIV> </DIV> <DIV>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</DIV> <DIV> </DIV> <DIV>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</DIV> <DIV> </DIV> <DIV>for one number you have:</DIV> <DIV> </DIV> <DIV>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.<U>6</U>% chance.</DIV>

Eirgo

10-07-2005, 12:36 AM

<BR> <BLOCKQUOTE> <HR> TheRealDude wrote:<BR> <DIV> </DIV> <DIV> </DIV> <DIV>to match two numbers thats a 6 : (36*36) or 6:1296 ... a 0.462% chance</DIV> <DIV><BR> </DIV> <HR> </BLOCKQUOTE><BR> <DIV>Ok I think we are both off on the match 2, 3, 4 and 5</DIV> <DIV> </DIV> <DIV>On match 2 - it can be any of the 2 so the probablity would be:</DIV> <DIV> </DIV> <DIV>1/36 x 1/36 x ( 6! / (4! x 2!)) = 1/36 x 1/36 x 15 = 1.157%</DIV> <DIV> </DIV> <DIV>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.</DIV> <DIV> </DIV> <DIV>So Match3 would be:</DIV> <DIV> </DIV> <DIV>1/36 x 1/36 x 1/36 x ( 6! / (3! x 3!) ) = 0.0429%</DIV> <DIV> </DIV> <DIV> <DIV>and Match4 would be:</DIV> <DIV> </DIV> <DIV>1/36 x 1/36 x 1/36 x 1/36 x ( 6! / (3! x 3!) ) = 0.00119%</DIV> <DIV> <DIV> </DIV> <DIV>and Match5 would be:</DIV> <DIV> </DIV> <DIV>1/36 x 1/36 x 1/36 x 1/36 x 1/36 x ( 6! / (2! x 4!) ) = 0.0000248%</DIV></DIV></DIV> <DIV> </DIV> <DIV>so please disregard my previous statements, the odds are slightly better than I first thought (still a rip though hehe)</DIV>

Dyse

10-07-2005, 01:10 AM

<DIV>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 <img src="/smilies/3b63d1616c5dfcf29f8a7a031aaa7cad.gif" border="0" alt="SMILEY" /> Ya, I live in a Lottery state and a casino city, and I know that'd never happen... but alas, I can still dream.</DIV>

Eirgo

10-07-2005, 01:31 AM

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

AbsentmindedMage

10-07-2005, 01:42 AM

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. <div></div>

lisasdarr

10-07-2005, 02:13 AM

<span><blockquote><hr>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. <div></div><hr></blockquote>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.</span><div></div>

Moorgard

10-07-2005, 04:11 AM

<BR> <BLOCKQUOTE> <HR> lisasdarren wrote:<BR><SPAN><BR>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.<BR></SPAN> <BR> <HR> </BLOCKQUOTE><BR> <DIV>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:</DIV> <DIV> </DIV> <DIV>You can also watch an episode of <EM>My Name is Earl</EM> to gauge the effects of karma on your character's chances to win. Have fun! :smileywink:</DIV>

Furyslay

10-07-2005, 05:21 AM

<DIV><BR></DIV> <BLOCKQUOTE> <HR> AbsentmindedMage wrote:<BR>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.<BR><BR>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.<BR><BR>The odds of matching 3 numbers correctly would be (1/36)(1/36)(1/36) = 1/46656<BR><BR>The odds of matching n numbers correctly would be (1/36)^n = 1/36^n<BR><BR><BR>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.<BR><BR> <BR> <HR> </BLOCKQUOTE> <DIV>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. <BR><BR></DIV>

Furyslay

10-07-2005, 05:33 AM

<P>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.</P> <P>Odds:</P> <P>match 1= 1/2.28</P> <P>match 2= 1/4.74</P> <P>match 3= 1/23.99</P> <P>match 4= 1/298.50</P> <P>match 5= 1/10,821.07</P> <P>match 6= 1/1,947,792.00 </P>

Eirgo

10-07-2005, 05:57 AM

<BR> <BLOCKQUOTE> <HR> AbsentmindedMage wrote:<BR>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.<BR><BR>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.<BR><BR>The odds of matching 3 numbers correctly would be (1/36)(1/36)(1/36) = 1/46656<BR><BR>The odds of matching n numbers correctly would be (1/36)^n = 1/36^n<BR><BR><BR>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.<BR><BR> <BR> <HR> </BLOCKQUOTE> <P>Your reasoning is abit off but I see what you are talking about. </P> <P>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)</P> <P>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.</P> <P>BUT</P> <P>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</P> <P>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. </P> <P>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</P> <P>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:</P> <P>ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF BCD, BCE, BCF, BDE, BDF, BEF CDE, CDF, CEF DEF<BR></P>

AbsentmindedMage

10-07-2005, 06:19 AM

<span><blockquote><hr>Furyslayer wrote:<div></div> <blockquote> <hr> 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. <div></div> <hr> </blockquote> <div>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. </div><hr></blockquote> 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.</span><div></div>

AbsentmindedMage

10-07-2005, 06:37 AM

<span><blockquote><hr>Eirgorn wrote: <blockquote> <hr> 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. <div></div> <hr> </blockquote> <p>Your reasoning is abit off but I see what you are talking about. </p> <p>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)</p> <p>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.</p> <p>BUT</p> <p>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</p> <p>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. </p> <p>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</p> <p>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:</p> <p>ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF BCD, BCE, BCF, BDE, BDF, BEF CDE, CDF, CEF DEF</p><hr></blockquote>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</span><div></div>

Sol-the-Wi

10-07-2005, 07:55 AM

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%. <div></div>

Eirgo

10-07-2005, 08:09 AM

<P>I completely see where you are coming from and I think that we are both wrong and have been oversimplifying the matter!</P> <P>The way I see it is you have 4 ways of running this lotto.</P> <P>1. Each number can be used only once and order counts.</P> <P>2. Each number can be used only once and order doesnt matter</P> <P>3. Each number can be used as many times as there are draws and order counts.</P> <P><STRONG>4. Each number can be used as many times as there are draws and order doesnt matter.</STRONG></P> <P>I am assuming this game follows rule set #4.</P> <DIV>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</DIV> <DIV> </DIV> <DIV>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</DIV> <DIV> </DIV> <DIV>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)</DIV> <DIV> </DIV> <DIV>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.</DIV> <DIV> </DIV> <DIV>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</DIV> <DIV> </DIV> <DIV>But since order doesnt matter there are only 6 outcomes [1,1; 1,2; 1,3; 2,2; 2,3; 3,3]</DIV> <DIV> </DIV> <DIV>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</DIV> <DIV> </DIV> <DIV>So this is all more complicated than we thought because order does not matter and the items are being replaced.</DIV>

