I just added one more thing here, to maybe give some insight. Keep in mind that, as far as the lottery is concerned, the numbers are really just symbols to represent the balls. “1” could be replaced by a smiley face, “2” could be replace by a clover,…, “25” replaced by a diamond, and so forth. Thinking of it that way, it becomes clear that no mathematical pattern of numbers will have any relevance to the probability of that combination being drawn; they’re just drawing a collection of symbols.
Almost, but not quite, correct. If there were only one prize handed out, for the whole lottery, this would be correct. But then, nobody would play, because (virtually) nobody would ever win. Most lotteries also give out smaller prizes for matching smaller sets of numbers, e.g. ten bucks for getting three of the numbers or what-may-have-you. You have to add up the expectations of all the smaller prizes, too…
Generally, though, as the big jackpot goes up, so proportionally rises the number of tickets bought, so the added expectation generally amounts to precious little, perhaps a dime a ticket or less. Of course, there are exceptions. I mean, the jackpot may be $80 million, but your odds of winning it all by your lonesome, with not a single other soul having the same number combination, plummets pretty quickly.
What if you played the top numbers that came up in the last 200 lotteries? Some lottery mags keep track of them. I think the odds would be more in your favor.
Most of you talk about the probability theory. It’s still just a theory, not a law. If it were a law, then my idea would not give you a better chance.
Well, the mechanical ball picking systems MAY have some underlying tendencies that MAY be evident over time, i.e. certain balls may be lighter or whatever. Certainly if the number 8 cmae up every time that would be worth knowing. But even a truly random sample will show some “false” tendencies resulting from being in the tails of a normal distribution. Just like you may flip a fair coin a hundred times and get 70 heads. Or an even better example flip a hundred coins a hundred times and almost certainly one or two will be 70% heads or so, but that would not necessarily mean they are not fair coins.
Bear in mind, as well, that the lotteries themselves have a vested interest in keeping the game fair. If there were any sort of real bias in their number-picking scheme, the professional gamblers and number-crunchers of the world would be down on them like an anvil and they’d go out of business in a hurry. Most lottery corporations that use the billiard-ball approach, and there are a lot of them, have several sets of balls and intermix them frequently (ball #1 from this set, #2 from that set, #3 from the other set…).
Casinos actually do the same thing; if there’s anything mechanical that is used to generate odds (a roulette wheel, a pair of dice at a craps table, a ‘wheel of fortune’ or anything else), you can bet your mortgage that it has interchangeable parts, and that the parts are interchanged, frequently, like every day or two. There’s a story of a German gambler who made it rich on a roulette wheel in Monte Carlo, because the wheel was biased towards a certain number (not a lot, but just enough). The casino swapped around some parts that night, and the game was balanced again, or maybe just crooked in a different enough way that it was too tough to figure out from one day’s worth of data. And they’ve been swapping parts ever since.
So keeping track of what numbers have shown up most often recently is pretty much an exercise in futility. It’s usually the lottery companies themselves that publish these statistics, because many gullible souls believe there’s some sort of pattern in it and that they can ‘beat the system’. You can rest comfortable knowing that the lottery companies are not about to publish any information that would help you beat the system. And, if the system can be beat, there’s someone in Las Vegas with a big mother of a computer right now, working on beating it. And even he doesn’t find very many loopholes. Beside him, you don’t have a chance.
Remember my question?
Well, today I ran it by a smart computer scientist/programmer, and he confirmed my intuitive feeling. I am dumb, buy I try to repeat what he said.
Picking any set of numbers is less probable than picking any random numbers. The probability of picking any ball out of 50 is 1/50th. But if the task is narrowed like: “Pick only a number divisible by 7”, the probability would be 7 times less, because now you have onlky 7 balls to pick from, not 50! So, it becomes 1/50 : 7 = 1/350. And the overall probability, assuming that the ball is returned and the order is irrelevent, would be 1/350^6 for 6 balls.
I can’t calculate the actual probability for 1,2,3,4,5,6 sequence(in any order), but these numbers are in order, and, therefore, harder to pick.
The smart ass told me that the problem is beyond plain ststistics. Its a cross of statistics and something else, which is in effect, is a theory of order and chaos.
In general, a chance of something being in chaos (random) is much higher that a chance that something being in order.
Are any mathematicians there? Yahoo!
I don’t play the lottery, nor is this at all mathematically relevent. However, that in mind, here’s why I wouldn’t bet on a “lucky number”: once you picked a lottery number as your lucky number, you’d feel obligated to play it every day. If by some coincidence your number did happen to come up, imagine your disappointment if you didn’t buy a ticket that day. If you want to play, do quick-picks…that way, you’ll never really “lose” by not playing.
