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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)

thumbtackthief
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    You have about a $40%$ chance of winning at least once as you calculated correctly. That is to say, if we were to try this experiment several times, among about $40%$ of the times that we rolled fivehundred times, we will have rolled at least one $27$. This might sound like we somehow rolled a $27$ fewer than one onethousandth of the time though... until you recognize that in some of those attempts, in the 500 rolls we actually rolled two or more 27's! Indeed, In 500 rolls, the expected number of 27's is going to be $\frac{1}{2}$, as your intuition should have led you to. – JMoravitz Jun 27 '18 at 15:18
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    So., think of it like... if all of the "wins" were perfectly spread out among the attempts, we might have had half of the attempts with wins. But then, the wins start to move around and in some of the attempts we have multiple wins, leaving more than half of the attempts with no wins at all, some with one win, and some with multiple wins. – JMoravitz Jun 27 '18 at 15:21
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    A big part of this problem is that the "1 in 1000" event can happen multiple times within our attempt. Compare this to if you have a special deck of playing cards with 1000 cards in it, exactly one of those cards is the ace of spades. You draw 500 cards from the shuffled deck without replacement. Here in this problem, the probability of having drawn the ace of spades within those 500 cards will be exactly $\frac{1}{2}$. – JMoravitz Jun 27 '18 at 15:24
  • I'm still a little confused -- I would have thought that I had a greater than 50% chance of winning at least once, but I actually have fairly substantially less. I also don't understand your line: In 500 rolls, the expected number of 27's is going to be 1/2, as your intuition should have led you to.. – thumbtackthief Jun 27 '18 at 15:29
  • Also the average time it takes to roll a 27 is $1000$ so since you’re going for way less than the average it should probably be unlikely that you roll one. Of course this ignores the skew of the distribution (in fact the number at which the probability crosses 50-50 is about $1000\log(2)\approx 700.$ – spaceisdarkgreen Jun 27 '18 at 15:31
  • "I don't understand your line: in 500 rolls the expected number of 27's is going to be 1/2." The expected value of a random variable is a measure of a random process which is often intuitive and easy to grasp and unfortunately frequently confused for probability. It is what people are thinking of when they say "If I roll a die six times, i'm guaranteed at least one six" when in reality they are only expected at least one six. – JMoravitz Jun 27 '18 at 15:34
  • As for your problem, letting $X_i$ be the $0$-$1$ random variable denoting whether or not the $i$'th roll in your fivehundred is a 27, we have the expected number of $27$'s is $E[\sum\limits_{i=1}^{500}X_i]=\sum\limits_{i=1}^{500}(E[X_i])=\sum\limits_{i=1}^{500}Pr(X_i=1)=500\cdot\frac{1}{1000}=\frac{1}{2}$. I.e., having rolled the die 500 times, the expected number of 27's to be rolled is $\frac{1}{2}$. – JMoravitz Jun 27 '18 at 15:36

3 Answers3

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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 $50\%$.

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.

JMoravitz
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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 $50\%$.

Phil H
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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 $1-e^{-1/2}\approx 0.393$

Angina Seng
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