Eirgo

10-07-2005, 08:11 AM

<BR> <BLOCKQUOTE> <HR> Sol-the-Wise wrote:<BR>These aren't independent events. You have six chances to match one of six numbers from a set of thirty-six.<BR><BR>An easy way to figure this out is to use the complement.<BR><BR>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<BR><BR>1 - ~P(x) = P(x).<BR>or<BR>100% - (won't happen %) = (will happen %)<BR><BR>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.<BR><BR>Therefore:<BR>~P(x) = 0.335<BR>P(x) = 1 - ~P(x) = 1-0.335<BR><BR>The probability of rolling at least one match is 0.665, or 66.5%.<BR><BR> <BR> <HR> </BLOCKQUOTE><BR> <DIV>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.</DIV><p>Message Edited by Eirgorn on <span class=date_text>10-06-2005</span> <span class=time_text>09:12 PM</span>

Sol-the-Wi

10-07-2005, 08:15 AM

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. <div></div>

Eirgo

10-07-2005, 08:19 AM

<P>I havent been able to play since this lotto has gone live, could you explain how it works? apparently Im missing something.</P> <BLOCKQUOTE> <HR> Sol-the-Wise wrote:<BR>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.<BR> <BR> <HR> </BLOCKQUOTE> <P>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?</P>

Sol-the-Wi

10-07-2005, 08:20 AM

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. <div></div><p>Message Edited by Sol-the-Wise on <span class=date_text>10-06-2005</span> <span class=time_text>09:22 PM</span>

Eirgo

10-07-2005, 08:27 AM

<DIV>Bah!</DIV> <DIV> </DIV> <DIV>Thanks for the explanation Solnus!</DIV> <DIV> </DIV> <DIV>Well in this case the probability of matching x numbers would depend entirely on the jackpot numbers then wouldnt it?</DIV> <DIV> </DIV> <DIV>You odds are much worse if you have to match 1,1,1,1,1,1 (only one roll possible)</DIV> <DIV> </DIV> <DIV>opposed to if the numbers are 1, 2, 3, 4, 5, 6 (120 possible rolls would win this one - 6!)</DIV>

Sol-the-Wi

10-07-2005, 08:32 AM

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. <div></div>

Eirgo

10-07-2005, 08:41 AM

<P>Right I got that :smileyhappy:</P> <P>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.</P> <P>The jackpot consists of 2 numbers from a range of 1-3</P> <P>The jackpot could be either</P> <P>1,1 1,2 1,3 2,1 2,2 2,3 3,1 3,2 3,3 9 possibilities</P> <P>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</P> <P>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</P> <P>So the more repeat numbers there are, the less chance of substitution you have and therefore a lower probability of winning.</P> <P>I could be wrong though, it is getting late :smileyhappy:</P>

Eirgo

10-07-2005, 08:43 AM

<BR> <BLOCKQUOTE> <HR> Sol-the-Wise wrote:<BR><BR>The bottom numbers are all unique and do not change until someone jackpots. The top numbers are random and can be repeated.<BR> <BR> <HR> </BLOCKQUOTE> <P>Ack you mean the jackpot numbers NEVER have doubles?</P> <P>Ok then I see your point and I'll go to bed now. That simplifies things alot.<BR></P>

Sol-the-Wi

10-07-2005, 08:51 AM

If the set numbers were not unique (i.e. "had doubles"), then the whole scale would be shifted by an entire order. Instead of having to match all six, you would essentially only have to match five (if there was one double in the set). This definately should not happen, and if it does I would consider it a bug. <div></div>

Eirgo

10-07-2005, 08:52 AM

<DIV>I was thinking that doubles were allowed but if there were doubles you had to also have the doubles in your numbers.</DIV> <DIV> </DIV> <DIV>ie the jackpot is 1,1,2,3,4,5 </DIV> <DIV> </DIV> <DIV>to win you could have 2,3,4,5,1,1 but not 2,3,4,5,6,1</DIV>

Sol-the-Wi

10-07-2005, 08:57 AM

Ah, I see what you mean now; a single roll could not count for multiple places on the set, if there are doubles in the set. That depends on how they designed the system, and unless we get a dev response or someone wins the jackpot and this does happen, we won't know. The mathematics I posted was for a unique set; I will develop another for a repetitive set if I get around to it. <div></div>

Sol-the-Wi

10-07-2005, 09:05 AM

<div></div><div></div>This should hold for both repetitive and non-repetitive jackpot sets.n -- the number of sides on the dies -- the number of integers in the jackpot setS -- the number of unique integers in the jackpot setP(x) = 1 - ((n-S)/n)^sis the probability of matching at least one.P(x) = (1 - ((n-S)/n)^(s-1)) * (S-1)/n... two.P(x) = (1 - ((n-S)/n)^(s-2)) * (S-1)/n * (S-2)/n... three.P(x) = (1 - ((n-S)/n)^(s-3)) * (S-1)/n * (S-2)/n * (S-3)/n... four. P(x) = ( (S!) / ((s-S)!) ) / n^sto match them all (can be derived from patterns above).<hr>To match six unique numbers using a 36-sided die (approximates):1 match: 2/32 matches: 1/123 matches: 1/1254 matches: 1/1,8465 matches: 1/45,8086 matches: 1/3,023,309<p>Message Edited by Sol-the-Wise on <span class=date_text>10-06-2005</span> <span class=time_text>10:43 PM</span>

AbsentmindedMage

10-07-2005, 09:33 AM

Exactly Eirgorn. I have invested about 5g or so playing the game now. I have yet to see any double numbers so the probabilities and way I thought the lottery was setup were incorrect which does make a big difference. I did think they had it as just a random number generator for each slot. So, I thought it would be possible for the jackpot numbers to be all the same. From what I can see, we have the following conditions: 1) Order doesnt matter 2) No digits are repeated Assuming that no digits are repeatable then the outcome of one slot is in fact linked to the others. So, the probability of matching one number is in fact 6/36 and if we did have 36 slots then it would be unity. THe probability of getting all 6 numbers matched correctly would be (1/36)(1/35)(1/34)(1/33)(1/32)(1/31)= 1 out of 1402410240 <span>:smileyhappy:</span> <div></div>

Sol-the-Wi

10-07-2005, 10:03 AM

<span><blockquote><hr>AbsentmindedMage wrote: (1/36)(1/35)(1/34)(1/33)(1/32)(1/31)= 1 out of 1402410240 <span>:smileyhappy:</span> <div></div><hr></blockquote> You aren't taking into account that you can get any of the 6 on the first try, any of the 5 remaining on the second, etc.</span><div></div>

Kuulei

10-07-2005, 10:51 AM

<DIV>off at an angle..</DIV> <DIV> </DIV> <DIV>Power Ball is 205 million for saturdays draw!</DIV> <DIV> </DIV> <DIV>but maybe it'd be easier to win the EQ2 lottery :smileytongue:</DIV>

