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Topic: Winning strategy in the 5/90 lottery (Read 16497 times) |
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jollytall
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Winning strategy in the 5/90 lottery
« on: Mar 14th, 2006, 8:26am » |
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In my country the most popular lottery for the last 50 years is 5 out of 90. Weekly there are about 5 million tickets purchased. From the fee collected about 50% distributed (the rest is tax, profit, etc.): 20% for those guessed 2 numbers correctly, 10-10% for those guessed 3, 4, 5 numbers correctly. If there is nobody with 5 numbers met then that part of the money is rolled on to the next week(s).
I heard of someone who always played the numbers 1,2,3,4,5. Someone else told him that he was stupid, because it had never happened that 5 consequtive numbers were drawn, not even 4 (3 is very rare too). His reply was that it has the same probability as any other combination.
Someone else always plays the outcome of the previous week. The debate is the same.
Newbie question: Who is right?
Junior member question: Is the Newbie answer really correct (from a practical point of view)?
Full member question: Is there a winning strategy in this lottery (in terms of positive expected value)?
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towr
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Re: Winning strategy in the 5/90 lottery
« Reply #1 on: Mar 14th, 2006, 9:25am » |
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Every 5 numbers are, if everything is fair, equally probable. A chance of 1 in 90!/(5!*85!)
Assuming fairness, I very much doubt there is a winning strategy.
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jollytall
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Re: Winning strategy in the 5/90 lottery
« Reply #2 on: Mar 14th, 2006, 11:15am » |
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This is the perfect Newbie answer (sorry towr). I only put that question in, to start from somewhere. Even I - being a Newbie myself  - can calculate the probability of winning with 5 correctly guessed numbers (about 1:44 million), or with other (2-4) numbers.
But for all fairness, I would not spam this forum and waste your precious time with this if it were really so simple. That's why the other two questions.
Questions slightly rephrased:
Is the expected value of 1,2,3,4,5 or the last week's outcome is the same as any other combination?
Having answered this and looking the rules again, is there a positive expected value strategy?
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rmsgrey
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Re: Winning strategy in the 5/90 lottery
« Reply #3 on: Mar 14th, 2006, 11:30am » |
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In the UK National Lottery (pick 6 of 49, 50% of entry charge goes into prize fund, un-won jackpots roll over), there were at one point about 100,000 people playing 1,2,3,4,5,6
The net effect being: had the numbers come up, then each winner would be looking at less than 100 times their entry fee, rather than the usual several million
The expected return across all strategies is trivially 50% since if only one person enters, but buys all possible combinations, then they will get the entire prize fund - exactly 50% of what they put in
On the other hand, there are two things you can do to improve your expected return:
1) play unpopular combinations - particularly avoiding overlapping more than two numbers with 1,2,3,4,5,6 - the fewer people share your prize
2) Only play roll-over weeks - roughly doubles the value of the jackpot (a little less because other people are also preferentially playing on roll-over weeks) though that's obviously not going to double your expected return overall...
I suspect the only way to win overall though involves either a reliable crystal ball, or being managing director of the company running the lottery...
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towr
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Re: Winning strategy in the 5/90 lottery
« Reply #4 on: Mar 14th, 2006, 11:42am » |
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on Mar 14th, 2006, 11:15am, jollytall wrote:| But for all fairness, I would not spam this forum and waste your precious time with this if it were really so simple. That's why the other two questions. |
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There isn't enough information for anything else.
As rmsgrey points out, if a lot of people choose 1,2,3,4,5 Then the expected gain goes down. But if everyone thinks 5 consecutive numbers is nigh impossible and chooses something more seemingly random. Then its value goes up.
But without a demographic study, I wouldnt' hazard to guess what combinations people choose, and which thus I should avoid (You'd want the combination for which generally each subgroup is least chosen, to maximize winnings overal).
Still doesn't do anything for the chance of winning though.
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jollytall
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Re: Winning strategy in the 5/90 lottery
« Reply #5 on: Mar 22nd, 2006, 10:46am » |
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I'm a bit sad that this thread generated limited interest, although I think the topic deserves somewhat more. It must be because of my bad wording /I will work on improving my skills in that /. Let me try to summarise what I meant though:
Question 1 was really obvious and as towr clearly stated, if the draw is fair, the probability of any outcome is equal, including the strange ones, 1..5 or the last week's numbers.
Question 2 was intentionally left a bit open for discussion ("from a practical point of view") although Q3 included the hint "expected value", and in the rephrased question I used it again. From a practical point of view people are not interested in probability, but in expected value of any gambling. rmsgrey correctly stated that it does not only depend on the probability of the outcome, but also on how many people you have to share with. I.e. this was to note that while the draw is considered fair, the players' behaviour is far not.
Question 3 was supposedly the complex one. Again, rmsgrey covered the most important elements of it, but not fully.
The expected value of the lottery depends on three things. The probability of having 2,3,4 or 5 numbers met (Q1 killed this element), the number of people you have to share with and the total prize pool.
