Six random variables

Six variables x i x_i are chosen uniformly at random in the interval [ 0 , 1 ] [0,1] .

Find the probability that x 1 + x 2 + x 3 + x 4 + x 5 + x 6 < 1. x_1 + x_2 + x_3 + x_4 + x_5 + x_6 < 1. If the answer is of the form a b \frac{a}{b} , where a a and b b are coprime positive integers, what is a + b ? a+b?


Inspired by Brian Charlesworth


The answer is 721.

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4 solutions

Geoff Pilling
Apr 6, 2017

Consider the closed interval [0,1]. In general, we can solve this by nested integrations, where you integrate over all the possibilities for the first value being in this range. Then integrating over all the values where the second value is in this range (less the first value), then integrating over the third value being in this range (less the first two values) etc. until we have integrated over all the variables. The value of that integral will represent the final probability.

For example for six variables, x i x_i and u , v , w , x , y u, v, w, x, y and z z defined as this:

  • u = 1 x 1 u = 1 - x_1
  • v = 1 x 1 x 2 v = 1 - x_1 - x_2
  • w = 1 x 1 x 2 x 3 w = 1 - x_1 - x_2- x_3
  • x = 1 x 1 x 2 x 3 x 4 x = 1 - x_1 - x_2- x_3- x_4
  • y = 1 x 1 x 2 x 3 x 4 x 5 y = 1 - x_1 - x_2- x_3- x_4 - x_5
  • z = 1 x 1 x 2 x 3 x 4 x 5 x 6 z = 1 - x_1 - x_2- x_3- x_4 - x_5 - x_6

we would have something like:

P = 0 1 0 u 0 v 0 w 0 x 0 y d z d y d x d w d v d u P = \int_0^1 \int_0^u \int_0^v \int_0^w \int_0^x \int_0^y \, dz \, dy \, dx \, dw \, dv \, du

P = 0 1 0 u 0 v 0 w 0 x y d y d x d w d v d u P = \int_0^1 \int_0^u \int_0^v \int_0^w \int_0^x y \, dy \, dx \, dw \, dv \, du

P = 0 1 0 u 0 v 0 w x 2 2 d x d w d v d u P = \int_0^1 \int_0^u \int_0^v \int_0^w \dfrac{x^2}{2} \, dx \, dw \, dv \, du

P = 0 1 0 u 0 v w 3 6 d w d v d u P = \int_0^1 \int_0^u \int_0^v \dfrac{w^3}{6} \, dw \, dv \, du

P = 0 1 0 u v 4 24 d v d u P = \int_0^1 \int_0^u \dfrac{v^4}{24} \, dv \, du

P = 0 1 u 5 120 d u P = \int_0^1 \dfrac{u^5}{120}\, du

P = 1 720 P = \dfrac{1}{720}

P = 1 6 ! P = \dfrac{1}{6!}

1 + 720 = 721 1+720=\boxed{721}

Moderator note:

Solving this is equivalent to finding the volume of the unit 5-simplex in 6 dimensions (essentially, a "right tetrahedron" extrapolated into 6 dimensions).

A unit 2-simplex is 3 dimensions is shown below.

This is equivalent to finding the geometric probability via the volume of the 6-dimensional tetrahedron :)

Calvin Lin Staff - 4 years, 2 months ago

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Ah yes! :0)

(But not a regular tetrahedron, right? More of a "right" tetrahedron)

Geoff Pilling - 4 years, 2 months ago

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Yup. The heights of 1 tells us that the volume of the tetrahedron (calculated itertively over the dimensions) is 1 × 1 2 × 1 3 × 1 4 × 1 5 × 1 6 1 \times \frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} \times \frac{1}{5} \times \frac{1}{6} .

Calvin Lin Staff - 4 years, 2 months ago

Right, this scenario in particular is the standard n-simplex.

Jason Dyer Staff - 4 years, 1 month ago

While the multiple integration approach works, the formulae as stated are a little more complicated unless you define u-z a little more clearly.