DevinK

10-07-2005, 11:16 AM

<P>Let's just assume the statistics are moderately correct in the fact that there'd probably be well over 500 plat before anyone won the jackpot.</P> <P> </P> <P>..... why?</P> <P> </P> <P>Why not make it so the odds make it kick out from 25-100pp.. What's anyone going to do with 500pp? 1000? 20000? (based on some of your statitics =P)</P>

AbsentmindedMage

10-07-2005, 12:11 PM

<span><blockquote><hr>Sol-the-Wise wrote:<span><blockquote><hr>AbsentmindedMage wrote: (1/36)(1/35)(1/34)(1/33)(1/32)(1/31)= 1 out of 1402410240 <span>:smileyhappy:</span> <div></div><hr></blockquote> You aren't taking into account that you can get any of the 6 on the first try, any of the 5 remaining on the second, etc.</span><div></div><hr></blockquote>You are right. In order to get the jackpot, you have to match all 6 numbers. Well, that means every number that comes up has to be a correct match. So, on that first slot you are randomly pooling from 36 numbers. You must get a correct one. There are 6 possible correct ones. On the second slot you are randomly pooling from 35 numbers. You must get a correct one. There are now 5 remaining correct ones to select. Let's test what i am saying with a simple example. Let's say that the lottery was in fact drawing from number 1-4 and had two slots. I would say that the probability of winning would be 2/4*1/3=2/12=1/6 All the possibilities are below 1,4 1,3 1,2 2,1 2,3 2,4 3,1 3,2 3,4 4,1 4,2 4,3 Let's say that the winning numbers were 2,4. we see the probability is 2/12 = 1/6 since the combination 2,4 and 4,2 are equivalent, there are two winning combinations out of a total of 12. If each is equal likely then the probability is 2/12. So, for the goblin lottery it should be (6/36)(5/35)(4/34)(3/33)(2/32)(1/31)=6!/(36*35*34*33*32*31)= 1 out of 1,947,792 Again, this is all based on the assumption that the same number can not come up more than one time. So, no 2,2 or 1,1 or etc....</span><div></div>

Eirgo

10-07-2005, 03:14 PM

<BR> <BLOCKQUOTE> <HR> AbsentmindedMage wrote:<BR><SPAN><BR> <BLOCKQUOTE><SPAN>snip<BR></SPAN> <BR><BR>So, for the goblin lottery it should be<BR><BR>(6/36)(5/35)(4/34)(3/33)(2/32)(1/31)=6!/(36*35*34*33*32*31)= 1 out of 1,947,792<BR><BR>Again, this is all based on the assumption that the same number can not come up more than one time. So, no 2,2 or 1,1 or etc....<BR></SPAN></BLOCKQUOTE> <BR> <HR> </BLOCKQUOTE> <P>WOOT! I think given those assumptions(unordered non repetitive) that is exactly what I think the odds of winning the jackpot would be. Sorry for all my out of whack posts earlier, some were just wrong and some of the later ones were based on the wrong assumptions of how the game actually worked (thanks for the clarification Sol).</P> <P><BR> </P>

Xalmat

10-07-2005, 06:07 PM

The odds of winning are:20 silver prize - 1:63050 silver prize - 1:71402.5 gold prize - 1:58,90525 gold prize - 1:376,992Jackpot - 1:1,947,792 I figured this out a couple days ago while the servers were down <img src="/smilies/69934afc394145350659cd7add244ca9.gif" border="0" alt="SMILEY" />

BCasto21

10-07-2005, 06:55 PM

<div></div><div></div><div></div>Ok, here are the odds (assuming no doubles). If you want a page to reference how to do the calculations, here is a good one http://www.lottery.state.mn.us/hypergeo.html Odds of matching 6 = 1 in 1947792 = 0.000051% Odds of matching 5 = 180 in 1947792 = 1 in 10821 = 0.0092% Odds of matching 4 = 6525 in 1947792 = 1 in 299 = 0.33% Odds of matching 3 = 81200 in 1947792 = 1 in 24 = 4.17% Odds of matching 2 = 411075 in 1947792 = 1 in 5 = 21.10% Odds of matching 1 = 855036 in 1947792 = 1 in 2 = 43.90% Odds of matching 0 = 593775 in 1947792 = 1 in 3 = 30.48% If you were to invest 1 plat: Match 0 - 300.5 times for 0s Match 1 - 439.0 times for 0s Match 2 - 211.0 times for 2110s = 21.10g Match 3 - 41.7 times for 2085s = 20.85g Match 4 - 3.3 times for 825s = 8.25g Match 5 - 0.09 times for 225s = 2.25g So for a 1 plat investment, if you dont win the jackpot, you will walk out with 52.45g. In other words, you will on average lose 47.5% of your money. <p>Message Edited by BCasto21 on <span class="date_text">10-07-2005</span> <span class="time_text">08:05 AM (changed money rewarded for 2 matches) </span></p><p>Message Edited by BCasto21 on <span class=date_text>10-07-2005</span> <span class=time_text>10:06 AM</span>

legaleagle

10-07-2005, 06:59 PM

<P>Here are the actual odds of winning the jackpot:</P> <P>1 number (6/36) since all 36 numbers are in play, and you can match any of your numbers - even those in the last slot of your group. </P> <P>2 numbers (5/36) since all 36 numbers are still in play <STRONG>(see the Dev post on page 1 - there can be doubles</STRONG>) but you only have 5 slots of your group of numbers still open. </P> <P>3 numbers (4/36) since all 36 numbers are still in play, but you only have 4 slots of your group of numbers still open.</P> <P>etc, etc. </P> <P>Giving you a calculation of (6/36)*(5/36)*(4/36)*(3/36)*(2/36)*(1/36) = 720/2,176,782,336 or basically 3,023,308:1. Sol the Wise, was indeed, wise and correct above. </P> <P>If 1 = 10s, the Jackpot break even point is 302.33 platinum. </P> <P> </P> <P>Nimnh and Absentminded, your calculations are incorrect because there can be doubles, which adds a lot more combinations to the equation. </P> <P>Message Edited by legaleagle97 on <SPAN class=date_text>10-07-2005</SPAN> <SPAN class=time_text>08:04 AM</SPAN></P><p>Message Edited by legaleagle97 on <span class=date_text>10-07-2005</span> <span class=time_text>08:08 AM</span>

