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10-03-2005, 04:55 PM

http://www.ebtx.com/economy/doodah.htm

How much would each ticket have to cost to be profitable for the company producing it? What is the assumed typical variance on each card; meaning, how many card would have to sold a 'x' dollars to assure a 99% profitablity margin of win versus loss.

Whew, my math sucks, thanks in advance.

10-03-2005, 05:49 PM
Okay, as for my strategy is solving the minimum each ticket would have to sell for (pretending that the cost to manufacture the tickets are nill) was that I took a sample size of one million ticket sales, figured out the odds at each particular level and multiplied that by how much each person would make, added it together, and got the total payout....whew, that was one looong sentence.

For my 'imaginary' doo dah, I made the payouts as follows:

First set of squares: you scratch off your first square, there is no payout, but this is your 'winner square'

There are 10 more sets of squares with payouts as follows:

\$1 \$2 \$4 \$8 \$16 \$25 \$50 \$100 \$200 \$500.

At this schedule, each ticket would have to cost \$4.55 or so to be a zero sum game, no?

10-03-2005, 06:20 PM
Crap, I think I'm including duplicate winners.

For instance, the \$500 prize has odds of 1 in 1024.
With 1,000,000 players, I assumed 977 winners, for a
total payout of \$488,500, which means everyone that
played would have to play all the way to the end. But
most of the players would chicken out before the last
set of squares. How do I rid myself of these apparent
duplicate winners!?

LetYouDown
10-03-2005, 06:27 PM
Do you have a set of rules for how far people will go? Obviously if everyone stops if they win the first one, it's different than if everyone goes for broke.

Siegmund
10-03-2005, 06:34 PM
If you stipulate that you will only pay out tickets that have NO O's exposed ANYWHERE on the card, the value of a ticket is 50 cents.

Scratch the first space: it's either worth nothing, or it's worth \$1 now. Average of 50 cents.

Among tickets worth \$1, some will be turned in for \$1, while others will have a second space scratched, which will produce equal numbers of worthless tickets and \$2 tickets - but their expected value is not changed by scratching them, since they are equally likely to gain or lose a dollar in value.

Similarly, the \$2 tickets can either be cashed in for \$2, or double-or-nothinged again, which again does not change their expected value but does increase the variance.

----

BUT, on the example ticket, the implication is that as long as you have a series of Xes followed by one un-exposed square, the ticket is still cashable, even if there is an out-of-sequence O exposed. This causes a major change.

Suppose you have a ticket worth \$1 now. If you are allowed to expose the \$4 space risk-free, you can turn an extra profit. Half the time you will expose an O, and just turn in your ticket for your dollar. If you expose an X, you have a +EV gamble: scratch off a \$2 space, and either lose your dollar, or turn it into \$4.

How to maximize the value of a card if I am allowed to scratch out of sequence sounds like a fun way to spend the rest of the afternoon.

Siegmund
10-03-2005, 07:46 PM
Further to my previous post:

Suppose that a ticket is payable if it displays a sequence of X's, followed by one unscratched space, followed by anything else. (Or, of course, X's top to bottom.)

Scratch off the LAST space on the ticket.
If you reveal an X, scratch the previous space on the ticket.
If you reveal an O, skip over the previous space on the ticket, and scratch off the one before that.
Continue this process until you reach the top of the ticket.

A typical 14-space ticket, as pictured on the original site (except that he skipped from 128 to 512, which I believe is a typo), might be uncovered like this:

_____________X: so far so good
____________OX: OK, give up on the grand prize
__________O_OX: and on the thousand dollar prize
________X_O_OX: but the \$256 prize is looking good...
_______XX_O_OX
______OXX_O_OX: oops! Let's try for \$16...
____O_OXX_O_OX: oops again! Let's try for \$4...
__X_O_OXX_O_OX
_XX_O_OXX_O_OX
XXX_O_OXX_O_OX: win \$4.

Following this procedure, the value of a 14-space ticket with values doubling all the way is 20935/8192 ~ \$2.55.

More generally, let V(n) be the value of a ticket with n spaces marked for \$1, \$2, ... \$2^(n-1) prizes.

V(1)=1/2; S(0)=0

S(i) = S(i-1)/2 + V(i)
V(i) = S(i-2)/2 + 1/2

which approaches V(i) ~ i/6 + 2/9 for large i.
(For i=14, the exact value is 2.5555419 vs the asymptotic 2.5555555.)

