• @agamemnonymous
    link
    17 months ago

    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.