Eirgo

10-07-2005, 07:16 PM

<DIV><BR></DIV> <BLOCKQUOTE> <HR> Xalmat wrote:<BR>The odds of winning are:<BR><BR><STRONG>20</STRONG> silver prize - 1:630 <EM>thought it was 10s?<BR></EM>50 silver prize - 1:7140<BR>2.5 gold prize - 1:58,905<BR>25 gold prize - 1:376,992 <BR>Jackpot - 1:1,947,792 = (36x35x34x33x32x31) / 6!<BR><BR>I figured this out a couple days ago while the servers were down <img src="/smilies/69934afc394145350659cd7add244ca9.gif" border="0" alt="SMILEY" /><BR> <HR> </BLOCKQUOTE> <DIV>Also I think you calculated the odds wrong. Lets seperate these probabilities into the 6 you draw and the remaining 30 and then divide by the total number of possibilities to get the probability for each match.</DIV> <DIV> </DIV> <DIV>Match 6 = (30 choose 0) X (6 choose 6) / (36 choose 6) = 1 x 1 / 1,947,792 = 0.000000513 = 0.0000513%</DIV> <DIV>Match 5 = (30 choose 1) X (6 choose 5) / (36 choose 6) = 30 x 6 / 1,947,792 = 0.0000924 = 0.00924%</DIV> <DIV>Match 4 = (30 choose 2) X (6 choose 4) / (36 choose 6) = 435 x 15 / 1,947,792 = 0.00335 = 0.335%</DIV> <DIV>Match 3 = (30 choose 3) X (6 choose 3) / (36 choose 6) = 4,060 x 20 / 1,947,792 = 0.0417 = 4.17%</DIV> <DIV>Match 2 = (30 choose 4) X (6 choose 2) / (36 choose 6) = 27,405 x 15 / 1,947,792 = 0.211 = 21.1%</DIV> <DIV>Match 1 = (30 choose 5) X (6 choose 1) / (36 choose 6) = 142,506 x 6 / 1,947,792 = 0.439 = 43.9%</DIV> <DIV>Match 0 = (30 choose 6) X (6 choose 0) / (36 choose 6) = 593,775 x 1 / 1,947,792 = 0.305 = 30.5%</DIV> <DIV> </DIV> <DIV>So if you spent 1pp at 10s per game that would give you 1,000 shots at winning and on average you would win:</DIV> <DIV> </DIV> <DIV>Match 2 = 1000 x 0.211 = 211 times for 211 x 10s = 2110s</DIV> <DIV>Match 3 = 1000 x 0.0417 = 41.7 times for 41.7 x 50s = 2085s</DIV> <DIV> <DIV>Match 4 = 1000 x 0.00335 = 3.35 times for 3.35 x 250s = 837.5s</DIV> <DIV>Match 5 = 1000 x 0.0000924 = 0.0924 times for 0.0924 x 2500s = 231s<BR>Match 6 = 1000 x 0.000000513 = 0.000513 times for 0.000513 x JACKPOT = X</DIV> <DIV> </DIV> <DIV>X is going to be the breakeven point, or what the jackpot would need to be to give proper incentive to play for profit</DIV> <DIV> </DIV> <DIV>starting money =1pp = 10,000s - (2110+2085+837.5+231) = 4736.5s you are still out after deducting winnings not counting the jackpot</DIV> <DIV> </DIV> <DIV>So 4736.5 / .000513 = 9232943.47s is what needs to be in the jackpot = 923.29pp</DIV></DIV>

thedu

10-07-2005, 07:24 PM

Math aside, as I stated in a previous post, this is nothing but a money sink. <div></div>

Eirgo

10-07-2005, 07:25 PM

Well my above calculations were posted before I realized that in the first page of the thread the dev stated that there could be repeats. I also went off the assumption that Match 2 was worth 10s per the 10/6 patch notes instead of 20s as stated above.

Eirgo

10-07-2005, 07:44 PM

<BR> <BLOCKQUOTE> <HR> legaleagle97 wrote:<BR> <P>Here are the actual odds of winning the jackpot:</P> <P>Giving you a calculation of (6/36)*(5/36)*(4/36)*(3/36)*(2/36)*(1/36) = 720/2,176,782,336 or basically 3,023,308:1. Sol the Wise, was indeed, wise and correct above. <BR></P> <HR> </BLOCKQUOTE> <P>I thought with an unordered drawing with replacemet it would be:</P> <P>41! / (35!6!) = 4,496,388 total combinations</P> <P>Sorry if Im just not getting something, I used to be pretty good at this stuff<BR></P>

legaleagle

10-07-2005, 07:48 PM

<DIV>Its not a money sink because (I'm assuming all) the money that people play is ultimately returned into the game via the progressive jackpot. Money sinks are things like mending costs, travel tickets to Nek and TS, etc, where money leaves the game never to return. Here, eventually, the money will all be returned. </DIV> <DIV> </DIV> <DIV>Waste of money, yes, but not a moneysink. </DIV> <DIV> </DIV> <DIV>I'm also amused by people who say they won't play unless the jackpot is at least the size of the breakeven point. These are the same people, I'm sure, that kill every named mob they come across in the hopes of a master chest dropping, even though the upgrade from their adept 1 or adept 3 is minimal, and the chance of them receiving a Master is astronomical. Its the same concept. You kill mobs hoping for a big payday. When it doesn't come, you kill again... and again... and again. Along the way, you harvest for rares, hoping for a big payday. You harvest node after node after node. The odds of you receiving a rare are slim, but you keep doing it anyway. The whole game is gambling.. maybe not with money (real or virtual) but with your time. </DIV> <DIV> </DIV> <DIV> </DIV>

thedu

10-07-2005, 07:57 PM

Ok then, a-waste-of-time sink. <span>:smileytongue:</span> How exactly are we to know when someone wins the jackpot? Are there going to be an in game postings? <div></div>

Eirgo

10-07-2005, 07:59 PM

<BR> <BLOCKQUOTE> <HR> legaleagle97 wrote:<BR> <DIV>Its not a money sink because (I'm assuming all) the money that people play is ultimately returned into the game via the progressive jackpot. Money sinks are things like mending costs, travel tickets to Nek and TS, etc, where money leaves the game never to return. Here, eventually, the money will all be returned. </DIV> <DIV> </DIV> <DIV>Waste of money, yes, but not a moneysink. </DIV> <DIV> </DIV> <DIV>I'm also amused by people who say they won't play unless the jackpot is at least the size of the breakeven point. These are the same people, I'm sure, that kill every named mob they come across in the hopes of a master chest dropping, even though the upgrade from their adept 1 or adept 3 is minimal, and the chance of them receiving a Master is astronomical. Its the same concept. You kill mobs hoping for a big payday. When it doesn't come, you kill again... and again... and again. Along the way, you harvest for rares, hoping for a big payday. You harvest node after node after node. The odds of you receiving a rare are slim, but you keep doing it anyway. The whole game is gambling.. maybe not with money (real or virtual) but with your time. </DIV> <DIV> </DIV> <DIV> </DIV><BR> <HR> </BLOCKQUOTE> <P><BR>Hehe youd get along great with my wife! She does scratch off tickets maybe once every month or 2 just because she enjoys doing them and I just see it as a waste of money. Sure, they are no more of a waste than me spending money on a candy bar or something - I mean neither things are needs, they both just give a little fun for a little money.</P> <P>I personally just dont find playing the lottery to be fun or worth the price tag, my wife on the other hand does. She thinks its great when she buys 4 $1 tickets and wins 2 free tickets. I see it as taking time to lose $2.</P>

legaleagle

10-07-2005, 07:59 PM

<DIV>I don't remember all the formula's Nimmh, but I just worked it out logically. Do you see any problems with the logic or application in my post? Just trying to keep it simple. </DIV> <DIV> </DIV> <DIV>Since I can't remember your formula, does it make sense logically? Does it apply to this situation? </DIV> <DIV> </DIV> <DIV>36 potential numbers (doubles possible)</DIV> <DIV>6 numbers chosen for you (doubles possible)</DIV> <DIV>Order of selection irrelevant. </DIV>