Question is, would the public really pay more than that to buy one of these? There are lots of people who will fail to play anything remotely close to the proper strategy, of course... but there will also be a lot of people who produce fallacious arguments that the EV of all cards is \$1 or \$2. Perhaps a card with 10 spaces, whose true value is \$1.89, is the best compromise between big prizes and fair gaming.

10-03-2005, 09:38 PM
Dang, it didn't post my last response...which I did the math to produce the \$.50 value...as for the math that dictates a \$1.89 value (kudos to you, because I'd never be able to figure it out, never made it past pre-cal...lol). To me however, I think the sequential cards would be what would be more likely to be a 'real' ticket as the profit margin is much higher (think a ticket selling for \$3 and only being worth \$.50)...instant millionaire (\$.10 for the ticket and a small percentage to the convience store that sold it). So, a margin of over \$2 per ticket!

Hmm, on a side note, I wonder what the typical profit margin is on a 'real' scratch off? As I don't play them, I haven't a clue.

10-03-2005, 10:04 PM
[ QUOTE ]

which approaches V(i) ~ i/6 + 2/9 for large i.
(For i=14, the exact value is 2.5555419 vs the asymptotic 2.5555555.)

[/ QUOTE ]
I don't quite follow:

Let's say that the card has an unmarked value of \$1.
Then 1 empty square has a value of \$1
Two empty squares allow you to skip 1 so you get:
\$1 1/2 of the time
\$4 1/4 of the time
\$0 1/4 of the time
For an average of \$1.5

Now, let's say we have 3 empty squares.
Then if you mark off the last square, you get
\$1 1/2 of the time (The last mark is an 0)
\$1 1/4 of the time (The last mark is an x, and the second to last is an X)
\$0 1/8 of the time
\$8 1/8 of the time
For a total of \$1.75

Now, if we have 4 open spaces
1/2 of the time we get \$1.50 (the last spot is an O so this is equivlent to two spaces)
1/4 of the time we get \$1
1/8 of the time we get \$1
1/16 of the time we get \$0
1/16 of the time we get \$16
For an average of 1+3/4+3/8=2 1/8

With 5 we get:
1/2 of the time is the 3 case for 1.75
1/4 of the time is the 2 case for 1.50
1/8 of the time is the 1 case for 1
1/16 of the time is the 0 case for 1
1/32 is missed for 0
1/32 is hit for 32
So a total of 7/8+3/8+3/16+1=39/16=2 7/16

Wit 6 we we hit
1/2 of the time it's the 4 case for 17/8-&gt;34/32
1/4 of the time it's the 3 case for 7/4-&gt;14/32
1/8 of the time it's the 2 case for 3/2-&gt; 6/32
1/16 of the time it's the 1 case for 1 -&gt; 2/32
1/32 of the time it's the 1 case for 1 -&gt; 1/32
1/64 is the jackpot for 64 -&gt; 32/32
For a total of 89/32 -&gt; 2.78... which is clearly larger than 2.555

I expect that this will turn out to be unbounded.

Siegmund
10-04-2005, 12:46 AM
[ QUOTE ]

I don't quite follow:

[/ QUOTE ]

You came very close to following - you've gotten the same sequence of numbers I have, just with a factor of two and offset of one difference from my squence. (My sequence begins 1/2, 1/2, 3/4, 7/8, 17/16, 39/32; yours begins 1, 3/2, 7/4, 17/8, 39/16.)

[ QUOTE ]

Let's say that the card has an unmarked value of \$1.
Then 1 empty square has a value of \$1

[/ QUOTE ]

I understood the original question to say that a card with one X on it was worth \$1 - asking us to calculate a fair price for the sale of an unmarked card.

Your answers for 1, 2, 3 ... empty spaces are equivalent to the answers I would get for a card that started with 2, 3, 4... spaces and has already had an X exposed on line for the \$1 wager. These happen to be exactly twice the values you get if you have to gamble on actually exposing that X on the first line.

[ QUOTE ]
For a total of 89/32 -&gt; 2.78... which is clearly larger than 2.555

I expect that this will turn out to be unbounded.

[/ QUOTE ]

The 13th term in your sequence will turn out to be \$5.11, corresponding to my value of \$2.55 for my 14th term. Your 14th term should be \$5.44.

And yes, it's unbounded -- but no, the EV doesn't increase exponentially with the number of spaces on the ticket; it goes up linearly, since the doubling of prize sizes and halving of chances of winning sequences exactly cancel out, and we just add one new winning sequence for each new space on the ticket.