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Infinite variables over infinite time

  1. Jan 11, 2013 #1
    Hi There,

    This is my first post in Math, and I can't read a formula beyond arithmetic, so please bear with me.

    I've always been curious about infinity and how it affects probability. I was reading on Wikipedia about the Infinite Monkey Theorem, and understand (I think) that with a finite set of variables (e.g. typewriter keys) over infinite time, the probability of an occurrence or reoccurrence is almost sure, or probability one, which I have taken to to mean 9.99999999_ (repeated).

    My question is twofold:

    1. Am I getting that first part right?
    2. What happens with infinite variables over infinite time? What is the probability of a specific (but arbitrary) outcome occurring?

  2. jcsd
  3. Jan 11, 2013 #2


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    I believe you mean ##0.\bar{9}## rather than ##9.\bar{9}##. In any case, ##1 = 0.\bar{9}##; trying to treat them differently will probably only lead to further confusion.

    The first part is more or less correct: with a finite keyboard, the monkeys will eventually type out any given work. There was a website that attempted to randomly simulate Hamlet or something, but I don't remember the site anymore and a very quick google search didn't turn it up.

    For second question, I'm not sure of the answer. I would guess it may depend on the specific set of variables you pick. It certainly depends on the probability density for hitting a given 'key'/variable, because with an infinite set of single-key hit outcomes you cannot make them all equally likely. I suspect this could result in some (if not all) possible outcomes occurring almost never (with probability zero, but still some chance of occurring). Again, I don't know for sure, though; perhaps someone else here knows more definitely.
  4. Jan 13, 2013 #3
    Thank you.

    Now that I think of it, I'm curious how probability works with infinite variables and infinite time as well as with infinite time. I'm just not sure how infinite variables affects probability at all.

    Any other takers on this question?
  5. Jan 13, 2013 #4
    Adding more typewriters just increases the probability over any finite amount of time, so using the Squeeze Theorem tells us that as we add more and more typewriters, the probability of typing any given work will still go to 1 as time progresses. Taking the limit just means we'll still have a probability of 1.
  6. Jan 14, 2013 #5


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    Often when dealing with things that happen with probability one it is important to consider the expected time required. Two event may occur with probability one, but if there expected times are one second and 10^1000 years, the situations are quite different. Also it is important to consider the distribution as Mute points out there are problems with choosing equally from an unbounded set.
  7. Jan 14, 2013 #6


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    This is actually not true as stated. There are some additional mathematical assumptions which we need to make. A crucial assumption is "independence". Intuitively that is: a previous outcome cannot affect the presents outcomes.

    Another assumption is that the variables are "identically distributed". That is: the distribution does not change over time. This condition can be weakened however.

    Finally, as you remarked, you must have a finite set of possible outcomes. The theorem is not true with infinitely many outcomes.
  8. Jan 14, 2013 #7


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    It doesn't. At least not in the sense you seem to mean. Rather than treating "infinity" as if it really were a number in the same sense that "1" or "10000000" are, we change from "discrete" to "continuous". And one result of that is that there may be a number of different probability distributions on the same interval. You have to specify which probability distribution you are using.
  9. Jan 14, 2013 #8
    Thanks for all of the responses.

    I don't think I asked my correction correctly. The responses here are based on the idea that I'm adding typewriters to the equation, but I'm actually adding keys to the typewriter.

    Let me start over:

    With a 6-sided die and one roll, the probability of me rolling a 1 is 1 in 6. With two rolls, it becomes 3 in 6, or 1 in 2. With infinite rolls it becomes probability one, correct?

    Now, what if I have an infinite-sided die? What is the probability of rolling a 1 on one roll, two rolls, or infinite rolls?

    This would be the "infinite outcomes" that micromass mentioned, I believe. Let's say that there is independence and identical distribution. Thanks for helping me with those factors!
    Last edited: Jan 14, 2013
  10. Jan 14, 2013 #9
    So there is a problem with how you state your question. And there is a problem with your math. Lets start with your math and that will lead into how you state your question. The probability or rolling a 1 in 2 rolls is not 1/2. it is 1 - (5/6)^2 = 0.30556

    You see there is a sequence. The probability of rolling a one in X number of rolls is 1 minus the probability of NOT rolling a 1 in X number of rolls.

    So there is a formula

    Probability of rolling a 1 in X number of rolls = 1 - (5/6)^X

    Here we can let X get bigger and bigger. As you know that is what we call letting x go to infinity. And eventually it becomes clear that the term (5/6)^X gets closer and closer to 0 as X gets bigger and bigger.

    Now you say, let the dice have an infinite number of sides. Mathematically what does that mean?

    let Y = the number of sides a die has. The probability of not rolling a certain side is then (Y - 1) / Y.

    So that means:

    The probability of rolling a 1 in X number of rolls is 1 - ((Y - 1) / Y) ^X

    Now you want to let Y go to infinity and X? This in a sense is the exact same thing as "Addiung sides to a die"

    The die can not inherently have an infinite number of sides, we have to formulate it somehow. And the only way that makes sense is to think of the number of sides as increasing.

    You see it is difficult to formulate actual math from what you have stated. And there is no way for us to answer you unless you formulate it yourself.
  11. Jan 14, 2013 #10
    Yep. Well, with two rolls, it's actually 11 in 36. But you have the right idea.

    With one roll, the probability of rolling a 1 is 0. With 2 rolls, it's, likewise, 0. Taking the limit tells us that, after an infinite number of rolls, it's still 0.

    Mathematically, if we let [itex]P_x\left(y\right)[/itex] represent the probability of rolling a 1 in at least one of our x rolls on a y-sided die, we could say

    [tex]\lim_{a\to\infty}\left(\lim_ {b\to\infty}\left(P_a\left( b\right)\right)\right)=0[/tex]

    (Note that complementary counting gives us that [itex]P_x\left(y\right)=1-\left(\dfrac{y-1}{y}\right)^n[/itex]

    From the first quote, we'd write

    [tex]\lim_{b\to\infty}\left(\lim_ {a\to\infty}\left(P_a\left( b\right)\right)\right)=1[/tex]

    Confusing as this is, this means it's meaningless to ask "what's the probability of getting a 1 on a die with infinite sides after an infinite number of rolls?" One of these limits tells us it's 0, and one tells us it's 1, and so if we assume the answer to that question exists, we get that it's two different values.
  12. Jan 14, 2013 #11

    He is right, and I tried to say the same thing in laymen's terms because the OP admittedly "can't read a formula beyond arithmetic"
  13. Jan 14, 2013 #12
    Thank you both very much!
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