Eirgo

10-07-2005, 08:04 PM

<P>Somethings making me think you are oversimplifying it and leaving out some logic dealing with the effect doubles/triples etc have on probability.</P> <P>This is where I got the formula for a combination for unordered replacements (I dont remember all the formulas either)</P> <P><A href="http://www.public.coe.edu/~gcross/probstat/prob/prob-22.html" target=_blank>http://www.public.coe.edu/~gcross/probstat/prob/prob-22.html</A></P>

Oumana

10-07-2005, 08:06 PM

<P>Let's be honest folks....SOE couldn't even get the group XP modifier correct with the recent expansion/CU. Do you think that they've programmed the gambling game correctly? I don't think so...</P> <P>lottery = money sink (in-game and IRL)</P> <P> </P>

legaleagle

10-07-2005, 08:06 PM

<P>LOL.. does your wife find your playing a computer game for hours on end as much of a waste as you find her lottery tickets? (by the way, I haven't played the lotto in about a decade.. I hate it. I also hate playing slot machines when I go to the casino, which is basically what this is). </P> <P> </P> <P>It all boils down to what value an individual places on entertainment. </P> <P>For your wife: The chance to win something big may have a ton of value to one person (your wife) and $2 + 30 seconds of her day is worth the adrenaline rush. </P> <P>For you: The slow progression of a virtual, dragon-killing character on a computer has a ton of entertainment value to you, so not only are you willing to invest $15 per month + high speed internet + cost of game + cost of expansions + cost of adventure packs + cost to upgrade computer to play the game, you're also willing to invest lots and lots and lots of hours per month compared to her 30 seconds at a time. </P> <P> </P> <P>(this post assumes you don't eventually sell your char on an Exchange Server, but even if you do, all the hours you put into character development would hardly give you a decent return on any profit you made from the char :smileyhappy<img src="/smilies/3b63d1616c5dfcf29f8a7a031aaa7cad.gif" border="0" alt="SMILEY" /> </P>

Eirgo

10-07-2005, 08:10 PM

<BR> <BLOCKQUOTE> <HR> legaleagle97 wrote:<BR> <P>LOL.. does your wife find your playing a computer game for hours on end as much of a waste as you find her lottery tickets? (by the way, I haven't played the lotto in about a decade.. I hate it. I also hate playing slot machines when I go to the casino, which is basically what this is). </P> <P> </P> <P>It all boils down to what value an individual places on entertainment. </P> <P>For your wife: The chance to win something big may have a ton of value to one person (your wife) and $2 + 30 seconds of her day is worth the adrenaline rush. </P> <P>For you: The slow progression of a virtual, dragon-killing character on a computer has a ton of entertainment value to you, so not only are you willing to invest $15 per month + high speed internet + cost of game + cost of expansions + cost of adventure packs + cost to upgrade computer to play the game, you're also willing to invest lots and lots and lots of hours per month compared to her 30 seconds at a time. </P> <P> </P> <P>(this post assumes you don't eventually sell your char on an Exchange Server, but even if you do, all the hours you put into character development would hardly give you a decent return on any profit you made from the char :smileyhappy<img src="/smilies/3b63d1616c5dfcf29f8a7a031aaa7cad.gif" border="0" alt="SMILEY" /> </P><BR> <HR> </BLOCKQUOTE> <P><BR>HAHA! You hit the nail on the head!</P> <P>She does NOT get EQ2, she even tried to play EQ1 way back and made a character and ran around for about 3 minutes and was bored.</P> <P>When she checks in on me she always makes fun and asks if Im killing a dragon for treasures.</P> <P>I dont know why I like it so much, Im just glad Im not alone :smileyhappy:</P>

Eirgo

10-07-2005, 08:14 PM

<BR> <BLOCKQUOTE> <HR> legaleagle97 wrote:<BR> <P>Here are the actual odds of winning the jackpot:</P> <P>Giving you a calculation of (6/36)*(5/36)*(4/36)*(3/36)*(2/36)*(1/36) = 720/2,176,782,336 or basically 3,023,308:1. Sol the Wise, was indeed, wise and correct above. <BR></P> <HR> </BLOCKQUOTE> <P>Ok I think I figured out what is hanging me up on this. You are double counting when doubles/triples etc are concerned.</P> <P>Your factor is 720 or 6! which is used if all numbers are unique.</P> <P>That means you are counting 1,1, 2, 3, 4, 5 and 1,2, 3, 4, 5, 1 as seperate combinations when they are in fact the same since we are looking at unordered strings.</P> <P><BR> </P>

Eirgo

10-07-2005, 08:29 PM

<DIV>With 6 jackpot numbers that can be repeated you have 36^6 possible combinations</DIV> <DIV> </DIV> <DIV>Since they are unordered:</DIV> <DIV> </DIV> <DIV>If they are all unique, you divide that by 6! or 720 as you did because you have 6 spots you could put the first number, leaving 5 for the second etc etc as you stated</DIV> <DIV> </DIV> <DIV>If you have a double then you have 6 places for the first, 5 for the second, 4 for the third, 3 for the fourth and the last two dont matter because they are the same and can go in either of the remaining spaces. So for these circumstances you have a 360 divider.</DIV> <DIV> </DIV> <DIV>If you have a triple you have 6x5x4, but you double counted the doubles and have to add them back in</DIV> <DIV> </DIV> <DIV>blah blah blah....</DIV> <DIV> </DIV> <DIV>Does this make sense? It gets even hairier with 2 sets of triples, or 3 sets of doubles etc, but you just have to make sure you count each scenario only once.</DIV><p>Message Edited by Eirgorn on <span class=date_text>10-07-2005</span> <span class=time_text>09:40 AM</span>

