Wikipedia:Reference desk/Archives/Mathematics/2008 February 8

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February 8

Ever any lotteries with an expected value greater than the ticket price?

I live in the UK and I like to think I'm rational. While it could be fun to gamble, I am only prepared to gamble where the expected value of the lottery is greater than the ticket price. Does this ever happen with any lottery available in the UK please? Unfortunately it seems to be impossible to get hold of the statistics to calculate this (and possibly the lotteries are set up to make this impossible - I'm not sure). I've no idea how many "rollovers" it would take. 80.0.102.226 (talk) 00:02, 8 February 2008 (UTC)

Yes, it occasionally happens. To make it simple, let's consider just the main (UK) Lotto game, and just the jackpot. The probability of a given ticket winning the jackpot is in fact 1 in 13,893,816. So if the predicted jackpot is in the region of £14 million or above, it makes mathematical sense to buy a ticket (the National Lottery website gives the predicted jackpot for the next draw - it has to be a double rollover to get to that kind of figure). Incidentally, always buy your ticket on Friday afternoon or Saturday - if you buy it any earlier in the week, you're more likely to be dead by Saturday evening than to win the jackpot! AndyofKent (talk) 02:51, 8 February 2008 (UTC)

But any calculation of payout would have to include the possibility that more than one person wins the jackpot, thereby splitting the amount of prize money. If you assume that each person chooses their lottery numbers independently and randomly, then presumably you could estimate this as something like (# of people buying tickets) / (# of combinations), but as many people have "methods" of selecting lottery numbers, if you can find a method of your own that selects a less-popular combination, you will greatly increase your expected payout. Given that the average lottery is designed to make more money than it pays out in jackpots, it would require fairly specific circumstances to have such an occurence. Confusing Manifestation(Say hi!) 03:52, 8 February 2008 (UTC)

Plus, even when this does happen, the expected utility of buying a lottery ticket is still negative. —Keenan Pepper 04:45, 8 February 2008 (UTC)

If the utility is negative, how come millions of people buy tickets? 80.0.121.236 (talk) 12:23, 8 February 2008 (UTC)

The utility is only negative if you base utility solely on net worth. By that argument it would be even more irrational to go to the movies, where the ticket price is higher and the chance of winning the jackpot essentially nil. For many people the attraction of a movie is that it lets them forget their own lives for a while and vicariously live the life of someone else who has amazing adventures and does great things. I think the people who buy lottery tickets do it for the same reason. Instead of £5 for a few hours' fantasy they spend £1 for a week's fantasy. It's a cost-effective way of satisfying a basic human need, and I think it's perfectly rational, though also somehow sad. -- BenRG (talk) 12:57, 8 February 2008 (UTC)

If you ignore rollovers, the expected value of a £1 ticket is 50p, simply because that's the amount that goes into the prize fund (the rest goes to good causes, admin, tax and, of course, profit), and all tickets have an equal chance of winning. So you expect to lose half your money. For the expected net gain to be positive, you would need enough rollovers for the total prize fund to be double what it would normally be - that's a pretty easy calculation to check, I think, as they give estimated prize funds and jackpots, so it's just basic arithmetic once you've found the right numbers. --Tango (talk) 11:32, 8 February 2008 (UTC)

There are also more minor prizes in addition to the jackpot - these would affect the expected value a lot. Very good point that the expected value of a ticket must be 50p on average for the Lotto - is it the same for Euromillions? And are these the only UK lotteries that have rollovers? And what is the normal prize fund - I think, maybe I'm wrong, that the lottery TV rollover adverts only quote the forecast jackpot size, rather than the total prize fund. I'm not sure if extra ticket sales would be reflected in a bigger jackpot for that draw or if the extra sales money goes to make the prize-money in the next draw. And do the smaller prizes vary in value, or are they fixed? Sorry, so many questions. 80.0.121.236 (talk) 12:13, 8 February 2008 (UTC)