The first Pick 3 draw in Connecticut after TWA flight 800 (it was 800, wasn’t it?) was 800. They lost $7 million, if I recall correctly, on that one draw. You lottery addicts are a morbid lot.
I wouldn’t be so sure–economists generally don’t play the lottery, except when the expected value of the ticket is greater than the cost. I’m sure there are plenty of mathematicians who are as smart as economists. And you.
Speaking of which,
I don’t know about pretty quickly. An $80 million jackpot doesn’t generate sales like it used to. It’s exceedingly unlikely that more than 80 million tickets are sold for the Big Game or PowerBall when the jackpot is $80 million. When the jackpot gets to $160 million, it’s even more unlikely that 160 million tickets are sold (that would be one ticket for roughly three-quarters of the adult population). Thus, the expected value still remains above $1.
Even if it falls slightly below $1, that’s where the value of the dreams kicks in, for me anyway. I always go for the quick-pick, or whatever they call it, to minimize the chances of having to split the pot. Obviously no guarantees, though.
Sheesh, and here in Australia I’m surprised when the jackpot goes as high as 20 mil… must be the lower population, right? Less buyers = less payout?
And given the exchange rates, that only comes out to about $11 million USD. The jackpot is based on the number of tickets sold; the reason Powerball jackpots regularly go so high is that it’s a multi-state lottery. Ergo, tons mo’ money in the pot. This is also why the jackpot increases precipitously after it reached a certain point (after it hits $100 many more people buy tickets, quickly pushing the payout ever higher).
If this is anything close to what your friend told you, he is stunningly poor at mathematics for a guy you say is a really smart programmer. No offense; I include an explanation below.
Your friend knows nothing about statistics. The odds of picking a 7 out of 50 numbers is 1 in 50. It can’t be 1 in 350.
The odds of picking 7, 14, 21, 28, 35, and 42 out of a 50-number lottery are not 1/350^6. They’re 1 in(50)(49)(48)(47)(46)(45) divided by 6!, or 720, assuming you can pick them in any order (e.g. 7-21-42-35-28-14 is just as good.) You’re friend’s answer is about 1 in 1.8 quadrillion; the REAL answer is 1 in 15.9 million.
I don’t know where this “1 in 350” comes from; the number 350 has no bearing on this. Take it through a thought exercise:
You need to get 7, 14, 21, 28, 35, and 42 out of 50 balls.
The odds of getting any one of those six balls on the first draw are 6 in 50.
The odds of getting any one of the five remaining balls on the second draw are 5 in 49, so the odds of doing these first two things are 30 in 2450.
The odds of getting any one of the four remaining balls on the third draw are 4 in 48. So the odds of doing these first three things are 120 in 117600.
The odds of getting any one of the three remaining balls on the fourth draw are 3 in 47. So the odds of doing these first four things are 360 in 5,527,200.
The odds of getting either of the two remaining balls on the fifth draw are 2 in 46. So the odds of doing these first five things are 720 in 254,251,200.
The odds of getting the remaining ball on the sixth draw are 1 in 45. So the odds of doing all six of these things, consecutively, are 720 in 11,441,304,000.
720 in 11,441,304,000 is 1 in 15.9 million (roughly.)
Funny how that worked out. As others have pointed out, the numbers on the balls are just symbols. They could as easily be pictures of the St. Louis Rams, or playing cards, or the names of SDMB posters.
I can. The odds of getting 1,2,3,4,5,6 in a 6-out-of-50 draw are 1 in 15.9 million.
The odds of getting 2-4-6-8-10-12? 1 in 15.9 million.
The odds of getting 5-10-15-20-25-30? 1 in 15.9 million.
Of getting six of your relatives’ birthdays? 1 in 15.9 million.
Of getting any one specific sequence of unrelated numbers? 1 in 15.9 million.
ROFL! He’s been reading “Jurassic Park” too much.
He’s wrong. Trust me; I have done more studies on the lottery than I care to mention. If your buddy does programming, have him write a program that randomly selects six balls out of a pool of 50 with a really, really, REALLY good random number generator and watch as 1,2,3,4,5,6 comes up just as often as 3,11,12,25,32,36.
My boss once asked me to print out the possible drawings of the “Cash 5” game in Virginia. There are 278,256 of them! He thought he had some system that would be facilitated by having a printout.
I created an ASCII text file of them and imported it into WordPerfect. It was going to take a lot of paper, so I reduced the font size to 3 point and made it newpaper-column format. The first page crashed the memory on my poor little printer.