legaleagle

10-07-2005, 08:48 PM

<DIV>You are right, because my calculations break down <STRONG>if</STRONG> there are repeaters. </DIV> <DIV> </DIV> <DIV>Triples: </DIV> <DIV> </DIV> <DIV>|__|__|__|__|__|__|<BR>|15|15|15|32|04|06|</DIV> <DIV> </DIV> <DIV>Odds of me getting one match? 4/36. (4 unique numbers to select from 36)</DIV> <DIV> </DIV> <DIV>|04|__|__|__|__|__|<BR>|15|15|15|32|04|06| </DIV> <DIV> </DIV> <DIV>Odds of me getting two matches? 3/36 (3 unique numbers to select from 36)</DIV> <DIV> </DIV> <DIV>|04|32|__|__|__|__|<BR>|15|15|15|32|04|06| </DIV> <DIV> </DIV> <DIV> </DIV> <DIV> </DIV> <DIV>Doubles: Or.. we can go this route:</DIV> <DIV>|__|__|__|__|__|__|<BR>|15|15|19|32|04|06| </DIV> <DIV> </DIV> <DIV>Which would make the numerators start with 5 and decrease instead of the starter of 4 above. Both of these examples kill my calculation (with the numerators starting at 6). </DIV> <DIV> </DIV> <DIV>Here's what I think, because I just can't apply that formula to this situation: The odds of winning the jackpot will vary considerably depending on the presence or absence of doubles or triples. You cannot give exact odds of winning it unless you actually check to see if there are actually any doubles or triples. I'm going to claim victory in this first instance because the reality of the matter is that there are no doubles in the winning set of numbers (on Kith at least). Since we are dealing with a known set of numbers, an exact calculation can be performed. I think the odds have to recalculated if there are ever repeaters in the winning numbers. </DIV> <DIV> </DIV> <DIV>I don't think your calculation is correct either though, because of this scenario:</DIV> <DIV> </DIV> <DIV>|__|__|__|__|__|__|<BR>|20|20|20|20|20|20| </DIV> <DIV> </DIV> <DIV>Odds of matching one number? 1/36</DIV> <DIV>Odds of matching your second number? 1/36</DIV> <DIV>Third? 1/36</DIV> <DIV>Fourth? 1/36, Fifth 1/36 and Sixth 1/36. </DIV> <DIV>(1/36)*(1/36)*(1/36)*(1/36)*(1/36)*(1/36) = 1 / 2,176,782,336 </DIV> <DIV> </DIV> <DIV>Or...</DIV> <DIV>|__|__|__|__|__|__|<BR>|15|20|20|20|20|20| </DIV> <DIV> </DIV> <DIV>Odds of matcing one number? 2/36</DIV> <DIV>Odds of matching two numbers? can either be 2/36 or 1/36 depending on the first number drawn. </DIV> <DIV>Odds of hitting your third number? also either 2/36 or 1/36. </DIV> <DIV>fourth and fifth are the same either 2/36 or 1/36</DIV> <DIV>sixth number is 1/36. </DIV> <DIV> </DIV> <DIV>That calculation will certainly give a different outcome than the (1/36)^5 above. If these two calucations give different outcomes, how can anyone give a probablity of hitting the jackpot with one single number? </DIV> <DIV> </DIV> <DIV><BR> </DIV>

AbsentmindedMage

10-07-2005, 08:50 PM

I really do not think that repeated numbers are possible. I have tried a number of times to see if I can get a repeat and have yet to see anything like that occur. That is why I reduced the number of numbers to pool from on each slot. <div></div>

legaleagle

10-07-2005, 10:02 PM

<P>hmm.. maybe everyone mis-interpreted Moor's post incorrectly? Maybe he meant that by upholding old lottery traditions, numbers DON'T repeat? </P> <P> </P>

Eirgo

10-07-2005, 10:36 PM

<P>How about this?</P> <P>Here is a table I made that shows all the combinations considering the jackpot numbers can be repeated: (SRY this forum hates tables or I at least dont know how to make one that it likes)</P> <P></P> <DIV> <P><SPAN>Set<SPAN> </SPAN>Ways to Order<SPAN> </SPAN>math<SPAN> </SPAN># of ways of Jackpot being like this</SPAN></P> <P><SPAN>111111<SPAN> </SPAN>720<SPAN> </SPAN>36x35x34x33x32x31 / 720 <SPAN> </SPAN>1947792</SPAN></P> <P><SPAN>21111<SPAN> </SPAN>360<SPAN> </SPAN>36x1x35x34x33x32 / 360<SPAN> </SPAN>125664</SPAN></P> <P><SPAN>2211<SPAN> </SPAN>180<SPAN> </SPAN>36x1x35x1x34x33 / 180<SPAN> </SPAN>7854</SPAN></P> <P><SPAN>222<SPAN> </SPAN>90<SPAN> </SPAN>36x1x35x1x34x1 / 90<SPAN> </SPAN>476</SPAN></P> <P><SPAN>3111<SPAN> </SPAN>180<SPAN> </SPAN>36x1x1x35x34x33 / 180<SPAN> </SPAN>7854</SPAN></P> <P><SPAN>321<SPAN> </SPAN>60<SPAN> </SPAN>36x1x1x35x1x34 / 60<SPAN> </SPAN>714</SPAN></P> <P><SPAN>33<SPAN> </SPAN>20<SPAN> </SPAN>36x1x1x35x1x1 / 20<SPAN> </SPAN>63</SPAN></P> <P><SPAN>411<SPAN> </SPAN>30<SPAN> </SPAN>36x1x1x1x35x34 / 30<SPAN> </SPAN>1428</SPAN></P> <P><SPAN>42<SPAN> </SPAN>15<SPAN> </SPAN>36x1x1x1x1x35 / 15<SPAN> </SPAN>84</SPAN></P> <P><SPAN>51<SPAN> </SPAN>6<SPAN> </SPAN>36x35x1x1x1x1 / 6<SPAN> </SPAN>210</SPAN></P> <P><SPAN></SPAN><SPAN>6<SPAN> </SPAN>1<SPAN> </SPAN>36x1x1x1x1x1 / 1<SPAN> </SPAN>36</SPAN></P> <P><SPAN><SPAN> </SPAN>Total possibilities<SPAN> </SPAN>2092175</SPAN></P> <P><SPAN>So this means that when the jackpot initially draws "its" numbers there are 2,092,175 distinct options it could choose from. Once it chooses, you can match what pattern it made with the above table to figure out your odds of hitting the jackpot.</SPAN></P> <P><SPAN>If the jackpot consists of all unique numbers instead of all repeat numbers, your odds are much higher (720 times higher I think).</SPAN></P> <P><SPAN>Anyone see obvious holes in this thought process?</SPAN></P> <P><SPAN></SPAN> </P></DIV><p>Message Edited by Eirgorn on <span class=date_text>10-07-2005</span> <span class=time_text>11:47 AM</span>

Johaan

10-07-2005, 10:50 PM

<DIV> <P>Wow, it's been a loooong time since I used this stuff.<SPAN> </SPAN>How about this?</P> <P> </P> <P>Start with an unpopulated lottery field, and an unpopulated ticket field.</P> <P>What is the probability of <SPAN> </SPAN>winning using unordered replacement of the same 36 number population for both fields?</P> <P> </P><SPAN>My head hurts. LOL!</SPAN></DIV>