I don't know about anything other than Lotto. With that, the prize fund for any given week is 50% of the total ticket sales for that week, plus any rolled over jackpots. Extra ticket sales due to there being a rollover would be included, but would actually reduce the expected net gain, since the effect of there being more tickets out there is more significant than the extra prize money. I haven't really watched the lottery for a while, but I know they used to give both the total prize fund and the jackpot fund before the draw - I would assume both numbers are available somewhere in advance. The only fixed prize is £10 for 3 numbers, all the others are done in terms of percentage - I think it's x% split between all the people that get 4 balls, y% split between all the people that get 5 balls, etc., with the x, y... being fixed (and published on the official website somewhere, as I recall). --Tango (talk) 14:03, 8 February 2008 (UTC)

"Extra ticket sales due to there being a rollover would be included, but would actually reduce the expected net gain, since the effect of there being more tickets out there is more significant than the extra prize money." Does this mean its impossible in practice to get an expected value greater than the ticket price? I'm curious to see any statistics that demonstrate that rollovers decrease the expected value due to the increase in ticket sales. Does anyone know where I can find the stats to calculate the expected value - I need a) total prize fund, and b) number of ticket sales - both figures now seem to be kept secret. 80.3.47.25 (talk) 23:23, 8 February 2008 (UTC)

The extra tickets will reduce the expected gain, but not by as much as the rolled over jackpot increases it, I would expect. Rollovers are still better value, I would think. --Tango (talk) 13:34, 9 February 2008 (UTC)

Actually, that's obviously the case, I don't know what I was thinking when I wrote that. It can't decrease it by more than the rollover increased it, since the reason it's decreasing it is because the extra value is shared between more people - the extra value per person can obviously never go below 0.--Tango (talk) 00:00, 12 February 2008 (UTC)

It isn't enough for the expected value to be greater than the ticket price. Imagine if, for one pound, you could by a ticket that would automatically win a £1.01 prize. Would you do it? Would you go through the work of buying a ticket for one measly pent? Also, AFAIK it's common for lotteries to pay in increments. It's generally excepted that money later is worth less than money now. — Daniel 23:45, 8 February 2008 (UTC)

It isn't enough for the cost of the ticket to be less than the chance of winnings multiplied by the winnings amount. The number of people playing becomes an important part too, as its more likly you will have to share the prize. I know that in Australia on new years eve there is usually a $31 million dollar draw. The amount you get for winning is only about 2 million because of the huge numbers of winners. While lotteries of $9 - $13 million have average wins of 4 to 5 million. Unfortuently most places don't post the number of people who entered, you have to go through there public tax records and work out how many tickets they must have sold etc. The best way to win lottery is just to pick a sequence that is unlikely to be picked by people because its seemingly rediculous like 1 2 3 4 5, that way if you win, you have more chance at being the only winner.--Dacium (talk) 22:26, 10 February 2008 (UTC)

You can quite easily find out the number of tickets sold - take the total prize fund and divide by how much of each ticket goes into it (after minusing off any rolled over jackpots). As for the best way to win the lottery: Don't play. I win 50p twice a week on the lottery using that method. (Incidentally, I think I read somewhere that a very large number of people buy a ticket with 1,2,3,4,5,6 on the UK lottery each week, so if it ever comes up they won't win much at all.) --Tango (talk) 23:44, 10 February 2008 (UTC)

$\lambda$ calculus

I was reading the article on lambda calculus, and it appeared to me that it is used almost exclusively for programming. Is this true, or does it have real mathematical applications? Thanks, Zrs 12 (talk) 03:27, 8 February 2008 (UTC)

