b
...@gibdo.engr.washington.edu (Bob) writes:
>thomp
...@atlas.socsci.umn.edu writes:
>| By the way, since the odds of getting all six right in any order are 1
>| in 15,890,700, a ticket is worth more than one dollar only if the
>| jackpot goes above $15,890,700. (And this is true only if no one else
>| holds the same number.)
> What is the general equation to calculate the expected value that
> takes the number of people playing and the possibility of sharing
> the prize into account?
> In other words, calculate the expected value of a $1 ticket given
> a prize of $P, where n numbers (6 in this case) are chosen out of
> a possible N numbers (50 in this case) and there are m people
> playing.
> Assume that everyone picks randomly or that you pick your ticket
> randomly (does it matter?).
the general population precisely because some people use "systems"
that supposedly identify favored numbers. This introduces
correlations among the numbers selected by different people.
To answer your question: Suppose that m people play, that each chooses
one n-number combination out of N independently from everyone else,
and that everyone chooses each n-number combination with equal
probability.
The probability that any given n-number combination will be chosen
(out of N) by any given individual is
(N-n)! n!
p = ---------
N!
Now we can forget about N and n and work with p instead.
A ticket is assumed to pay J/W if it wins and nothing otherwise, where
J is the size of the jackpot (assumed fixed) and W is the number of
winning tickets. Obviously W is an integer between 0 and m inclusive.
All tickets have equal likelihood of winning. So the lottery is
equivalent to a procedure where (a) the number of winning tickets W is
determined first, and (b) the actual winners are then selected
randomly from among the players.
The probability that there are exactly W winning tickets is given by
the binomial probability
m!
b(W) = -------- p^W (1-p)^(m-W).
(m-W)!W!
The probability that a given player will be selected at random to be
one of the W winners out of m is W/m.
So, the probability that there are W winners and that a given
individual is one of them is b(W)*W/m, or
(m-1)!
q(W) = ------------ p^W (1-p)^(m-W).
(m-W)!(W-1)!
Putting it all together, the expected return to each player (which is
the same as the expected value of any given ticket under the
assumptions made here) is
m m (m-1)!
E = sum q(W) * J/W = J * sum --------- p^W (1-p)^(m-W)
W = 1 W = 1 (m-W)! W!
This is the exact expected value. As m -> infinity the odds that
there will be more than one winner given that there is at least one
winner becomes very small. So E is very well approximated by p*J for
large m, which is what the expected value would be if prizes were
duplicated for each winner instead of shared. The issue of multiple
winners only makes a substantial difference if the number of players
is small or if players' selections are correlated.
--
T. Scott Thompson email: thomp...@atlas.socsci.umn.edu
Department of Economics phone: (612) 625-0119
University of Minnesota fax: (612) 624-0209
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