Eirgo

10-07-2005, 10:59 PM

<P>/aspirin</P> <P>/take2myself</P> <P>Thats how I thought it worked closer to the beginning of the thread but without repeats.</P> <P>I think the answer to your question would depend on what the 6 numbers picked were (assuming of course you won and all 6 numbers matched).</P> <P>The lotto field and the player field both being 1,1,1,1,1,1 is alot more rare than the lotto field being some combination of 1,2,3,4,5,6 and the ticket field being some other or the same combination of 1,2,3,4,5,6.</P> <P>That sound right?</P>

Johaan

10-07-2005, 11:21 PM

<P>I agree totally. -and thanks for the asprin.</P> <P>That is the answer for one round of the lottery. Now what the cumulative percentage of winning for multiple rounds? That's what I was trying to set up. How do we propagate that through all combinations of Lotto Numbers and Tickets? How do we get the combined probability? Aside from Monte Carlo simulations? I know there is a way to do this analytically, but its not jumping up and down in front of me at the moment.</P> <P> </P>

thedu

10-07-2005, 11:21 PM

Well, you guys might to re-adjust some of your math. Today's update: - The Gigglegibber goblins are now a bit more forthcoming about the fact that they sell tickets for 10 silver each, not 1 silver. <div></div>

legaleagle

10-07-2005, 11:38 PM

<P>/boggle</P> <P>/confused</P> <P> </P> <P>odds of winning: A very large number to 1. Required jackpot size to make playing an even money or better play. Very very high (like enough to make you the #1 richest player on your server high). </P> <P> </P> <P>/quit</P> <P> </P> <P>:smileytongue:</P>

Eirgo

10-07-2005, 11:58 PM

<DIV>Does the jackpot size grow 10s everytime you play without winning? And does it lower everytime you win by matching some numbers?</DIV> <DIV> </DIV> <DIV>If so the title of the thread still stands - The Goblins are Stupid! They never keep any for themselves (and actually lose the initial 10pp starter pot)</DIV> <DIV> </DIV> <DIV>If not, then this may be an ok way to take some of the money out of the economies.</DIV>

Ertoo

10-08-2005, 02:28 AM

<div></div><span><blockquote><hr>Eirgorn wrote:<div></div> <div>If not, then this may be an ok way to take some of the money out of the economies.</div><hr></blockquote> What money? I've been playing since initial release and only have 5pp in the bank. I feel as if our RL gas prices have bled over into game and I'm riding a race horse....need to start looking for a goat with a saddle I guess. </span><div></div><p>Message Edited by Ertoocs on <span class=date_text>10-07-2005</span> <span class=time_text>05:30 PM</span>

-Aonein-

10-08-2005, 03:05 AM

<P>I really dont want to thrwo this subject off the road its taking but i have a question.</P> <P>What happens when the person who wins the jackpot comes up on the SoE possible Bot / IGE bought money List and GM procced to Ban this person because he has all of a sudden aquired a lump sum of plat?</P> <P>Please dont take this the wrong its a very legit question.</P>

Carroth

10-08-2005, 05:55 AM

<P>The Odds of winning Jackpot are close to nil.. Add in the fact that the soo called *Progressive Jackpot* is being reset to next to nothing every morning, wiping out what little chance that the coin dumped into the lotto will ever makes it's way back into the economy. What part of *Progressive Jackpot* does SOE not get? Instead of watching *My Name is Earl* they should be watching Las Vegas if they going to be implementing a Casino element to the game.. :smileyvery-happy:</P> <P>Ok /end rant</P> <P>Carroth </P>

Astatine

10-08-2005, 08:52 AM

<DIV><BR> <BLOCKQUOTE> <HR> AbsentmindedMage wrote:<BR><SPAN><BR> <BLOCKQUOTE> <HR> Furyslayer wrote:<BR> <DIV><BR></DIV> <BLOCKQUOTE> <HR> AbsentmindedMage wrote:<BR>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.<BR><BR>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.<BR><BR>The odds of matching 3 numbers correctly would be (1/36)(1/36)(1/36) = 1/46656<BR><BR>The odds of matching n numbers correctly would be (1/36)^n = 1/36^n<BR><BR><BR>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.<BR><BR> <BR> <HR> </BLOCKQUOTE> <DIV>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. <BR><BR></DIV><BR> <HR> </BLOCKQUOTE><BR>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<BR><BR>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.<BR></SPAN> <BR> <HR> </BLOCKQUOTE><BR>The really ironic thing about the probability discussion is that probability has only an indirect relation to the EQ2 lottery.</DIV> <DIV> </DIV> <DIV>You correctly set out what is usually called the "gambler's fallacy"(note 1) and point out that just because a fair coin came down tails last does not make it more likely to come up heads next time... but EQ2 doesn't use a fair coin, it uses a sequence of pseudorandom numbers. The funny thing about pseudorandom numbers is that they aren't "picked at random", the sequence is very specific for each generating algorithm and seed and will always be the same. That one fact means that none of these probability theory calculations actually apply(note 2).</DIV> <DIV> </DIV> <DIV>What a jackpot winner needs is a particular sequence of twelve numbers in the list of numbers the game will use as lottery numbers, with those numbers being related to each other. Most probably that will be a sequence of A A B B C C D D E E F F, if the game is picking the "ticket" and the "match" numbers one after the other, but it could be a sequence A B C D E F A B C D E F if the game picks first the ticket numbers then the match numbers. </DIV> <DIV> </DIV> <DIV>Finally, the whole thing may be complicated immensely if the sequence of numbers used is also used for other things such as to-hit rolls for combat - which puts the problem back in touch with probability theory.... but if the sequence of numbers wasn't ever used for anything other than lottery games then not only would there be no guarantee that a jackpot winning sequence would come up, the possibility might exist of proving in advance that the jackpot was unwinnable (by demonstrating some property of the method for generating the sequence of numbers that would rule out a jackpot sequence). That's why real-world lotteries won't touch a pseudorandom number generator with a barge pole, because if someone proved the game unwinnable there might be talk of fraud.. so a real random element is essential.</DIV> <DIV> </DIV> <DIV>(note 1) Although I'm not so sure Furyslayer had actually been caught by the gambler's fallacy.</DIV> <DIV>(note 2) Very similar techniques from the field of statistics do apply of course, and for practical purposes the distinction is just a quibble.</DIV> <DIV><SPAN class=time_text></SPAN> </DIV><p>Message Edited by Astatine on <span class=date_text>10-07-2005</span> <span class=time_text>10:03 PM</span>