Definitely has "real mathematics" applications. The lambda calculus predates electronic computers, and was used by its inventor Alonzo Church and his students to derive important results in mathematical logic in the 1930s. Although our lambda calculus article says it "can be thought of as an idealized, minimalistic programming language", this is a reversal of the actual historical development - programming languages such as Lisp were designed around the lambda calculus, which already existed. So it would be more accurate to say that programming languages are implementations of the lambda calculus. Gandalf61 (talk) 07:49, 8 February 2008 (UTC)

I think you've engaging in a bit of historical revisionism there. Lisp definitely was not designed around the lambda calculus — I'm not sure McCarthy had even heard of the lambda calculus when he started working on Lisp. The driving idea behind Lisp was symbolic computation. At the time electronic computers were seen as replacements for human computers, and the idea that programs might work with words instead of numbers was new. All of Lisp's other innovations were accidental byproducts. The lambda keyword was introduced to the language very early on, but Lisp did not at the time contain anything resembling Church's lambda calculus. It used dynamic scoping, not by choice but because nobody involved understood scoping. It had destructive update. It had statement labels and goto. It had car, cdr, ctr and cpr. The funarg problem was noticed by an outsider. McCarthy saw it at the time as a bug, not the fundamental design flaw it actually was, and Steve Russell fixed it with a trick called the "funarg device", now known by the less embarrassing name of "lexical closures". Lexical closures are a great implementation trick, but unfortunately they've permanently contaminated the semantics of the language because of their interaction with destructive update of bindings. Church understood the trickiness of lexical scoping, but McCarthy's team doesn't seem to have benefited from that. I don't think Church's work had any influence beyond the name lambda, which has pointlessly frightened generations of students. They should have called it function. It wasn't until lexical scoping finally became standard in Lisp that you could reasonably claim that it even contained the lambda calculus as a sublanguage, and it's still a major hassle to program in that style in Common Lisp; you have to write #' and funcall all over the place and you can't rely on tail recursion. The language was pretty clearly not designed for that kind of programming. Even Scheme's embedded lambda calculus isn't very faithful to the original, with its eager evaluation order and lack of automatic currying (and logical inconsistency, but that's a different story). I'm sorry to rag on Lisp so much; I actually like it a lot, but it gets romanticized these days in a way that has little bearing on reality. Historically speaking, Lisp is very thoroughly in the Perl school of "design". -- BenRG (talk) 19:17, 8 February 2008 (UTC)

Very interesting. The second paragraph of our Lisp (programming language) article says "Lisp was originally created as a practical mathematical notation for computer programs, based on Alonzo Church's lambda calculus" - so maybe it needs some expert attention to correct what appears to be an urban myth. Gandalf61 (talk) 11:09, 9 February 2008 (UTC)

Why does mathematics use the idea of 'truth'?

I don't get why mathematics uses the idea of statements being 'true' or 'false' rather than just being 'consistent' or 'unconsistent' or 'unattainable' commpared with the system in question.

What's truth got to do with it? What's truth but a second-hand emotion? —Preceding unsigned comment added by 212.51.122.27 (talk) 17:28, 8 February 2008 (UTC)

Well, to a Platonist, mathematical objects exist just as other objects do, and mathematical statements are statements about these objects which are true or false just as statements about regular objects are. From the point of view of other philosophical positions (such as formalism), 'true' is a defined term like any other which is used because it is useful. Results such as Gödel's completeness theorem tell us that the semantic notion 'true in every model of a theory T' coincides with the syntactic notion 'provable in T', and so to a certain extent we can equate truth with provability (is this what you mean by being 'consistent' or 'unconsistent'?). I have no idea what you mean when you call truth a second-hand emotion. Algebraist 17:46, 8 February 2008 (UTC)

What's Love Got to Do with It (song) --LarryMac | Talk 17:49, 8 February 2008 (UTC)

