Setting aside idea of playing a lottery, if you insist on playing it at least do it properly. Don't choose any pattern of numbers as you are exposing yourself to having to share the win with other people. Don't choose a date as it is more probable somebody will also chose a valid date than a completely random set of numbers. Don't choose digits from your favorite irrational number, for the same reason.
Ideally, choose numbers that no other person is likely to choose. Not exactly random. Maybe generate a random sequence and check if you can see a pattern, do a search on the Internet, etc.
This all makes sense, except for "Don't choose digits from your favorite irrational number"... You can choose whatever combination of numbers you want, and find it in your favorite irrational number if you search long enough :)
> This all makes sense, except for "Don't choose digits from your favorite irrational number"... You can choose whatever combination of numbers you want, and find it in your favorite irrational number if you search long enough
You're not reading it right. You don't disqualify digits for occurring consecutively in the expansion of an irrational number. You disqualify digits if the method you used to pick them was to think of an irrational number and extract some of the digits. That is a method that other people can also use.
Yeah I guess I don’t. Why don’t you explain it to me? You seem to be quite a smart guy who picks up on everything. Please tell me more about how you are right and I am wrong.
> Please tell me more about how you are right and I am wrong.
Sure. So far you've claimed that (1) the smile at the end of rfonseca's comment should be viewed as good-natured and not mocking. This is not true of such expressions in general, but you've also claimed that (2) you have special knowledge, external to the thread, indicating that (1) is true.
You haven't bothered to support either claim, except to the extent that (2) supports (1); (1) is far from certain and (2) seems extremely unlikely, so unlikely that I surmise you didn't realize what you were saying.
The reason is that there isn't that large number of named/popular irrational numbers and you are likely to choose some digits close to the start (say within couple thousand digits). Good chance somebody else will do the same. I know personally two people who do this (chose consecutive digits from an irrational number).
We can prove that almost all irrational numbers are normal, which is a quite large set. It's just when we pick out a particular irrational number that we have difficulty proving normality (unless that irrational number was intentionally constructed to be normal).
Indeed not every irrational number has this property, for instance you could create an irrational number using only the digits 1 and 2, but some irrational numbers do have this property.
Until I did some research a few minutes ago, I thought this property was the irrational number being a "normal number"; however, that is not the case. That all said, a Normal number definitely has this property and I don't think having this property implies normality, but I don't know either way.
An interesting fact though is that almost all real numbers are normal, which means pretty much every irrational number has this property, though not every irrational number. However, we still don't know if pi, e or square root 2 are normal.
> I don't think having this property implies normality, but I don't know either way.
Having this property cannot imply normality. Imagine an irrational number z which has this property, and another number z' constructed from z by taking the first 1 digit of z, appending 1 "2", appending the first 2 digits of z, appending 2 "2"s, appending the first 3 digits of z, appending 3 "2"s, and so on.
Using e as an example, our z' would begin 2.7 2 71 22 718 222 7182 2222 71828 22222...
z' is irrational and shares the property that every sequence of digits can be found in its decimal expansion. But it is obviously not normal; over half of its digits are "2".
Interesting, my first inclination is to be skeptical that the property would still hold, since we're "potentially" slicing up a necessary sequence and adding 2s in the middle.
However infinities are weird* and I think you could construct a proof by contradiction making use of the fact that a sequence of N digits is embedded in infinitely many longer sequences most of which that won't have been broken up by the inserted 2s.
* I'm always skeptical when dealing with infinities and probabilities. Human intuition doesn't gel well with either concept.
Shoring up places where I thought my proof was weaker:
- I rely on the assumption that an irrational number with this property exists. (If it didn't, then the property would imply normality.) This is easy to fix; Champernowne's constant has this property and so the assumption is valid.
- I assert without proving that z' is irrational. We can prove this using the definition of a rational number as one whose decimal expansion repeats after some index. Since z is irrational, somewhere in its decimal expansion there is a digit not equal to 2. (Otherwise, every digit of z would be 2, and z would be the rational number 2/9.) Since z' successively repeats larger and larger stretches of z, this suffices to show that, for any index i into z', there is a higher index j > i such that the jth digit of z' is not 2.
- But we also know that a sequence of n "2"s in a row can be found within z' for any positive n. Assume that the jth digit of z' is not 2. Since we know that a sequence of 2j "2"s occurs later within z' -- it can't occur earlier because not enough digits have yet occurred -- any cycle in the digits of z' cannot yet have begun by index j.
