Probability of at least one random number out of 3 being greater than 3 other random numbers?

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There are 6 random numbers (drawn from arbitrarily large pool). What is the probability that biggest number lies in second half? Answer is $1/2$ but I tried to solve it with combination. Can you show me where I went wrong?

I thought, for one particular number in second half, the probability that it is greater than all three points in first half is $1/8$ and likewise probability that it is not greater than all three points in first half is $7/8$, so

$${3 \choose 1} \left(\frac18\right)^1 \left(\frac78\right)^2$$ is probability that only of them is bigger than first half

$${3 \choose 2} \left(\frac18\right)^2 \left(\frac78\right)^1$$ is probability that two of them are bigger than first half

$${3 \choose 3} \left(\frac18\right)^3 \left(\frac78\right)^0$$ is probability that all three of them are bigger than first half

Total is $\dfrac{169}{512}$ not $\dfrac{256}{512}$. So what is my mistake?

Edit: I think I figured out, biggest number, rank 6 must be in second half; so I can choose other two numbers

(5 choose 2) different ways * 3! configuration for right side * also 3! configuration for left side;

divided by 6! equals to 1/2. now i can generalize it to solve secretary problem, : )

• I think my argument is not correct but suppose that the 6 randomly drawn numbers are $\{a_1,\dots,a_6 \}$. Suppose $a_1$ is the largest of them. Then $a_1$ can be placed in 2 baskets in 2 ways and $a_1$ can be placed in basket 2 in 1 way. So the probability is $1/2$. – user96343 Mar 1 '15 at 17:06
• yes, argument is wrong but i want to know exactly where it is wrong. I am really stuck in this basic :) question. i know answer is 1/2 with simple logic – xcvbnm Mar 1 '15 at 17:08
• I meant my argument. :) – user96343 Mar 1 '15 at 17:08
• Hahaha. Easy to answer. Hard to justify. Um, all the numbers have an equal chance of being the biggest number...the biggest number is either in one or the other...damn idk :)) I'd like to know the explanation too! – BCLC Mar 1 '15 at 17:09
• I don'see the definition of "random number drawn from an arbitrarily large pool." Please, explain. – zoli Mar 1 '15 at 17:12

Your problem is that the probability that the fourth number is greater than the preceding three is not $\frac 18$, it is $\frac 14$. Your calculation of $\frac 18$ assumes that the fourth number being greater than the first and the fourth number being greater than the second are independent. They are not. If the fourth number is large, it has a good chance of being greater than any of the first three.

• thank you, it is interestingly so, i had written a small script and it confirms it. but how can i show it with formula? thanks again – xcvbnm Mar 1 '15 at 17:27
• I don't think there is a graceful way in the approach you describe. There are too many correlations between the chances that one number in the second set is greater than all the first three and the chances for another. If the six numbers are distinct, you can just change them into $1$ through $6$ by using their ranks and you have a permutation of the numbers. – Ross Millikan Mar 1 '15 at 17:33
• Thank you, i solved with your hint – xcvbnm Mar 2 '15 at 12:49

The probability is $1/2$. Here is a possible explanation.

Let's suppose we have two groups containing three random numbers each.

Group 1: $\{x_1,x_2,x_3\}$ Group 2: $\{x_4,x_5,x_6\}$

Assume that these numbers are uniformly distributed and in [0,1]. All the random variables are also independent of each other.

$$p_x(X) = \left\{\begin{array}{rl} 1, & \textrm{if X \in [0,1]}\\ 0, & \textrm{otherwise} \end{array}\right.$$

Defining $z_1 = \max(x_1,x_2,x_3)$ and $z_2 = \max(x_4,x_5,x_6)$, we want to obtain $P(z_2 > z_1)$.

This probability can be obtained by integrating the joint density $p_{z_1z_2}(Z_1,Z_2)$ of the random variables $z_1,z_2$, over the region where $z_2 > z_1$. Since $z_1$ and $z_2$ are independent because all the $x$ are independent, we have that $p_{z_1z_2}(Z_1,Z_2)=p_{z_1}(Z_1)p_{z_2}(Z_2)$.