I think "true/false" and "consistent/inconsistent" are different. "True" means "provable" (these kinds of statements will always be of the form "A=>B", it's generally meaningless to say simply "B is true" in maths, although we don't always explicitly state the assumptions). "False" means "the negation is provable". "Consistent" means "does not contradict", it's not necessarily true, but there is nothing stopping you assuming it's true (for example, the axiom of choice is consistent with the ZF axioms, you can't use them to prove it or it's negation [ie. it's independent], so it's not "true" or "false", but you don't get a contradiction if you simply assume it's true). "Inconsistent" means you do get such a contradiction - A is inconsistent with B if A=>C and B=>not C for some C. That doesn't say anything about the truth or A, B or C, just their relationship with each other, a priori, either A or B can be true (given appropriate assumptions), just not both. --Tango (talk) 18:05, 8 February 2008 (UTC)

No, "true" does not mean "provable". The most important take-home message from the Gödel incompleteness theorems is that you cannot identify provability and truth (at least without losing properties of truth that we would like it to have, such as excluded middle).

Pre-Gödel formalists tried to identify provability with truth. Gödel showed that that sort of formalism was non-viable. Post-Gödel formalism simply does away with the notion of truth. The only major school that still identifies provability with truth is intuitionism, but it gets away with it by declining to fix a single formal notion of provability. --Trovatore (talk) 18:15, 8 February 2008 (UTC)

While you are of course correct, I feel I ought to defend my post by pointing out that modern formalists still admit truth (as a defined term) in the sense of true-in-a-model, just not in the sense of true-in-a-theory. Algebraist 18:31, 8 February 2008 (UTC)

I think this is getting a bit off track. Head-in-the-skies philosophy is important, but keep in mind, there is such a thing as being actually true. It's hard to dispute a winning Nim strategy, or the fact that every square is a sum of consecutive odd numbers. One artificial construction may be as good as another in some sense, but that doesn't change what's real. (This is not, BTW, Platonism. I'm not saying circles "exist", I'm saying that what exists exists.) Both those examples are discrete math with small numbers, which is the area of math most easily accepted as not made up. a(b+c) = ab+ac, no matter who you are. Black Carrot (talk) 18:40, 8 February 2008 (UTC)

Small detail - the name of the link references the scrollover text of the image. Black Carrot (talk) —Preceding comment was added at 18:41, 8 February 2008 (UTC)

BC, you make a good point, but it's surprisingly hard to draw the line. Where do you put Goldbach's conjecture, for example? The probabilistic evidence for it is extraordinarily convincing, yet it's entirely possible that it will never be proved from any "foundationally relevant" theory. That's a case where, if there were a counterexample, it could be checked by computer. If you're willing to take that as part of "reality", then what about the twin prime conjecture, where there's no such thing as checking an example or counterexample by computer? Before you know it you're having Woodin cardinals for breakfast and expressing your opinion on the truth or falsity of the continuum hypothesis before lunch. --Trovatore (talk) 18:45, 8 February 2008 (UTC)

Oh come on, it's not a slippery slope. There are obvious limits. I'm not saying the line is easy to draw, or even that there could ever be a single well-defined boundary, just that it's silly to paint everything with the same brush. Some things are confusing and uncertain, sure, but most of them are things nobody's heard of and nobody cares about. Number theory is uncertain, but arithmetic isn't. Black Carrot (talk) 18:58, 8 February 2008 (UTC)

By the way, you can't use probabilistic evidence for the Goldbach conjecture, no matter how you slice it. Within the range of numbers the vast bulk of humanity actually uses, it's been proven beyond reproach. Beyond that, it hasn't been shown probable at all. Black Carrot (talk) 19:00, 8 February 2008 (UTC)

No, that isn't true; the probabilistic argument is not simply the absence of a counterexample within numbers that can be checked. It's the fact that the bigger the numbers get, the harder it is for there to be a counterexample, because you have more primes available from which to make sums. This argument is so convincing that in my opinion the truth of the Goldbach conjecture is settled, to the point that a formal proof would not change its epistemic status much -- after all, the theory from which you prove it could also turn out to be inconsistent.

