4 ms·

$100 Invested in 100 $1 Lottery Tickets

17 points by magikbum 14y ago

anomonym 14y ago
haha -59% is a horrible rate of return. not an investment i'll be making any time soon.
JonLim 14y ago
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 14y ago
  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 ^_^
csense 14y ago
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 14y ago
  In most jurisdictions I think you can claim your losses against such taxes. Something you'd certainly want to consider beforehand.

17 points by magikbum 14y ago

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

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 14y ago
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 ^_^

csense 14y ago

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 14y ago
In most jurisdictions I think you can claim your losses against such taxes. Something you'd certainly want to consider beforehand.