The number of people you have to share with depends on the probability others played some of the same numbers (since 80-100% of the prize pool is distributed for 2-4 found-out numbers, not only full match is important). There was a big study in Hungary analysing the tickets of years (not the result of draws but the individual tickets played, so the sample is much larger and it reflects player behaviours). They found that individual tickets depend on three factors: the frequency an individual number is choosen, the use of specific graphical combinations and some other special combinations. The most significant is the value of individual numbers. Supersticious numbers (1,3,7,13,21) appear much more often than others, numbers 1-12 (months) and 1-31 (day) are again very frequent. Combinations that are looking strange on the ticket (vertical lines, horizontal lines, diagonals) are also more frequent (the effect is much smaller especially for 2-3 matches), while finally complete matches to earlier outcomes is also more frequent than random combinations (but it is only important for jackpot). There is a cool site at http://www.sg.hu/lottotitka.php?proc=program based on the study, unfortunately in Hungarian, but it is pretty self explanatory. Mark five numbers and check how much you win with a 2,3,4 or 5 result. Since the probability of them is independent of the numbers choosen, the very significant difference between different numbers played is only because of other players. To maximise your expected value, you have to choose large, random numbers without any pattern. I do not have the figure for how it overall changes the expected value (it could be fun to figure out), but can definitely compensate a part of the 50% tax, etc.
The other is the size of the prize pool. Since there are about 43M combinations and 5M tickets (not random, some are repeated ones), the expected value of roll-on is about 8 weeks (this information was in the riddle but not used before), although we actually saw 40 weeks too. After 6 weeks (assuming the same number of players every week) the prize pool is 20% for 2, 10-10% for 3-4 and 60% for 5 numbers, i.e. 100% of the total amount played might be potentially distributed (later even more).
The last question left (not very difficult though): When does it worth joining the rush (taking into account only the roll-on of the prize pool)? On the 6th week, when potentially more distributed then played or only after the 48th week when even taking into account the 1:8 probability of the jackpot being paid out, the total expected pay-out is still more than 100% of the played amount?
All in all, my best estimate to get a positive expected would be: join after more than 6 but less that 48 weeks (e.g. 25-30 weeks) with large random numbers.
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| « Last Edit: Mar 22nd, 2006, 10:55am by jollytall » |
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Grimbal
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Re: Winning strategy in the 5/90 lottery
« Reply #6 on: Mar 23rd, 2006, 2:47am » |
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Thanks! A really interesting summary.
I once heard that when the Jackpot is high, the increase in number of players is more than the increase in the prize, making it it fact less interesting to play when there is a high jackpot that is well advertized. But it probably changes the odds only for those who play "predictable" numbers.
Else, I wonder if you should really only play large numbers. Those who play dates play numbers 1..12 and 1..31. Those who play "random, large" numbers play only 32..90. Wouldn't it make sense to play just one number in the 1..12 or 1..31 range to make a difference from all the people who don't play small numbers?
Last, if the probability to win the jackpot is lower than the probability to die on the way to buy the ticket, (and I believe it is), then it is probably better to stay home.
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JocK
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Re: Winning strategy in the 5/90 lottery
« Reply #7 on: Mar 23rd, 2006, 4:02am » |
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on Mar 14th, 2006, 8:26am, jollytall wrote:Is there a winning strategy in this lottery (in terms of positive expected value)?
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No.
An optimum strategy has the property that if all participants follow this strategy, it does not pay off for an individual to deviate from this strategy. The only strategy that satisfies this Nash equilibrium requirement is the simple strategy to select the numbers uniformly at random. However, this strategy would lead to a negative expected return.
The above assumes rational behaviour of all participants. In practice this is obviously not the case. So, the stated problem boils down to determining a winning strategy in the presence of irrational behaviours of (some of the) participants. To determine such a strategy would require one to understand the behaviours of the irrational participants. For that, analysing historical data indeed seems the right approach.
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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jollytall
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Re: Winning strategy in the 5/90 lottery
« Reply #8 on: Mar 23rd, 2006, 9:50am » |
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Grimbal, some interesting comments you raised:
I thought of the increasing jackpot vs. increasing number of players. If every week approx 1:8 is the ratio of players vs. combinations then cca 1/8 (what can be the exact number?) is the probability of someone hitting the jackpot, i.e. the expected total distributed prize is 40%+#weeks*10%/8, i.e. the break-even is at #week=48.
If on one week the number of players doubles then the probability of hitting the jackpot is cca 1/4 (obviously the function is not linear, but I assume in this range it is not too far). On that week the played amount is 2, the prize pool is 0.8+((#weeks-1)*0.1+0.2)/4, so if it happens on week 47 then it is at break even. 47 or 48 is roughly the same, so I do not see the effect to be significant in this range. If the increase is much more, then it is a different story (so later).
Regarding, using only large numbers at random, I am not sure. I would assume that since a large part of the return comes from 2, 3 numbers matched, it is more important what the probability a given number is played than the composition of the whole set of five numbers.
Actually entering 5 large numbers or 4 large numbers and #7 into the quoted link, I found that the latter was only half prize for the jackpot and also a significantly reduced prize for 2-4 matches.
Regarding staying at home or not: Since I value my life infinite higher than any prize or salary, I should not even go to work .
