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^n is a positive, non-

If you toss the coin long enough, yes, S will appear. If you=20 continue tossing the coin, S will appear again, and again, ad=20 nauseam.

Does this have anything to do with the die problem? =20

dg (domain=3Dccwebster)

zero, number.

The coin being tossed infinitely often is a mathematical coin, needless= to say. =20 Let's take the complete works of ... result, then with= probability one the string S will appear on the record. And not just=20 once but infinitely often.

If you toss the coin long enough, yes, S will appear. If you=20 continue tossing the coin, S will appear again, and again, ad=20 nauseam.

Does this have anything to do with the die problem? =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 decimalapproximationof 1/3 which becomes more accurate as you add more digits, but will never be equal to 1/3.

No, "0.333..." is an exact representation. The "..." is part of the representation. It's a representation of an infinite series, whose value is exactly 1/3. (The original formulation was "naught point three recurring", which is even more transparently an infinite series.)

Evan Kirshenbaum + HP Laboratories >It is a popular delusion that the

1501 Page Mill Road, 1U, MS 1141 >government wastes vast amounts ofPalo Alto, CA 94304 >money through inefficiency and sloth.

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Adding one more 3 where?

At the end. It's like the old problem of adding a guest to a full hotel with an unbounded number of rooms: you move guest 1 to room 2, guest 2 to room 3, and so on. Then the new guest moves into room 1. You can always add one more to an unbounded set of integers.

The original of this was told using the word "infinite" but I prefer to avoid that widely misunderstood term.

dg (domain=ccwebster)

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.

Sure you can. Throw it ten times in the first minute, then 10 times in the next thirty seconds, and so on. You'll be done in two minutes.

Okay,

**I* can't do it, and *you**can't do it, but that says more about us than about the mathematics of the situation.

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1501 Page Mill Road, 1U, MS 1141 >to a Q-Tip yeah, it's somethingPalo Alto, CA 94304 >you stick in your ear, but there

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Let's take the complete works of Shakespeare and digitize them ... on the record. And not just once but infinitely often.

If you toss the coin long enough, yes, S will appear. If you continue tossing the coin, S will appear again, and again, ad nauseam. Does this have anything to do with the die problem?

It's actually identical. In the die case, the sequence (S) is simply "6". The point is that in an infinite number of trials,

**any**finite sequence of outcomes (each of which has a non-zero probability) will occur with probability one. The probability of it not occurring will be zero. In any finite number of trials, there will be a non-zero probability of the sequence not occurring.

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No, "0.333..." is an exact representation. The "..." is part of the representation. It's a representation of an infinite series, whose value is exactly 1/3. (The original formulation was "naught point three recurring", which is even more transparently an infinite series.)

The sequence 0.3, 0.33, ..., 0.333... has a limit of 1/3 as the number of digits grows without bound. That's not the same as saying it *is* 1/3.

dg (domain=ccwebster)

But 1/3 = 0.333..

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

If 0.333... is taken to be a notation for 0.3 repeating then it is just another notation for 1/3.

BTW, is the notation with dots over the repeating decimals still used? As in

.

0.3 = 0.333.. .

0.12345 = 0.123452345...

Jan

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.

No disagreement on that, and indeed you cannot distinguish the two on basis of a finite number of experiments.

But the whole notion of proabability measure

is not needed in the finite, where you can just count cases.

Nevertheless, a probability distribution on the infinite sequences {1,4,2,5,3,4,...}

in which different sequences have different chances of being drawn is not a good model for throwing a die infinitely many times.

You can disagree with that of course,

Jan

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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.

"Mathematics has nothing to do with actualities" (now who said that?)

Indeed, that's what we have got mathematics for.

And math gives us a good method

for dealing with infinite sequences of integers.

Modelling the set of all possible infinte sequences of die throws is just an application,

Jan

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