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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