Phoxtrot

10-08-2005, 12:50 PM

<P>My first post simply assume that the 6 numbers drawn are differents numbers. </P> <P>(Which is how it is in RL for all lotteries with a high number of well, numbers. The lottery usually consists of picking out numbers out of a urn and not replacing them until all needed balls are drawns. Doing otherwhise would also increase by 6 times the size of the sheet of paper used by the player to pick his numbers or have him write down the numbers with all the problems of image recognition you can imagine...)</P> <P>You have to realise that under these conditions, having for exemple your second picked number matches one of those drawn by the lottery IS dependant on wheter or not your first picked number was a match. This is easy to see, if your 1st number is one of the 6 lucky ones, this leave only 5 possibilities for your 2nd number to be right since all numbers drawn are different.</P> <P>Now for the maths, there are C(36;6) differents ways to draw 6 different numbers out of 36. (Considereing the order in which you draw or pick them is not important which is true). Where C(36;6) == 36! / ( 6! * 30!). This gives you directly the jackpot chance.</P> <P>Now, the chance of getting *exactly* one match is : C(6;1) * (C30;5) / C(36;6)</P> <P>C(6;1) = 6 and represents the different choices you have to pick exactly one of the winning numbers. </P> <P>This leaves you to pick your 5 remaining numbers out of the 30 loosing ones</P> <P>C(30;5) = 30! / (5! * 25!)</P> <P>And C(36;6) is like above the total number of possible draws.</P> <P>For those who don't know, n! = n * (n-1) * (n-2) * .... * 2 * 1 or if you prefer : n! = n * (n-1)! with 0!=1</P> <P> </P> <P> </P> <P> </P><p>Message Edited by Phoxtrot on <span class=date_text>10-08-2005</span> <span class=time_text>02:03 AM</span>

Phoxtrot

10-08-2005, 01:18 PM

<BR> <BLOCKQUOTE> <HR> legaleagle97 wrote:<BR> <DIV>You are right, because my calculations break down <STRONG>if</STRONG> there are repeaters. </DIV> <DIV> </DIV> <DIV>Triples: </DIV> <DIV> </DIV> <DIV>|__|__|__|__|__|__|<BR>|15|15|15|32|04|06|</DIV> <DIV> </DIV> <DIV>Odds of me getting one match? 4/36. (4 unique numbers to select from 36)</DIV> <DIV> </DIV> <DIV>|04|__|__|__|__|__|<BR>|15|15|15|32|04|06| </DIV> <DIV> </DIV> <DIV>Odds of me getting two matches? 3/36 (3 unique numbers to select from 36)</DIV> <DIV> </DIV> <DIV>|04|32|__|__|__|__|<BR>|15|15|15|32|04|06| </DIV> <DIV> </DIV> <DIV> </DIV> <DIV> </DIV> <DIV>Doubles: Or.. we can go this route:</DIV> <DIV>|__|__|__|__|__|__|<BR>|15|15|19|32|04|06| </DIV> <DIV> </DIV> <DIV>Which would make the numerators start with 5 and decrease instead of the starter of 4 above. Both of these examples kill my calculation (with the numerators starting at 6). </DIV> <DIV> </DIV> <DIV>Here's what I think, because I just can't apply that formula to this situation: The odds of winning the jackpot will vary considerably depending on the presence or absence of doubles or triples. You cannot give exact odds of winning it unless you actually check to see if there are actually any doubles or triples. I'm going to claim victory in this first instance because the reality of the matter is that there are no doubles in the winning set of numbers (on Kith at least). Since we are dealing with a known set of numbers, an exact calculation can be performed. I think the odds have to recalculated if there are ever repeaters in the winning numbers. </DIV> <DIV> </DIV> <DIV>I don't think your calculation is correct either though, because of this scenario:</DIV> <DIV> </DIV> <DIV>|__|__|__|__|__|__|<BR>|20|20|20|20|20|20| </DIV> <DIV> </DIV> <DIV>Odds of matching one number? 1/36</DIV> <DIV>Odds of matching your second number? 1/36</DIV> <DIV>Third? 1/36</DIV> <DIV>Fourth? 1/36, Fifth 1/36 and Sixth 1/36. </DIV> <DIV>(1/36)*(1/36)*(1/36)*(1/36)*(1/36)*(1/36) = 1 / 2,176,782,336 </DIV> <DIV> </DIV> <DIV>Or...</DIV> <DIV>|__|__|__|__|__|__|<BR>|15|20|20|20|20|20| </DIV> <DIV> </DIV> <DIV>Odds of matcing one number? 2/36</DIV> <DIV>Odds of matching two numbers? can either be 2/36 or 1/36 depending on the first number drawn. </DIV> <DIV>Odds of hitting your third number? also either 2/36 or 1/36. </DIV> <DIV>fourth and fifth are the same either 2/36 or 1/36</DIV> <DIV>sixth number is 1/36. </DIV> <DIV> </DIV> <DIV>That calculation will certainly give a different outcome than the (1/36)^5 above. If these two calucations give different outcomes, how can anyone give a probablity of hitting the jackpot with one single number? </DIV> <DIV> </DIV> <DIV><BR> </DIV><BR> <HR> </BLOCKQUOTE> <P>Not even. </P> <P>If there ARE repeating numbers (which would surprise me), each of the numbers you pick must of course match specifically one of the numbers drawn and the ORDER is important else either some draws made by the goblin would be impossible to match perfectly or some picks made by the player would have an objective lesser chance of winning than others. </P> <P>Now the actual chance under those conditions of getting say *exactly* 2 matches:</P> <P>(1/36)^2 * (35/36)^4 * C(6;2) = 1,034%</P> <P>I could fill in the rest if you want and compute average the return and all that...</P> <P>You have to remember that getting exactly 2 matches is just that, 2 matches and NOT 3 or 4 (where those 35/36 come from). And you have to pick which 2 numbers are right (where the C(6;2) comes from).</P> <P> </P> <P> </P> <P> </P> <P><BR> </P> <p>Message Edited by Phoxtrot on <span class=date_text>10-08-2005</span> <span class=time_text>02:22 AM</span>

Renvhoek

10-08-2005, 04:53 PM

<div></div>this is more of a slot machine than it is a lottery. the problem i see is that a typical slot machine, you go up and down a lot, you can spend an hour and break even, where in this game, you constantly lose, and lose, and lose. the only chance you really have at making money at all is either matching 5 or matching 6. (just curious, anyone out there got 5 yet?) they should increase the awards a bit imo to at least make it an enjoyable waste of money !! <div></div><p>Message Edited by Renvhoek on <span class=date_text>10-08-2005</span> <span class=time_text>05:59 AM</span>

10-08-2005, 05:53 PM

While all your number theroy and postulating is entertaining... it's most likely a moot point. Im sure everyone has seen firsthand the effects of the random seed in group loot lotto, and even tradeskilling criticals. Random in Norrath seems to be based on something sticky. I can't begin to count the number of times I have been in a group and the same person would continuously win every loot draw. With this in mind... it seems to me that the odds of you getting a particular draw on the lottery really depends on if the random seed is sticking in some or all of the slots. If a particular slot stops generating the full number range, and sticks on a smaller set of numbers, then your odds change. Take into account the fact that this can happen six times... six pseudo-random numbers some sticking to smaller number ranges some not, you end up with a subset of generated numbers that will not adhere to conventional statistical analysis. I personally have not played the lottery yet, but I most likely will not simply due to the fact I don't trust the random number generation that the EQ2 engine uses. Also I dont trust goblins. <div></div>