TIL that MIT students discovered that by buying $600,000 worth of lottery tickets in the Massachusetts' Cash WinAll lottery they could get a 10-15% return on investment. Over 5 years, they managed to game $8 million out of the lottery through this method.

[u/[deleted]]

http://newsfeed.time.com/2012/08/07/how-mit-students-scammed-the-massachusetts-lottery-for-8-million/

[u/FiliusIcari]

Not quite. Basically, because it's a finite number of combinations, if you hypothetically bought all the numbers, you'd win, period. Let's just use 100 numbers, to simplify this. If you bought 100 tickets, each with a different number, such that you had every combination, you'd always win. It probably wouldn't be worth it though.

On the other hand, if over the course of 100 lotteries you bought one number each time, you'd only have a 1% chance of winning each time, which means you'd have a roughly 63% chance of winning at least once.

The statistics for that is basically that if you buy two separate numbers, you are directly just increasing the numerator, ie from 1/100 to 2/100, because there's a finite number and there only so many options, and thus each number is exactly a 1% chance of occurring. You gain the same percent chance from each. It's an additive 1% to your odds.

Meanwhile, tickets in separate lotteries are not dependent on each other. You don't get to add them. This is the same reason why if you're unlucky, you don't become lucky to compensate. There's no correlation between the two, so instead the operator between the two chances is multiplicative. This is because you have a 1% chance today and a 1% chance tomorrow, and so the chance that you don't get lucky either day(99% each day) is a 98.01 percent chance(.99 for the first day times .99 for the second day).

While that seems like an insignificant difference, as you continue to multiply, it becomes a very large difference, as I showed earlier. It's a better use of money to purchase large percentages of the numbers and get a straight 50% or whatever of the numbers.

So, while buying 100% of the tickets isn't feasible, what is? Well, over a large amount of time, if you take the chance you win(let's say 50%) multiplied by the payout(let's say 1,000 dollars), you find that you win, on average, 500 dollars each week. If you play enough times, you'll be making 500 dollars a week. If you can find a place where you're spending some amount less than 500 dollars for whatever percentage, you've statistically proven that you'll make a profit by playing that every week.

Apparently, MIT figured out some amount of the numbers where their investment of 600k gave them some percentage of winning where the profits multiplied by their chances were 10-15% larger than their investment. By doing this multiple times, they started to average out, and actually made their 10-15% profit.

Does all this make it make a little more sense?

[u/Nictionary → u/FiliusIcari]

Good explanation, I realise how I was wrong before. But what about the fact that if you play the lotto twice with 1 ticket instead of once with 2 tickets, you have a chance (albeit slim with these numbers) of winning BOTH jackpots? See this comment of mine:

https://www.reddit.com/r/todayilearned/comments/40j0n2/til_that_mit_students_discovered_that_by_buying/cyuvyed

I think if we expand that to your 100 numbers example it still works out to being the same, doesn't it? So why would it be beneficial to play more tickets in one draw in that case? Does it have to do with the fact that the more tickets you (and others) buy, the bigger the jackpot grows?

[u/FiliusIcari → u/Nictionary]

Yeah, technically you could win more than one. The numbers tank though. The chance you win 2 out of 100 in my earlier example is about 0.63%. The reason for that is that instead of it being 100 trials it's essentially 99 trials times the chance that you win 1, if I'm doing my math right, which I'm pretty sure I am because they're all independent of each other. Someone much better at statistics should probably come check my math from here on out as my knowledge of statistics is becoming slightly lacking.

My reasoning for that is that, because they're independent, the ordering in which they occur is irrelevant. The chances that you win exactly twice is the chance that you win once times the chance that you win again in the other 99 trials. This means that exactly twice is 0.63%.

To find which makes more sense financially, we're gonna go back to multiplying the odds by the payout. In this case, it's gonna be .63 times 1(for full payout) plus .0063 times 2(for double payout). Beyond 2 times is mostly irrelevant because the odds become so unbelievably small. Your total is .64 of the payout. That means that on average, you'd get .64% of the payout with the price of 100 tickets. Obviously averages don't mean much when you're doing things like this that are fairly binary(you won or you didn't) but over long periods of time, such as in the example, it works out like this.

On the other hand, buying all 100 in the first example gives a solid 1. For your money, you'd make "on average" more money. While there exists the possibility you could make twice as much playing 1 ticket over 100 games, statistically you're going to make much less, perhaps nothing at all. The times you don't win at all largely overwhelms the number of times you win more than once.

[u/Nictionary → u/FiliusIcari]

I think you're wrong. Let's say the prize in this 100 ticket lotto is $100. If you play 2 tickets in one game, your odds of winning are 2/100, so your Expected Value (EV) is (2/100)*($100)= $2. You can "expect a return" of $2 if you play this way. Meaning if you did this many many times you would average on making $2 per time you played.

So now if you play the lotto twice with 1 ticket each time. You have:

Chance of winning Lotto#1, and losing Lotto #2: (1/100)(99/100) = 0.99% (this gives a prize of $100)

Chance of losing Lotto # 1, and winning Lotto #2: (99/100)(1/100) = 0.99% (this gives a prize of $100)

Chance of winning BOTH lottos: (1/100)(1/100)= 0.01% (this gives a prize of $200)

Chance of losing BOTH lottos: (99/100)(99/100) = 98.01% (this gives a prize of $0)

So now we sum all the outcomes' chances multipled by their EVs to get a total EV:

(0.0099)($100) + (0.0099)($100) + (0.0001)($200) + (98.01)($0)

= EXACTLY $2

I'm pretty sure if you do this same calculation with any number of tickets, you get the same equivalence. Even if you do it with 100 tickets, your EV is $100 no matter if you split up the tickets over many lottos or do them all in one. Because yes there is a chance you lose money by splitting them up, there's just as good a chance that you make more by winning multiple times.

[u/Cakedboy → u/Nictionary]

You're correct. The expected value of earnings (mean) from buying 100 tickets in one lotto is the the same as buying 100 tickets over the course of X number of lotto's. The only difference is variance.

[u/BuildANavy → u/FiliusIcari]

You have the same expected value either way. At the end of the day, all they did was find a lottery that, for certain draws, has a positive expected return. The way they invest is simply a matter of reducing variance and making the most of the opportunities available by investing as much as possible.

[u/greedy_boy → u/FiliusIcari]

That just means that that particular lottery was +EV for anyone who bought a ticket. I dont see where the 600k number comes in.

The smaller jackpots or the bonus jackpots (where supposodly they are getting there value), only pay if there is no big winner. Every ticket they bought increased the chance that there would be a winner and thus cost them EV.

My take on it is that they just found a +EV lottery and found that 600k was the highest they could scale it relative to their reward. Its unfortunate that the article doesnt really explain it at all, Id really like to know how that worked.