Ok, i cant find out what this is called.. in mathematics. so here goes. 

There is a raffle, or draw - to win an object. say the object cost 100 dollars.
an arbitrary number of possible raffle tickets are created - say, 250. one of
these tickets is randomly chosen the winner at the outset of the
raffle, and  the object is won if the player is sucessful in buying
the right ticket.

however, each time a player chooses incorrectly, the pool of possible
winning tickets is reduced, meaning the probability of winning is
increased for the next player.

the raffle co-ordinator wants to make the raffle ticket cost match the
probability of winning the raffle, as each player loses - until the
winning ticket is chosen. so, as the raffle draws onwards.. and the
prize isnt won, the tickets become more expensive. the co-ordinator
wants to earn a particular profit margin on each draw prize (say for
arguments sakes 50%), so on average he achieves the desired margin on
multiple raffles of objects. also, rather than have every ticket
separately priced to reflect the probability of winning the prize, the
co-ordinator would rather draw up price-bands based on the probability
of winning. so in this case, perhaps 10 price
bands, of 25 tickets each. once each price band is exhausted, the
player must purchase from the next price-band up. (Obviously this
means that tickets at the beginning of a price band are slightly less
favourable to buy than the tickets at the end of a price band).

Question 1: how does the co-ordinator calculate the cost of each
ticket? given the follwing data: 1. cost of prize, 2. number of
tickets to be created3. percentage margin required on the prize by the
co-ordinator. 4. the
number of price bands required by the co-ordinator.

Question 2: how many raffles, on average would the co-ordinator have
to create, in order to obtain, on average, the required margin on his
prizes, within a 5% variance?

Cheers!

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Clarification of Question by angurio-ga on 24 Mar 2006 22:27 PST

ok, let me clarify... its not a *normal* raffle. imagine that there is
a grid, of squares, say, (as above) 250 - each one has a piece of
paper on it. on one of the pieces of paper there is the winning
ticket.

the first player, has a 1-250 chance of winning, when they choose
thier ticket. once they have chosen, and lost... the next player has a
1-249 chance of winning and so on. its a game where, each players
action of purchasing a ticket reduces the odds for the next players.

thither-ga - yr quote: 

"Your (better) odds of winning when there are less tickets are
counterbalanced by the odds of having those picking ahead of you not
winning."

is not correct. if 247 tickets have been bought (unsucessfully) from
the grid... then there will be only 2 left. that is a 1/2 chance of
winning.. since the previous ones drawn were not the correct tickets.

All the raffles I have experienced avoid this problem by insisting
that all tickets are sold before the drawing takes place, which makes
the raffle fair to all at one price.
I know someone who won the car in two raffles because her husband
generously bought a lot of tickets to empty the ticket basket.

I'm sure a researcher will provide a comprehensive answer but I'd like
to point out that (unless I'm mistaken or overlooking something) that:

"however, each time a player chooses incorrectly, the pool of possible
winning tickets is reduced, meaning the probability of winning is
increased for the next player."

is not a true statement.

Image there were 5 raffle tickets:
1st person's odds of winning: 1/5
2nd person's odds of winning: (4/5)*(1/4) = 1/5
3rd person's odds of winning: (4/5)*(3/4)*(1/3) = 1/5
4th person's odds of winning: (4/5)*(3/4)*(2/3)*(1/2) = 1/5
5th person's odds of winning: (4/5)*(3/4)*(2/3)*(1/2)*(1/1) = 1/5

Your (better) odds of winning when there are less tickets are
counterbalanced by the odds of having those picking ahead of you not
winning.

This is the whole idea behind drawing straws to see who gets the
unfortunate task (or whatever).

I hope this helps.

Have a good day.

Angurio-ga,
No doubt than there may be a formula for what you propose, but it
presupposes that  the the winning ticket will be drawn towards the
end.  If you have already drawn the winning number in secret and
announce the winner when that ticket is selected, you have the problem
of an early participant winning, thus no one else buying tickets. 
Bingo!  The raffles is over, except for possible consolation prizes.

Your proposal cannot deal with this.  Theoretically, the ticket price
for the earlier participants would have to cover the cost of the
prize.

angurio,

While we really need a statitician to explain this properly...

In response to your clarification: Yes, if you came to me when there
were two squares left to choose, I would have a 50% chance AT THAT
SPECIFIC TIME of winning but I WOULD NOT have a better OVERALL chance
of winning than the 248 people who already chose, or the one who would
chose last if I didn't win. (capitals were just for emphasis, not
yelling).

I'm not sure I can say this any better, except that it is true - the
odds of me getting to that specific point where there are only 2
tickets left is very very small (it's very likely that someone would
have already won) that it counterbalances what would be my good odds
of winning at that point.

Think of it this way: Let's say 249 people picked and were wrong.
Sure, the last person has a 100% chance of winning AT THAT POINT. Does
that mean that his OVERALL odds of winning the raffle were 100%?
Obviously not.

The math -
Let's say there were x squares:

If I choose first, I have a 1/x chance of winning.

If I choose second, in order for me to win, the first chooser would
have had to be wrong (x-1)/(x) and I would have to choose right
(1)/(x-1). You multiply these to get the probability of this happening
and it is again 1/x.

If I choose third, in order for me to win, the first chooser would
have had to be wrong (x-1)/(x) and the second chooser would alse have
had to be wrong (x-2)/(x-1) and I would have to choose right 1/(x-2).
Multiply through and once again you'll see it's a 1/x probability.

And so on...



I hope this helps. I' sure I didn't explain it very well but I'm quite
certain that a mathematician would back me up on this.

Have a good day.

angurio,

Please disregard my previous comments - they were foolishly based on
the idea that people would be picking ahead of the drawing which you
clearly indicated wouldn't be the case.

I apologize for the mistake.

On average, the prize will be won when half of the tickets have been
purchased.  This being said, you should price the tickets at

Prize Value / (Total # of Tickets / 2) = Price per ticket

In this case:

$100 / (250/2) = Price
$100 / 125 = $0.80 per ticket

So after half of the tickets are purchased (on average this is when
the prize will be won), you will have collected $100 and the prize is
valued at $100.  It is very unlikely (1 in 250) that this middle
ticket will actually be the winner, but if you run this raffel a
million times, you will average $100 in ticket sales per raffel.