

$100 Invested in 100 $1 Lottery Tickets - magikbum
http://www.altinvestments.org/investments/23/100-invested-in-100-1-lottery-tickets

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csense
You should look at EV (expected value), not odds of winning.

Here are some cautions:

(1) A decent amount of the EV for lottery tickets is likely in large prizes,
and your return will be quite variable unless the number of tickets you buy is
about the same magnitude as the odds of winning the rarest prizes

(2) Win tables don't count taxes, but your actual winnings will be taxed if
large; so your effective EV (including taxes) will be less

(3) +EV lottery tickets are theoretically possible for "jackpot" lotteries
like powerball if you only play when the jackpot is sufficiently large, but
see (1) and (2) before you conclude it's a way to make free money.

~~~
jeza
In most jurisdictions I think you can claim your losses against such taxes.
Something you'd certainly want to consider beforehand.

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JonLim
Something I wish they took into account was diversity in locations. I'm not
sure it would matter, but buying ALL 100 tickets from the same store is
probably not the same odds as buying them in chunks from different stores.

My thinking is: lottery ticket producers will never produce 100 winning
tickets in a row, but they will most likely produce many losing tickets in a
row because they expect, at most, 3-5 being bought at once.

But I could be just out of my element.

~~~
csense
You could design an experiment to figure out whether manufacturers
deliberately don't print blocks of a certain length.

Suppose the probability of winning on a single ticket is p.

We suspect that blocks of k winners are never manufactured.

The null hypothesis (the skeptic's belief before we do our experiment and
gather evidence) is that each ticket is an independent random variable.

Assuming the null hypothesis, the following hold:

1\. Probability of buying a block of k tickets where all are winners is p^k
(where ^ is the exponent operator, double asterisk in Python, pow() function
in many other languages).

2\. Probability of buying a block of k tickets with at least one loser is thus
1-p^k.

3\. If we buy n blocks, the probability that all of them contain at least one
loser is (1-p^k)^n.

If our hunch is correct, how many blocks of k do we have to buy and scratch,
verifying each and every block contains at least one loser, to prove it to
confidence level c?

1-c = (1-p^k)^n

-> log(1-c) = n log(1-p^k)

-> n = log(1-c) / log(1-p^k)

For p=1/5, k=4, here are the numbers:

Confidence Blocks

90% 1438

95% 1871

99% 2876

99.5% 3309

99.9% 4314

So if 4-blocks are never produced, and tickets cost $1, you'd have to spend
roughly $6000-$18000 to prove it (the more expensive testing gives more
convincing evidence).

For 5-blocks, those numbers go up to $36000-$108000.

For 6-blocks, the numbers are $44000-$130000.

These numbers are initial outlay for the tickets, and don't account for the
fact that you'll have winnings which you'll use to recoup costs.

You'd have to use more sophisticated modeling to account for drawing-with-
replacement effects if you knew that manufacturers produce the same number of
winners in each print run. (I believe they actually do this.)

Chi-squared tests are a more sophisticated tool you might use, but they're
sort of a black box magic formula, whereas it's easy to understand the
fundamentals of the method above.

Our protocol is rather weak in that it only recognizes a rejection of your
hypothesis when a group of k winners is aligned on a k-ticket boundary in your
purchase history. By considering a better protocol which takes into account
unaligned groups of k winners, you could probably substantially reduce the
cost of the experiment, but I'm now out of _my_ depth ^_^

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anomonym
haha -59% is a horrible rate of return. not an investment i'll be making any
time soon.