So then I changed tactics: I wrote the combinations to an AutoCAD script file and plotted the results to our big plotter. Its memory crashed. Then I wrote a new font that was as simple as possible. Success! So I plotted out his results on 26 sheets of paper that were 3 feet wide and 12.5 feet long (limit of our plotter). His eyes popped out on Monday when he saw his output. :D:D
I feel caught between you two,
Rick. This is your logic:
You need to get 7, 14, 21, 28, 35, and 42 out of 50 balls.
The odds of getting any one of those six balls on the first draw are 6 in 50.
It looks to me that the odds of any one ball out of 50 is 1/50th. The odds of getting ball #7 is much lower. Do a mental experiment. It’s easy to get ANY ball out of the drum. It will take you some time to produce ball #7, although chances are that you hit it at the very first try. Right?
peace, if you’re still confused I suggest you reread RickJay’s post, several times if necessary. His analysis is both clear and correct.
You guys will never win this way. These players are good because they’re sucking up the good luck from their numbers. You need to pick the numbers of crappy players.
BTW, there’s a site on the web giving away 1 BILLION dollars. It’s free to enter.
That’s the good news. Here’s the bad.
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They pay it off over 40 years in unequal payments (Starting at $5 mil).
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You can only enter once.
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You have to get on a junk mail list to enter.
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There’s a 1 : 2.4 billion chance you’ll win (Technically 2.404808340, but who’s counting?)
You guys have gone down the probability rat-hole and are in a dead end. You need lateral thinking.
The big question at the root of this is, how can I win the lottery? Aside from any issues of probability on any one number, you need a strategy that will give you the winning number. And there is precisely ONE strategy that will give you the winning number every time.
In order to pick the winning combination of numbers, you merely need to purchase one ticket for EVERY set of numbers. I haven’t worked out how many possible different tickets there are in a given lottery (i.e. Powerball) but someone did, and calculated he could buy one ticket for every set of number with something like 12 million dollars. Then you just have to get people together to help you jam your 12 million ticket purchases through lottery counters across the state. As long as the jackpot is above $12mil, you are guaranteed a profit. A group of people actually raised money to try this and won the jackpot. Then they tried it again, and someone ELSE also picked the jackpot and they had to split it. It didn’t pay off as richly, they barely covered their bet. Then the lottery office changed the rules so ticket sales took more time, delaying mass purchases, and making the scheme ineffective.
There are two problems. One is that the number of tickets you need to buy are the same as the odds to win. It’s been a while since I looked at the Powerball or Big Game odds (which were increased last year, by the way), but it’s in the neighborhood of 1 in 80 million (likely much lower for most state lotto games, but still 1 in 40 million or so). The reason those are the odds is that there are 80 million (or however many) possible combinations, and each ticket obviously only contains one set of numbers.
The other problem is that you are not guaranteed a profit because, as you note, the jackpot can be split. And of course, if the jackpot gets large enough to make this worthwhile, many more people play, increasing the chances of a split.
I have often thought, though, that the thing to do (if you have a load of money), would be to try to cover half of the numbers. If you did this repeatedly (with enough money), and quit while you were ahead, I think you could do rather well. I suppose this could work with Pick 3 or Pick 4 as well–get every number from 1 to 500, or every even number, or whatever, and, though on average you’d break even (except for all the other types of plays that I know nothing about), you could put together a string of winners and come out ahead. If you can withstand the string of losers before you get there.
No. The odds of getting ball #7 are 1 in 50.
The odds of getting ANY ball out of the drum are 50/50, unless I’m so weak I can’t pick up a ping pong ball. 
The odds of getting a particular ball out of the drum are 1/50. That’s the odds for #7, #11, #33, or any other ball. #7 is just as likely as any other number… 1/50.
Let me see if I can help.
Peace, do you agree that you have a 100% chance of getting one of the balls if you reach in and grab one? Assume a fully functional human body for this portion of the experiment.
If you agree that you are definitely going to get one of the balls when you reach in, then that ball has to be one of the 50 balls. You can’t reach into a can of 50 balls and pull out a ball that wasn’t there.
So, if the odds of getting #7 are less than 1 in 50, what are the odds of getting #6? #37? They all must be identical to the odds for #7, and less than 1 in 50 by your logic. Let’s say the odds are 1/350, just for fun. The odds of getting any of the other balls must be the same, because how can one ball be favored over another? So, the chances of reaching in and grabbing ball #23 are 1/350. The chances of reaching in and grabbing ball #37 are 1/350. Add all the chances up, and you have a 50/350, or 1/7 chance of getting any ball.
But, you agreed (I hope) that there is a 100% chance of getting a ball if you reach in! Therefore, your assumption of the chance of getting #7 must be incorrect.
Does that make any sense to you?