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this post was submitted on 01 Oct 2023
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I know they give you the odds of winning with one ticket but what are the odds of winning with 10 tickets? It is not as simple as multiplying by 10, is it?
Long ago I was part of a lottery pool and we never really won anything. Just wondering if this has better odds than going on your own.
Yes, it’s as easy as multiplying by 10 for this kind of lottery. Let’s say you have to pick three numbers correctly. There’s 1000 possibilities (000 - 999). A random 3 digit string has a 1/1000 chance of winning. That’s close enough for most purposes. If you want to get really pedantic about it, you can say that with one ticket picked from a pool of 1000 (eg 123) you have a .999 chance of losing. Your second ticket is now picking from a pool of 999 possibilities, so the chance of picking the winner slightly increases to 1/999. The difference between doing the math that way and just multiplying the chance of a winning ticket by the number of tickets you buy is very small, though. There’s other lottery styles (eg, where the winning number is picked from a group of sold tickets, so a winner is guaranteed), where the math changes.
What you really want to do is figure out the expected value of a ticket. If there’s 1000 possible numbers and a $1000 prize, the tickets are worth &1. If they’re selling for $2 each, it’s not worth it. If they’re selling for $0.50 each, it’s worth buying them all. There was an investment group of friends that used to do this by searching the internet for lotteries. You just have to realize that it’s the kind of gamble that will pay off over time and not martingale yourself.
It’s 1:292.2 million, so 10 tickets is 1:29.22 million. There are theories about buying all 292.2 million numbers, but the issue always comes back to not being alone in the pool; you will win, but if someone else wins (or did the same thing), you will lose a few million for all your hard work.
Edit: also the tickets are now $2, so buying all numbers is still more expensive than the prize after taxes.