Is there a way of solving this using characteristic functions? As a rule they make these sorts of problems much easier to resolve - though I have yet to do that in this case

Malcolm Rich - 4 years, 1 month ago

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Good point... I should probably update the solution to be a bit more clear.

Geoff Pilling - 4 years, 1 month ago

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There's probably a substitution that turns this into the six-time integration of 1. Integrate (A-x)^n between 0 and A still gives (1/n)*A^(n+1). And if A is successively x1, x1+x2, x1+x2+x3,.. then we get the same result

Malcolm Rich - 4 years, 1 month ago

is there uhhhh an easier way to do this without calculus

John Smith - 4 years, 1 month ago

I think the proposed solution is the answer of; the probability of chosing numbers obtaining the given start up condition, in order. If you choose 6 different numbers between 0 and 1; all chosen set will obtain that condition. It doesn't need to be in order to obtain the given condition. The problem is not; what's the probability of chosing the numbers from a bag with exact order?

What is the probability of total sum being between 3 and 4? Is it same as being smaller than 1?

M H - 4 years, 1 month ago

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Interesting follow up question... I believe the probability of being between 3 and 4 will be much higher!

Geoff Pilling - 4 years, 1 month ago

"In general, we can solve this by nested integrations, where you integrate over all the possibilities for the first value being in this range. Then integrating over all the values where the second value is in this range (less the first value), then integrating over the third value being in this range (less the first two values) etc. until we have integrated over all the variables. The value of that integral will represent the final probability." I didn't understand why this would give the final probability. We want the sum of the 6 variables to be less than 1. How is that condition is equivalent to the above condition? Thanks.

Pranav Rao - 4 years, 1 month ago

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I agree with you. Apologies for not formatting, since I'm not sure how to on mobile, but it seems to me that you just calculated the probability that u <v <w <x <y <z. So it seems like 1/6! Should be the denominator ofthe true probability, while the numerator is the probability that the sum is less than 1 given that the variables are ordered by magnitude.

Kyle Coughlin - 4 years, 1 month ago

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I'm not sure why you think that u < v < w < x < y < z. How about if u is 0.9 and the rest are 0.01?

Geoff Pilling - 4 years, 1 month ago

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@Geoff Pilling Sorry, I got the inequality backwards, it's actually u >v>w>x>y>z. However, the issue here is that the case u = 0.9, v = 0.8, and the rest 0 is included in the integral above, while that case clearly doesn't satisfy the conditions of the problem since the sum is greater than 1. Or to dramatize the problem further, u=v=w=x=y=z=1 is also part of the domain of the integral, but those numbers sum up to 6, which is quite obviously larger than 1. Again, the integral above is the probability that u>v>w>x>y>z, and you need to also find the probability given u>v>w>x>y>z that their sum is less than 1 in order to solve this problem

Kyle Coughlin - 4 years, 1 month ago

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@Kyle Coughlin Yes. You are correct that I need to clarify the solution... Will work on that.

Geoff Pilling - 4 years, 1 month ago

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@Geoff Pilling Thanks, I understand it now.

Pranav Rao - 4 years, 1 month ago
Sergio Alvarez
Apr 21, 2017

The sum of n independent random variables with uniform distribution is the Irwin-Hall distribution, whose cumulative distribution function is: Let X be the sum of the 6 random variables, then: P(X<1)= 1 6 ! ( ( 1 ) 0 ( 6 C 0 ) ( 1 0 ) 6 + ( 1 ) 1 ( 6 C 1 ) ( 1 1 ) 6 ) \frac{1}{6!}*((-1)^0*(6C0)*(1-0)^6+(-1)^1*(6C1)*(1-1)^6)

P(X<1)= 1 720 \frac{1}{720}

a+b=720+1= 721

Not a creative solution thought :(

Oh, so a triangular distribution is a special case of this distribution? Thanks, I learned something new!

Pi Han Goh - 4 years, 1 month ago
Aryan Goyat
Apr 19, 2017

let us consider no line in 0 to 1 is made up of n points (mark them 1 to n) choosing any 6 pts we have way n^6 choice now let us find no of ways x1+x2+x3+x4+x5+x6<n(marked values) we get it as (n-1)C6 so divide and find limit n tends to infinity. :P

does it equal to the probability 5<x+y+z+u+v+w<6 ?