What I'm saying about the "slippery slope" is that it's extremely hard to do anything in mathematics without accepting, in one way or another, the real existence of (at least some) abstract objects. And once you've done that, the case against even-more-obviously-abstract objects is harder to make. --Trovatore (talk) 19:17, 8 February 2008 (UTC)

I don't find that argument convincing, though. As certain as it may be that almost all of them are sums of pairs of primes, even that almost all of them are sums of pairs of primes in many ways, that doesn't rule out, or even make less likely, that one or a few somewhere up the line won't be. It's not a matter of how many of them there are, it's how they fit together, and that's what's hard to prove.

Yes, it's hard to draw a line, but that's true of anything. In any area of study, there are some things that are too convincing for argument and a whole lot of fringe ideas that are just silly, and everything in between. I'm just saying that you don't have to ignore one end of the spectrum in favor of the other. Black Carrot (talk) 19:27, 8 February 2008 (UTC)

I think it does make it less likely. So much less likely that, as I say, the question is effectively settled. A counterexample to Goldbach would require an unbelievably massive "conspiracy" of prime numbers "avoiding" summing to the counterexample. A similar conspiracy resulting in a proof of 0=1 from Peano arithmetic is just as believable.

I think there is no "silly" end to this spectrum, because full-on hardcore realism about large cardinals is just fine, and it allows you to decline to draw an arbitrary line somewhere in the middle. --Trovatore (talk) 19:33, 8 February 2008 (UTC)

I have no problem with the prime numbers conspiring. They've done it before, and they'll do it again, they're fickle like that. There's nothing fundamental to my view of the world that says they're well-behaved. You might view them differently. Peano arithmetic, though, is abstracted barely an inch and a half from its roots in bean-counting. If there's something wrong with that, the universe is broken. The only claim it makes that isn't obviously true is that there are infinitely many whole numbers. If a paradox comes out of that, fine. Maybe there's a cap on how far you can push the pattern, but for sufficiently small numbers it's impossible for it to be wrong.

Every crackpot has a reason. I think most people would agree that a system with more patches than an old sweater, that reasons glibly about imaginary objects nobody could have conceived of a hundred years ago, is a bit fringy. I'm not saying it can't be justified philosophically, in the same way that you can justify any crime by a sufficiently extreme plea to moral relativism, I'm saying that that's what it takes to justify it. It doesn't take mental gymnastics to find a foundation for arithmetic. Black Carrot (talk) 20:07, 8 February 2008 (UTC)

There are no "patches" in set theory. That's a misunderstanding of the antinomies. The antinomies did not come from an informal notion of set, but from the wrong informal notion (the conflation of the extensional and intensional concepts). --Trovatore (talk) 20:15, 8 February 2008 (UTC)

Fair enough. I'm sorry, bad example. I think you should reread the discussion that prompted my first comment, though, and compare it to our discussion since then. I don't think you actually disagree with what I was trying to say. I was only arguing for a more balanced viewpoint. Especially in front of the OP, who needs it. Black Carrot (talk) 21:15, 8 February 2008 (UTC)

There is a big difference between finding a counterexample for something that all evidence points towards being true and finding a counterexample for something that's been proven true. One is very unlikely, the other is impossible, that being what "proven" means. By the standard definitions of 0, 1 and equals, $0\neq 1$. The only way you could change that would be to change the definitions, and that's moving the goalposts. --Tango (talk) 21:26, 8 February 2008 (UTC)

No, sorry, Tango, you're laboring under a common misconception here. Just because something has been proved does not imply that we know apodeictically that it is true. Possibly the assumptions we used to prove it were false. In the case of sufficiently simple arithmetic statements like Goldbach, for the assumptions to prove such a statement true even if it were actually false, the assumptions would have to be more than false--they'd actually have to be mutually inconsistent. But we do not know beyond all doubt that that doesn't happen. --Trovatore (talk) 21:30, 8 February 2008 (UTC)

