What is the chance that a 1 in 1000 event will happen within the first 500 tries?

Question

Say you have a thousand-sided die. You win if the die lands on 27, so a one in one thousand chance. I was curious as to what the chances are of winning if you roll the die 500 times. My gut told me that the odds were in your favor. I computed the probability as:

1 - (999/1000) ^ 500

which gave me 0.3936210551388153.

Did I solve this correctly? And if so, can anyone give me an intuitive explanation as to why you have less than a 40% chance of winning? The only explanation I can think of is that if you rolled 1000 times, you certainly wouldn't have a 100% chance of winning, but I'm wondering if there's any other angle to think of.

Edit: I understand basic probability concepts and am good at math, but so far most of the answers and comments are beyond my level of understanding. I'm really looking for a more intuitive understanding than complex formulas explaining this... while I appreciate the effort I don't really understand how the more complicated formulas better explain the probability than my simplistic formula above (assuming it is correct in the first place, which it seems to be based on responses)

Answer 1

Okay... for the time, ignore the phrase expected value (though it is in effect what makes the following argument work). This is just an attempt to try to build intuition.

Imagine for the time being that we have 100 boxes and exactly 50 balls. We will choose to distribute these balls randomly among the boxes.

Notice that in this scenario, since we have 50 balls, at most 50 of the boxes will have at least one ball in it. This implies that at least 50 of the boxes will have no balls in.

Now... if all 50 of the balls happened to be in different boxes, then sure, there will have been exactly 50 boxes with balls and exactly 50 boxes without balls.

Notice though that as we distribute the balls randomly, it is possible for some box or boxes to receive more than one ball. In the case that there is some box or boxes with strictly more than one ball then we will have strictly fewer than 50 boxes with at least one ball and strictly more than 50 boxes with no balls.

Now... suppose we ask the question, "what is the probability that we picked a box with at least one ball in it", the answer will be strictly less than $50\%$.

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Recognize that this is a metaphor (albeit an imperfect one) for your scenario in your question. Rather than boxes we have "experiment trials" and rather than balls we have "27"'s.

Answer 2

This is $1-(1-1/n)^{n/2}$ for $n=500$. For large $n$, $(1-1/n)^{kn}$ is approximately $e^{-k}$. So here, you expect the answer to be approximately $1-e^{-1/2}\approx 0.393$

Answer 3

Your equation is correct and is the simplest form of calculating this probability. It may help to look at a simpler version of this problem to see how it works. Say we have a $4$ numbered die. Intuitively you might think that the probability of getting a single number, say $1$, would be $50\%$ for two rolls (half the numbers on the die) but lets look at all the possible outcomes of those two rolls. $(1,1)(1,2)(1,3)(1,4)(2,1)(2,2)$......etc $= 16$ equally probable outcomes with only $7$ of them with a $1$. Hence the probability of getting a $1$ in two rolls is $\frac{7}{16}$ which is less than $50\%$.