- But since there is no maximum index into z' beyond which all digits are not 2, a cycle in the digits of z' cannot have begun at any index into z'. This shows that z' is irrational.
> Since we know that a sequence of 2j "2"s occurs later within z' -- it can't occur earlier because not enough digits have yet occurred -- any cycle in the digits of z' cannot yet have begun by index j.
This is wrong -- the cycle might begin at j and continue into a huge series of 2s.
But we cannot yet have completed one cycle by index j, and this property can be extended -- there is no index into z' at which one cycle could have been completed, and hence the digits of z' never cycle.
A much simpler proof that there is no cycle goes like this:
Suppose there is a cycle of length n. We know that a stretch of 2n consecutive 2s will appear after the beginning of the cycle. This implies that the cycle consists entirely of 2s. We also know that a non-2 digit will appear after the beginning of the cycle. This is a contradiction; there cannot be a cycle of any length.
> my first inclination is to be skeptical that the property would still hold, since we're "potentially" slicing up a necessary sequence and adding 2s in the middle.
No. If the sequence you want occurs between places a and b of z, then it is a substring of the full sequence between places 1 and b of z, and all such sequences are included within the expansion of z'. (Going by example again, if you're interested in the sequence that occurs between decimal places 41,028 and 315,001 of z, then that sequence will occur within the part of z' that repeats places 1 through 315,001 of z.)
> However, we still don't know if pi, e or square root 2 are normal.
The math is somewhat beyond me - at least, without digging into the formal proof - but my understanding is that we can prove that pi cannot be represented as a ratio of two integers, and therefore cannot have a finite decimal representation.
We know all these numbers are irrational. That's been known for a long while. Normal numbers are something different, however.
You can look at the wikipedia definition[1], but that involves a few levels of definitions that I don't know, like density, but the gist is that every sequence of N digits occurs with equal frequency to every other sequence of N digits in the expansion of the number.
The definition is complicated due to dealing with infinity and multiple bases.
But that said even with this superficial understanding we can see two things:
* Rational numbers can never be normal, since the digits repeat after some period. (Just choose a sequence of numbers longer than the period and you can easily construct a sequence that doesn't appear)
* Normal numbers contain every sequence of N digits in their decimal expansion. So if we prove pi is normal (like we believe it is) then we know somewhere in its decimal expansion we can find any sequence of digits we want. Which is the property this comment[2] was referring to.
There's a fallacy in here somewhere. Let's say you pick a common sequence, and you win but have to share the winnings with several people. You say "Darn, I should have picked different numbers!" But then you would have won nothing.
Before the draw every set of numbers has the same probability of coming up. Your expected win differs on the uniqueness of the numbers you choose though, so you might as well choose unique ones.
Yes, of course you can say ex post that you should have picked the numbers that won, but that's not really useful.
Your comment implies you know which numbers are likely to be drawn, which you don't. When every number has an equal chance of being drawn and you don't know what is going to be drawn, it makes more sense to pick numbers that are less popular and thus you're less likely to share the winnings.
Also,instead of buying $10 in a weekly lotto every week, you save your money and wait for it to cross a threshold of value and then buy using the saved money.
50 entries into a $70 million lotto is better than 1 entry, 50 times onto an average lotto of $2 million. The expected payout of vastly bigger.
I've done the math on my local lotteries and I only buy tickets when the EV of the return is greater than the cost of a single ticket. Even then it's only a small amount of tickets. The odds may be in my favour but they're still really small to win.
> I've done the math on my local lotteries and I only buy tickets when the EV of the return is greater than the cost of a single ticket.
If you want to get really technical, you also need to consider the number of people playing. When the jackpot gets extra high more people tend to play, so the likelihood of having to split a jackpot (which happens all the time) becomes higher. Without doing the math I would guess EV of the return is rarely ever greater than the cost of a single ticket.
I factored that in when I did that calculations. How often are their multiple winners and when there are multiple, how many. A surprising amount of the return comes from the non-jackpot prizes, 58% or something. You'll see crowds lining up to buy tickets and yet no winner for a few weeks.
What the maths actually says is your $10 is more likely to make money if you don't do the lotto at all.
If you want to make money gambling then you need a game where you play against other people (like poker) rather than against the house because the house will always stack the odds in its favour.