Word. Could you point me to a good source on the words extensional, intentional, and antinomy? Our articles on them suck could be more detailed. Black Carrot (talk) 21:35, 8 February 2008 (UTC)

I'm afraid I'm caught short here -- can't think of a good source at the moment. --Trovatore (talk) 21:38, 8 February 2008 (UTC)

But mathematical theorems always take the form "A=>B", the assumptions are, by definition, true, because that are assumed to be. If those assumptions don't actually hold for what we intuitively think of as numbers, then the theorem isn't very useful, but it's still true. The statement explicitly said "Peano arithmetic" - that Peano arithmetic implies $0\neq 1$ is not disputable. That Peano arithmetic is an accurate description of what we intuitively know about arithmetic is possibly disputable, but that's another matter. --Tango (talk) 22:38, 8 February 2008 (UTC)

No, that's the shallow sort of formalism, and is quite demonstrably wrong. Peano arithmetic does not define what is true about the naturals; it's a collection of statements that we believe are true about the naturals, and from which we can derive others that -- assuming we were correct in the first place -- must also be true. But maybe we were wrong in the first place.

I agree that Peano arithmetic implies 0≠1, but that isn't the question -- the question is whether it implies 0=1. It does not follow, merely because it implies the first, that it does not imply the second. --Trovatore (talk) 22:53, 8 February 2008 (UTC)

Has Peano arithmetic not been proven consistent? If not, then I need to do some more reading on the subject before continuing this discussion. --Tango (talk) 23:10, 8 February 2008 (UTC)

Ok, after a quick bit of reading, it seems it depends on what you actually mean by "proven consistent"... I think this is all getting a bit too deep for this time of night... (it's 2313hrs here). --Tango (talk) 23:13, 8 February 2008 (UTC)

Depends on what you mean by "proven". Sure, it's been proved, but not without assumptions. Maybe those assumptions are wrong.

Gödel's second incompleteness theorem is taken by some to eliminate all hope of proving that PA is consistent "by finitistic methods", as Hilbert wanted. However there has never been a good definition of "finitistic", so this is a little hard to pin down absolutely. As I recall Gödel himself disclaimed this interpretation of his theorem. From the other side of the question, Gentzen proved that PA is consistent by analyzing possible proofs of a contradiction and performing induction -- but it was transfinite induction up to a certain infinite ordinal number. Taking the two results together, the status of Hilbert's second problem is particularly muddled -- has it been resolved positively, resolved negatively, or is it as yet unresolved? --Trovatore (talk) 23:16, 8 February 2008 (UTC)

I can't help thinking of "That that is is that that is not is not that that is is not that that is not that that is not is not that that is is that not it".  :) -- JackofOz (talk) 23:23, 8 February 2008 (UTC)

Power reduction writ large

Does anyone know of a relative of the addition theorem on circles rather than spheres? In particular, I'd like to have the coefficients in the expansion
$P_{l}(\cos \theta )\equiv \sum _{j=0}^{l}a_{j}\cos j\theta$
The extension to $\cos(\theta -\phi )$ would then be easy — apply the sum formula to each term and get a different sum of cosines and a sum of sines. Of course, one way to approach this would be to apply the power reduction rules repeatedly, but I'm not sure how (or if it's even always possible) to write $\prod _{i=1}^{n}\sin ^{p_{i}}j_{i}\theta \prod _{i=1}^{n}\cos ^{q_{i}}k_{i}\theta$ as a sum of only first-power sines and cosines. --Tardis (talk) 20:46, 8 February 2008 (UTC)

It occurs to me that it would be sufficient to know how to evaluate $\int _{0}^{2\pi }\cos(j\theta )\cos ^{k}\theta \,d\theta$ for $j,k\geq 0\in \mathbb {I}$. Any thoughts on that? --Tardis (talk) 05:12, 9 February 2008 (UTC)