Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site utcsstat.UUCP Path: utzoo!utcsstat!anthony From: anthony@utcsstat.UUCP (Anthony Ayiomamitis) Newsgroups: can.politics Subject: Lotteries Message-ID: <2228@utcsstat.UUCP> Date: Fri, 12-Jul-85 14:45:56 EDT Article-I.D.: utcsstat.2228 Posted: Fri Jul 12 14:45:56 1985 Date-Received: Fri, 12-Jul-85 15:21:18 EDT Organization: U. of Toronto, Canada Lines: 60 In <5784@utzoo.UUCP> henry@utzoo.UUCP (Henry Spencer) writes: > It is worth remembering that probability-based arguments indicating that > lottery tickets are a losing proposition have a serious flaw: the > sample set is not large enough to be a statistical universe. If I buy Probability based arguments use what mathematicians/statisticians call expected values - i.e. what is the AVERAGE return given a decent number of chances. Expected values are completely independent of the sample set since Exp Value = sum (x times p(x)) where p(x) is the probability that event x will occur and x is simply the event's value (for example, $). Thus, if some proposition has an EV of 10 units and it will cost 5 units to participate, given completely random conditions, I should expect that in all likelihood I will come out ahead. Similarly, if I were to participate in such a lottery daily, I will expect to be nicely in the black at some future date (the longer away this "future date" is the better I can be sure of my being in the black!). With lotteries, the converse is of course true (i.e. EV <<< cost of partic.) since they must make money to be worthwhile to the organizers. > one $10 ticket for a lottery that has a $1M grand prize and nothing else, > and spends 1/2 of its take on the prize, my expected return is *not* > $5. It is "zero or $1M". Speaking of an average expected return is not No way!! Average expected returns mean that if you were to continously bet, IN THE LONG RUN, you will be averaging the expected value as your winnings. To compute the expected return, you have to know the probability of winning the $1M. For your return to be "zero or $1M" you are assumming that the probability of winning is 0.000000... or 1.0! In fact, since your example has only one prize, you are guaranteed that the expected return will be between $0 and $1M (and the probability of winning will determine how close you are to $0 or $1M). > $5. It is "zero or $1M". Speaking of an average expected return is not > meaningful for a single ticket. This is what makes lotteries interesting... It is meaningful in the sense that it tells you the odds of getting a return. If the expected value for Wintario is 99 cents, you would expect some sort of return almost every time out (even with 1 ticket). Conversely, if the expected return is 0.01 cents, whether you buy 1 or 100 tickets, don't bother worrying about expecting any winnings. IN ALL LIKELIHOOD, you will win nothing in the latter case. What makes lotteries interesting is the possibility of YOU being the 1 out of x thousands that gets the big pot only. However, an average expected return is as meaningful for one ticket as it is for 10,000 tickets since the expected return is not based on the number of tickets bought but the PROPORTION (in one sense) of the ticket's value. Thus, if the expected return on Wintario is x cents, whether you buy 1 or 10,000 tickets, your average return is going to be x cents per dollar. By buying more tickets the only thing that you are doing is increasing the chances of winning at least one prize. > Henry Spencer @ U of Toronto Zoology > {allegra,ihnp4,linus,decvax}!utzoo!henry -- {allegra,ihnp4,linus,decvax}!utzoo!utcsstat!anthony {ihnp4|decvax|utzoo|utcsrgv}!utcs!utzoo!utcsstat!anthony