The thing is, his maths isn't correct. Once upon a time some lottery jackpots would occasionally exceed the cost of buying every ticket. But then crime rings cottoned on to that and bought thousands of tickets. Now lotteries take that probability into account and cap the highest rollover amount. Some even have schemes where if there isn't a winning ticket the prize jackpot trickles down to the next highest tickets -- which is a great way to sell massively more tickets and keep the jackpot low enough that it doesn't exceed the ticket price.
Because I don't have the $50 million laying around to do it. The EV is only in my favour when I factor in the small prizes PLUS odds of main jackpot / value of main jackpot.
It doesn't take into account splitting the jackpot. Maybe the expected value is positive if you're the sole winner, but not when 3 other people picked the same numbers as you.
This is the reason why, rarely when I play, I avoid the quickpick and struggle to pick random numbers on the spot. They usually are in increasing order so I am aware I am a bad random number picker. Still, I’d rather share some win than nothing, so in the end it doesn’t make much of a difference
You might be right, but until i see the algorithm they use I’ll continue to have my own reservations. Their incentive to issue similar numbers and if not hit, the chances to increase the jackpot make a whole lot more people pour money into the game. The house always wins regardless of what numbers come out but if jackpot goes up the pool grows exponentially and so does their profit
Funnily, my math teacher always said that if she would play the lottery, she would just play the number 1, 2, 3, 4, etc. as nobody plays them and they have equal chances of being picked. I guess she was wrong.
How can a math teacher make such an obvious mistake? She's right it's picked less often than all other sequences combined, but that's not the same as being picked less often than every other sequence individually.
Best lottery advice I heard from a math teacher was only make bets that have a chance at a retire-early sized prize and preferably no chance of small winnings. The idea being that when he loses, he's donating to charity, which he would have done anyway.
In most of the US lottery proceeds go to schools. Not sure about other areas.
Of course, the fallacy here is that dollars in budgets are fungible, and when lotteries are established municipalities often redirect the same number of dollars away from the schools into whatever pork barrel projects they like.
That way you have your cake (technically the lottery money does fund schools) and can eat it too (in reality it funds other stuff under the table).
> In most of the US lottery proceeds go to schools.
I'm well aware. A government-run school system, however, is NOT a charity. And it's an enormously inefficient way to contribute to a cause -- more than half of the teacher's "donation" is kept by the lottery (distributed as prizes, vendor fees, admin costs)
Nor does much of the money actually impact kids. Most lottery-based education funding in the US is either misleading, or simply replacing (rather than adding to) other funding sources. For example, in New York:
“People think the money is going strictly for education, like for books, or schools, or to pay teacher salaries, but it’s not,” DiPietro told 2 On Your Side.
According to DiPietro, the money on occasion has been “pinched off” by the state, to pay for a variety of items, including attorney’s fees for construction projects and even to pave roads near schools.
“They could say there are school busses that are going to drive on this road so the spending would be ‘education based’ when it’s really not, to me that’s a stretch,” he said.
Oh yeah, the creative reallocation of school money is a problem. But this is the society we live in now; a significant part of the population believes that public schooling is a charity.
The drawing of these numbers is under question, but you would expect a significant proportion of players to select sequences such as this, even if it's not in their best interests to do so.
you have to share. A single person couldn't pull this off. A bunch of lottery officials got together and picked these numbers. They had to pick an obvious sequence, because it would not be credible for 20 people to pick a non-obvious sequence.
But why would they choose this incredibly specific set of numbers? If you can influence the draw, why influence it in the only way that riles up every human looking at the result? Why not just have it pick something that looks random?
Any combination of numbers has the same probability of showing up in a fair lottery. Why would you choose something like 1 2 3 4 5... that has basically same chance of happening but almost guarantees you will have to share the pot with a bunch of other people?
yes, but that's not the right framing of the problem.
if all outcomes are equally likely from the machine, but humans are more likely to choose numbers with personal significance.
Therefore more tickets will be sold with 1-31 in them vs. higher numbers, and the expected value of your ticket will therefore be higher if you include numbers above them.
No, you don't. If it's a split with one other person, you get 2/3 instead of 1/2, that's only 33% more. If it's a split with two other people, that's 2/4 instead of 1/3, which is only 50% more. etc. and always less than twice as much.
And in all the other cases where it's not a split (either you lose or you win alone), you paid twice as much.
Ideally, choose numbers that no other person is likely to choose. Not exactly random. Maybe generate a random sequence and check if you can see a pattern, do a search on the Internet, etc.