I believe that P(42) when choosing a number uniformly at random from Z is 0, and that this is not contradictory because 0 + 0 + 0 + ... is not necessarily zero. While the sum of any finite number of zeroes is clearly zero, I don't think we can extend that to infinite sets. If I've missed a proof of this earlier, someone can feel free to correct me, but...
Note that I'm obviously not saying that it's impossible to pick 42, simply that it almost never happens (in the mathematical sense of "almost never".)
I believe that P(42) when choosing a number uniformly at random from Z is 0, and that this is not contradictory because 0 + 0 + 0 + ... is not necessarily zero. While the sum of any finite number of zeroes is clearly zero, I don't think we can extend that to infinite sets. If I've missed a proof of this earlier, someone can feel free to correct me, but...
Note that I'm obviously not saying that it's impossible to pick 42, simply that it almost never happens (in the mathematical sense of "almost never".)
To put that into precise mathematical/statistical terms, we can formally prove that the event P(X != 42) will almost surely happen (and yes, "almost sure(-ly)" has a precise statistical definition).
Let us take a sequence of random variables {X[i]}, where i is indexed by the natural numbers, i is an element of N. For i < 42, P(X[i] = 42) = 0, and for i >= 42, P(X[i] = 42) = 1/i. So, the probability P(X[i] != 42) = 1 - 1/i, by properties of complementary events. As i -> infinity, simple calculus will demonstrate that P(X[i] != 42) -> 1, and that is the statistical definition which corresponds to exactly what we want to show, that when we draw a number from the set of natural numbers, it will almost surely not be 42 (or any other particular number).
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Went to a new shop from a friend's recommendation, DQ'ed for willful violation of CR 100.6b.
P{choose 1, or choose 2, or choose 2, ..., or choose n...} = 1, ie: if you sample you will select at least some integer by supposition
but they are disjoint so the that P{} = p(1) + p(2) + ... p(n) = 1, yet if each p(n) was identical to zero this sum would vanish and 1=0 which is not acceptable!
so they are non zero. if they were all zero then the law of total probability would be violated.
i suspect you need more information in order for this to be a well defined probability space.
Note that I'm obviously not saying that it's impossible to pick 42, simply that it almost never happens (in the mathematical sense of "almost never".)
To put that into precise mathematical/statistical terms, we can formally prove that the event P(X != 42) will almost surely happen (and yes, "almost sure(-ly)" has a precise statistical definition).
Let us take a sequence of random variables {X[i]}, where i is indexed by the natural numbers, i is an element of N. For i < 42, P(X[i] = 42) = 0, and for i >= 42, P(X[i] = 42) = 1/i. So, the probability P(X[i] != 42) = 1 - 1/i, by properties of complementary events. As i -> infinity, simple calculus will demonstrate that P(X[i] != 42) -> 1, and that is the statistical definition which corresponds to exactly what we want to show, that when we draw a number from the set of natural numbers, it will almost surely not be 42 (or any other particular number).
Went to a new shop from a friend's recommendation, DQ'ed for willful violation of CR 100.6b.
Have played duals? I have PucaPoints for them!
(Credit to DarkNightCavalier)
$tandard: Too poor.
Modern:
- GW Birthing Pod(?)
Legacy:
- UWR Delver
P{choose 1, or choose 2, or choose 2, ..., or choose n...} = 1, ie: if you sample you will select at least some integer by supposition
but they are disjoint so the that P{} = p(1) + p(2) + ... p(n) = 1, yet if each p(n) was identical to zero this sum would vanish and 1=0 which is not acceptable!
so they are non zero. if they were all zero then the law of total probability would be violated.
i suspect you need more information in order for this to be a well defined probability space.