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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

by 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 ^_^

by JonLim 14y ago

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 ^_^