Otherwise I think the probability of dieing in an accident is roughly the same as the jackpot. Probably you take in average 5 trips a day that are like going to play lottery. If you are active for 55 years or 20000 days, than you have 100k trips in your live. According to the statistics somewhat more than 1/430 of the population dies in accidents, but a large part of those are on the highways and not inside the cities.
Consider, playing on-line too.
JocK, what you raise is tricky. In the original riddle I assumed the number of players is fixed every week. So after 48 weeks it gets Expected Value positive.
Obviously if we release this condition than it will be different. On the first 47 weeks noone should play (Expected Value is negative), so there will be no roll over ever.
Should there be 47 weeks with x players on each week, the cummulative jackpot makes it EV=0 for x players. On week 49 it is EV>0 for x players, i.e. more tickets should be purchased. But as we increase the number of players the break even period gets longer, or with other words there is an optimal number of tickets. Every week ever after the optimal number of tickets increases and keeps the EV=0.
But as shown in many cases large population does never react logically.
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JocK
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Re: Winning strategy in the 5/90 lottery
« Reply #9 on: Mar 23rd, 2006, 12:12pm » |
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According to the above website, selecting large prime numbers would have worked quite well. For instance, one would have faired well on the combination 47, 59, 73, 79, 83... (Hope I don't misinterpret the website...  )
It seems likely that this website calculates your expected winnings based on the frequency with which individual numbers have been chosen in the past. (I.e. assuming all other forms are filled out randomly by selecting the individual numbers with the corresponding weighted probabilities.)
Jollytal, did you experiment with what is the optimal combination?
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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jollytall
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Re: Winning strategy in the 5/90 lottery
« Reply #10 on: Mar 23rd, 2006, 11:16pm » |
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Jock,
I could not figure out the optimal one. The paradox in it is that if there would be a best combination, probably more than one player would play that, so it would not be optimal any more.
Two more interesting problems regarding your same question:
What would be considered the optimal combination? I would vote for max sum(Pi*Wi) where i=2..5 and Pi is the probability of having i numbers match and Wi is the corresponding prize money you can win. I guess this is mathemetically the correct answer, but as seen above, in this game statistics is far not the only element. I'm sure many players would not agree with me. Some might say, that they would rather play a game where they can win a large enough jackpot and a decent amount in lower classes (where you have a fair chance to win that), than a game where you have a lower prize in smaller classes and a many times larger jackpot (imagine an extreme game, where only the jackpot pays but pays the whole prize money). Actually it sparkled heated discussions when the current roll-on jackpot structure started. Earlier if there was no class 5 ticket, that amount was distributed in the lower classes.
You might say that if a combination is "good" then it is good in all classes, but playing with the system it seems not to be the case.
The other problem how can one find the optimal version using this site. One way is obviously to try the 43M combinations, but there should be a better way.
I also played around with your numbers. I do not think primes are important, since most of the players do not know what they are. I actually changed 83 to different large numbers. Oddly enough the prize money for 4 and for 2 changed the opposite direction. I should have calculated the weighted prize total, to see which one is the best, but I was lazy.
On the other hand I changed 83 to 7, when the prize money for C5 remained (still assuming one ticket in the jackpot class), but all other classes dropped significantly.
As said above, the site does not only take into account the frequency the individual numbers played, but also e.g. their relative position on the ticket. I tried numbers 79..83, when the system assumed 7 jackpots, significantly reduced C4 prize (all those probably who chose consequtive numbers overlapping with this period) but increased C3 and C2 prizes.
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rmsgrey
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Re: Winning strategy in the 5/90 lottery
« Reply #11 on: Mar 24th, 2006, 7:00am » |
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The other issue, of course, is how up to date the site is with its information - suppose there's a similar lottery which runs for 10 years, and has the combination 1,2,3,4,5 come up 9 years in. All the people who played 1,2,3,4,5 faithfully for 9 years will probably learn a harsh lesson and switch to other combinations ASAP, at which point the tracking site's 9 years of accumulated small numbers is suddenly horribly misleading, and liable to remain that way for the following year...
That's an extreme example, but any recent shifts or trends are liable to get lost in the mass of accumulated data, making the site much less useful for identifying the optimum combination...
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jollytall
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Re: Winning strategy in the 5/90 lottery
« Reply #12 on: Mar 24th, 2006, 3:24pm » |
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In theory: YES. But I seriously doubt that more than x% (where x<1) reads these kind of sites. The remaining 1-x% follow their instinct and so that does not really change the overall scheme.
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rmsgrey
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Re: Winning strategy in the 5/90 lottery
« Reply #13 on: Mar 24th, 2006, 4:53pm » |
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on Mar 24th, 2006, 3:24pm, jollytall wrote:| In theory: YES. But I seriously doubt that more than x% (where x<1) reads these kind of sites. The remaining 1-x% follow their instinct and so that does not really change the overall scheme. |
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It depends how restricted the set of "best" lines is when you analyse from the site's data. For example, less than 1% of entrants in the UK lottery play 1,2,3,4,5,6...
A good test for your analysis before you start putting money on it would be to try anticipating each draw's data on lines played...
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