Except their chances are infinitely higher than yours. It’s miniscule, but miniscule and finite is infinitely bigger than zero. Math gets funky around the edge cases
The chances of finding a winning ticket on the street are many orders of magnitude lower. How often do you find unredeemed lottery tickets walking on the street? I never have, the most I’ve seen are losing scratch-offs.
Want to see it in numbers? Last year people in the US spend 105 billion on lottery tickets. Meaning, you have 1 chance in 403 846 153 to win. If you played every week, it would take you 7 766 272 year to win. And even that isn’t certain.
First, you fundamentally misunderstand how lottery winners are chosen. The odds are invariant with the number of players, the odds of winning are the same whether there’s no other players, or a billion. A sequence of numbers are chosen at random from a set range, and anyone whose ticket matches that sequence wins. You’re probably thinking of a raffle, where people purchase tickets which are shuffled together, and one is selected at random. The odds of winning a raffle do decrease with more players.
Second, the probability of two things happening is equal to the product of their individual probabilities, P(A and B) = P(A) × P(B). For example, let’s say I have 10 dice in a bag, and only one is red; the odds of pulling out the red die and rolling a 6 is equal to 1/10 × 1/6, for a total of 1/60. So the probability of finding a winning lottery ticket on the ground is exactly equal to the probability of finding an unredeemed ticket on the ground, P(A), multiplied by the probability of any individual ticket being a winner, P(B). If P(A) = 1/10, then you’re ten times more likely to win by buying a ticket yourself. If P(A) is 1/1,000, then you’re 1,000 times more likely. Considering that I’ve never found an unredeemed lottery ticket on the ground, P(A) is likely extremely small, so the odds are extremely higher if your purchase a ticket yourself.
Third, 1 does not equal 0.9999999999. 1 does equal 0.999…, which is very different. It only works with a literal infinite number of decimal places. Ten 9s is not sufficient, ten trillion 9s is not sufficient, no finite number of decimal places is sufficient to uphold that property.
The way I see it, is that by never trying, I have statistically about the same chances of winning as someone playing.
Except their chances are infinitely higher than yours. It’s miniscule, but miniscule and finite is infinitely bigger than zero. Math gets funky around the edge cases
Not really, the chances to find a winning ticket of a lottery walking on the street is almost comparable
The chances of finding a winning ticket on the street are many orders of magnitude lower. How often do you find unredeemed lottery tickets walking on the street? I never have, the most I’ve seen are losing scratch-offs.
The difference is still comparable.
That is how low you stand a chance of winning.
Want to see it in numbers? Last year people in the US spend 105 billion on lottery tickets. Meaning, you have 1 chance in 403 846 153 to win. If you played every week, it would take you 7 766 272 year to win. And even that isn’t certain.
In math: 1 = 0,9999999999
Couple things:
First, you fundamentally misunderstand how lottery winners are chosen. The odds are invariant with the number of players, the odds of winning are the same whether there’s no other players, or a billion. A sequence of numbers are chosen at random from a set range, and anyone whose ticket matches that sequence wins. You’re probably thinking of a raffle, where people purchase tickets which are shuffled together, and one is selected at random. The odds of winning a raffle do decrease with more players.
Second, the probability of two things happening is equal to the product of their individual probabilities, P(A and B) = P(A) × P(B). For example, let’s say I have 10 dice in a bag, and only one is red; the odds of pulling out the red die and rolling a 6 is equal to 1/10 × 1/6, for a total of 1/60. So the probability of finding a winning lottery ticket on the ground is exactly equal to the probability of finding an unredeemed ticket on the ground, P(A), multiplied by the probability of any individual ticket being a winner, P(B). If P(A) = 1/10, then you’re ten times more likely to win by buying a ticket yourself. If P(A) is 1/1,000, then you’re 1,000 times more likely. Considering that I’ve never found an unredeemed lottery ticket on the ground, P(A) is likely extremely small, so the odds are extremely higher if your purchase a ticket yourself.
Third, 1 does not equal 0.9999999999. 1 does equal 0.999…, which is very different. It only works with a literal infinite number of decimal places. Ten 9s is not sufficient, ten trillion 9s is not sufficient, no finite number of decimal places is sufficient to uphold that property.
I don’t think you can claim the win if you can’t show you’ve bought the ticket
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