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rec.puzzles |
To answer your question: Suppose that m people play, that each chooses The probability that any given n-number combination will be chosen (N-n)! 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 The probability that there are exactly W winning tickets is given by m! The probability that a given player will be selected at random to be So, the probability that there are W winners and that a given (m-1)! Putting it all together, the expected return to each player (which is m m (m-1)! This is the exact expected value. As m -> infinity the odds that b...@gibdo.engr.washington.edu (Bob) writes: Of course it matters. Certain numbers are more likely to be picked by
>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.
one n-number combination out of N independently from everyone else,
and that everyone chooses each n-number combination with equal
probability.
(out of N) by any given individual is
p = ---------
N!
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 binomial probability
b(W) = -------- p^W (1-p)^(m-W).
(m-W)!W!
one of the W winners out of m is W/m.
individual is one of them is b(W)*W/m, or
q(W) = ------------ p^W (1-p)^(m-W).
(m-W)!(W-1)!
the same as the expected value of any given ticket under the
assumptions made here) is
E = sum q(W) * J/W = J * sum --------- p^W (1-p)^(m-W)
W = 1 W = 1 (m-W)! W!
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