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Nought point three recurring is not a finite number.

Neither is 1/2, to base three,
Jan
And in the model I have in mind this set ... theory is the flexiblity one has in choosing a model.

That you can do, but then your model doesn't model independent throws with a die,

Of course it does. The altered model has the same "finite dimensional distributions" as your model, hence the tosses of the die are still independent.

J.
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But 1/3 = 0.333..

You're not serious, I hope! 1/3 is an exact fraction. 0.333... is a decimal approximation of 1/3 which becomes more accurate as you add more digits, but will never be equal to 1/3.
dg (domain=ccwebster)
=20 =20 Why doesn't it?

Consider the example I gave in another reply: throwing a die infinitely many times without a six coming up is possible (all throws being independent) but it has probability zero.

Sorry to disagree but you can't throw a die "infinitely many"=20 times, no matter how long you try. At every point in your=20 sequence of throws, there is a 5/6 probability of not throwing a=20 six. For any number of throws, n, (5/6)^n is a positive, non- zero, number.
=20
dg (domain=3Dccwebster)
But 1/3 = 0.333..

You're not serious, I hope! 1/3 is an exact fraction. 0.333... is a decimal approximation of 1/3 which becomes more accurate as you add more digits, but will never be equal to 1/3.

0.333 (= 333/1000) is an approximation of 1/3; so is 0.33..3(17 billion 3s). But 0.333.. (one 3 for each positive integer) is exactly equal to1/3.

J.
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Consider the example I gave in another reply: throwing a ... possible (all throws being independent) but it has probability zero.

Sorry to disagree but you can't throw a die "infinitely many" times, no matter how long you try. At every point in your sequence of throws, there is a 5/6 probability of not throwing a six. For any number of throws, n, (5/6)[/nq]^n is a positive, non-
zero, number.

The coin being tossed infinitely often is a mathematical coin, needless to say.
Let's take the complete works of Shakespeare and digitize them to get a long (but finite) string of 0s and 1s. Recode 1 as H(eads) and 0 as T(ails) to make a long (but finite) string S of Hs and Ts. If you toss the mathematical coin infinitely often and record the result, then with probability one the string S will appear on the record. And not just once but infinitely often.

J.
0.333 (= 333/1000) is an approximation of 1/3; so is 0.33..3 (17 billion 3s). But 0.333.. (one 3 for each positive integer) is exactly equal to1/3.

If that were true, then adding one more 3 would make it no longer exactly equal 1/3.

dg (domain=ccwebster)
0.333 (= 333/1000) is an approximation of 1/3; so is ... (one 3 for each positive integer) is exactly equal to1/3.

If that were true, then adding one more 3 would make it no longer exactly equal 1/3.

Adding one more 3 where?

J.
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...
}> If that were true, then adding one more 3 would make it no longer }> exactly equal 1/3.
}
} Adding one more 3 where?
At the infinitesimal end.

R. J. Valentine
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