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So I'm traversing this trivia book which has a question that has intrigued and widely confused me.

1.) Since we have $n$ number of elements, then the probability of it being the largest among the $n$ choices is $\frac{1}{n}$, which seems intuitive.

2.) The probability of it being larger than the second number $x_{2}$ is: either it will be larger or not, therefore $\frac{1}{2}$. Similarly, now this process is repeated $(n-1)$ times to check whether $x_{1}$ is larger than all other elements. Then the probability we will arrive at is : $\frac{1}{2^{n-1}}$

I equally believe in both processes, but of course, one is wrong. Apart from knowing which is right, it will be helpful to elucidate on why the other is wrong.

10Minhelp
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  • I'm not a specialist, but if you paint the smallest number red and all the others white, the probability of the red number being the first is 1/n, unless of course duplicates are allowed (e.g. $x_1 = x_7 = x_9$). – Ronald Oct 21 '22 at 08:02
  • Yes I understand that as my intuition was something similar as well, but it'd be great if you could explain the part I got wrong in my second approach. Thanks! – 10Minhelp Oct 21 '22 at 08:09
  • What you are missing is the information how you got those numbers. If they are from a book listing say polulation density of countries, then the probability is either 1 or 0, depending on if they start with the sparsest populated or densest populated. In the absence of anything more, the assumption would be that those are $n$ iid random variables, that means they are independent (the result you got for the first value is independet from what you got for any other value) and identically distributed. The latter means that if you throw a standard dice (sides 1-6) and I throw a "Dragon's Egg" – Ingix Oct 21 '22 at 08:16
  • (an almost round object that can show numbers from 1 to 100) than the results are still independent (knowing one result gives you no hint to the other), but the probability that your value is smaller than mine is much greater than $1/2$.

    If they are, as I said, indepent and identically distributed (iid), say you throw that Dragon's Egg 20 times, then your first answer ($1/n$) is correct. That's because while each throw's result $x_i$ is independent from each other, the comparison of results are not. Inituituvely, if you tell me that $x_1$ is smaller than $x_2,\ldots x_{19}$, then

     – Ingix Oct 21 '22 at 08:21
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    Well I think we can assume that all $x$ are uniformly distributed among [0,1], then it should be $\frac{1}{n}$ ? – 10Minhelp Oct 21 '22 at 08:21
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    I can conclude that $x_1$ is likely "pretty small", so that probability of it being smaller than $x_{20}$ as well is higher than $1/2$. – Ingix Oct 21 '22 at 08:22

1 Answers1

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The second argument is wrong.

Every event $$ `` \ x_1 \le x_k \ " , \qquad k \in \{ 2, ..., n\} $$ has probability $1/2$: you multiply these probabilities to get the probability of $$\bigwedge_{k=2}^n`` \ x_1 \le x_k \ "$$

But this process is allowed only if the events are independent.

Think about itin these terms: if you already knew that $x_1$ is smaller than $x_2, x_3, x_4, x_5, ... , x_{1000}$ then you would think that $x_1$ is a small number. Hence the conditional probability that $x_1 \le x_{1001}$ is not $1/2$. This means that those events are dependent, and the probability of the join event is not the product of probabilities.

Crostul
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  • I added a few edits, regarding the nature of all $x$ to make the problem more concise. – 10Minhelp Oct 21 '22 at 08:29
  • Yes, and considering (for example) the case of $N=3$, then the probability of $x_1<x_3$, conditional on $x_1<x_2$, is $\frac23$. So the probability of $x_1$ being smallest is computed as $\frac12\cdot\frac23=\frac13$, which is correct. – MJD Oct 21 '22 at 08:39
  • And generalizing it for n numbers would give us the desired probability of $\frac_{1}{n}$? – 10Minhelp Oct 21 '22 at 08:40
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    In my answer I focused on showing that the second argument doesn't work. The correct probability is $1/n$. – Crostul Oct 21 '22 at 08:49