The density $p_{z_1}(Z_1)$ is obtained by deriving the cdf $F_{z_1}(Z_1)$:

\begin{eqnarray*} p_{z_1}(Z_1) &=& \frac{d}{dZ_1}\,F_{z_1}(Z_1) = \frac{d}{dZ_1}\,P(z_1 \le Z_1) \\ \\ &=& \frac{d}{dZ_1}\,P(\max(x_1,x_2,x_3) \le Z_1) \\ &=& \frac{d}{dZ_1}\,P(x_1 \le Z_1, x_2 \le Z_1, x_3 \le Z_1) \\ &=& \frac{d}{dZ_1}\,P(x_1 \le Z_1)P(x_2 \le Z_1)P(x_3 \le Z_1) \\ &=& \frac{d}{dZ_1}\,Z_1^3 = 3Z_1^2, \quad \textrm{for $Z_1 \in [0,1]$}. \end{eqnarray*}

The same procedure is used for $z_2$ and we obtain $p_{z_2}(Z_2) = 3Z_2^2$, for $Z_2 \in [0,1]$.

The joint density is:

$$p_{z_1z_2}(Z_1,Z_2) = \left\{\begin{array}{rl} 9Z_1^2Z_2^2, & \textrm{if (Z_1,Z_2) \in [0,1] \times [0,1]}\\ 0, & \textrm{otherwise} \end{array}\right.$$

We now can obtain the desired probability $P(z_2>z_1)$:

\begin{eqnarray*} P(z_2 > z_1) &=& \int_{0}^{1}\int_{0}^{Z_2}p_{z_1z_2}(Z_1,Z_2)\,dZ_1dZ_2 \\ &=& 9\int_{0}^{1} Z_2^2\int_{0}^{Z_2}Z_1^2\,dZ_1dZ_2 \\ &=& 3\int_{0}^{1} Z_2^5dZ_2 \\ &=& \frac{1}{2}\,Z_2^6\bigg|_0^1 = 1/2. \end{eqnarray*}

• This is much more complicated than using the symmetry of the situation. I think OP wanted to know why his calculation was not correct. – Ross Millikan Mar 1 '15 at 20:11

If none of the numbers in the pool are equal, then the fact that six numbers are drawn from the pool is irrelevant. You are just dealing with six numbers, none of which are equal to any of the others. Since we are dealing with a set of six (an even number) numbers, then each number is either above the median or below it. So no need to do combinations or anything else fancy. The question you are asking is the same as asking what the probability of picking a red ball is from a set of 3 red balls and 3 blue balls.

• I think OP had understood this, but wanted to know why his detailed procedure failed. – Ross Millikan Mar 1 '15 at 17:30

We are given a nonatomic probability measure $P$ on ${\Bbb R}$, which on account of independence induces a probability measure $P^\otimes$ on ${\mathbb R}^6$. If $P$ can be realized as a probability density $x \mapsto f_X(x)$ then $P^\otimes$ is realized by $$P^\otimes (A)=\int _A f_X(x_1)\cdot\ldots\cdot f_X(x_6)\>{\rm d}(x)\ .$$ Let $$A:=\bigl\{x\in{\Bbb R}^6\>|\>\max\{x_1,x_2,x_3\}\geq\max\{x_4,x_5,x_6\}\bigr\}\ ,$$ and let $A'$ be the complement of $A$, defined by $\max\{x_1,x_2,x_3\}\leq\max\{x_4,x_5,x_6\}$. Then $P(A)=P(A')={1\over2}$ on account of symmetry. (Of course we knew this all along.)

If the numbers cannot be equal (if they can be equal the largest has to be defined with care), and are all equally likely, then proceed as follows.

Any collection of six numbers is equally likely - if they are chosen from $n$ possibilities the number of choices is $\binom n6$.

There are 6! possible arrangements of each possible six. Of these $5!$ have the largest number in the first place (and the same for the second, third etc). So the number of arrangements with the largest number in the first half is $3\cdot 5! \cdot \binom n6$. The number of arrangements with the largest number in the second half is the same.

• thank you, i solved myself just around this time, but thank you – xcvbnm Mar 2 '15 at 13:00