Mehdi K. - 4 years, 1 month ago

I don't understand what you're saying.

now let us find no of ways x1+x2+x3+x4+x5+x6<n(marked values)

What does this mean? What marked values are you referring to?

Limit n tends to infinity? Why is this needed?

Pi Han Goh - 4 years, 1 month ago

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I mean we have divided number line from 0 to 1 into n pts and mark them 0 to n with pt at x=1 marked as n so what we have to find no of ways any 6 pts can be selected=n^6 now since its provided that x1+x2++x3+x4+x5+x6<1 but (think using length) so marked value of x1 be its length and so so length of x1+x2++x3+x4+x5+x6< length of 1 ie x1+x2+x3+x4+x5+x6<n so we get (n-1)C6 ways P=(n-1)C6 /n^6 but since no line consists of infinite pt so n---->infinity we get P=1/61

aryan goyat - 4 years, 1 month ago

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Ah got it got it thanks

Pi Han Goh - 4 years, 1 month ago

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@Pi Han Goh Using the same method got:

for picking two numbers lim n i = 1 n 1 ( n i ) n 1 = 1 2 \displaystyle\lim_{n \to \infty} \frac{\sum_{i = 1}^{n - 1} (n - i)}{n ^ 1}= \frac{1}{2}

for picking three numbers got lim n i = 1 n 2 j = 1 n i 1 ( n ( i + j ) ) n 3 \displaystyle \lim_{n \to \infty}\frac{\sum_{i=1}^{n-2} \sum_{j = 1} ^{n-i-1} ( n-(i + j))}{n^3}

do you know how to generalize it for picking k numbers? or prove it equals to 1 k ! \frac{1}{k !} ? I got stuck there..

Mehdi K. - 4 years, 1 month ago

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@Mehdi K. Induction should take care of it...

Pi Han Goh - 4 years, 1 month ago

@Pi Han Goh pls upvote if u liked it!!

aryan goyat - 4 years, 1 month ago

this is the best solution for this question

space sizzlers - 4 years, 1 month ago
M B
Apr 19, 2017

Solving for all n n (here n = 6 n=6 ) with S n = i = 1 n x i S_n = \sum_{i=1}^n x_i :

i [ 0 , n ] 0 < x i < 1 , S n < 1 d x 1 . . . d x n = i [ 0 , n 1 ] 0 < x i < 1 , 0 < x n < 1 S n 1 , S n 1 < 1 d x 1 . . . d x n = i [ 0 , n 1 ] 0 < x i < 1 , S n 1 < 1 ( 1 S n 1 ) 1 d x 1 . . . d x n 1 = i [ 0 , n 2 ] 0 < x i < 1 , S n 2 < 1 ( 1 S n 1 ) 2 2 d x 1 . . . d x n 2 = . . . = 1 n ! \int_{\forall i \in [0,n] \, 0 < x_i < 1, \,S_n < 1} dx_1 ... dx_n \\ = \int_{\forall i \in [0,n-1] \,0 < x_i < 1, \, 0 < x_n < 1 - S_{n-1} , \, S_{n-1} < 1 } dx_1 ... dx_n \\ = \int_{\forall i \in [0,n-1] \, 0 < x_i < 1 , \, S_{n-1} < 1 }(1 - S_{n-1})^1 dx_1 ... dx_{n-1} \\ = \int_{\forall i \in [0,n-2] \, 0 < x_i < 1, \, S_{n-2} < 1 }\frac{(1 - S_{n-1})^2}{2} dx_1 ... dx_{n-2} \\ =... = \frac{1}{n!}

So the answer is n ! + 1 n! +1 , here 721 721 .

The sum of the variables on which you integrate. The trick is that you integrate them one by one, so Sn, Sn-1 etc. are constant when integrating for xn+1, xn etc.

m b - 4 years, 1 month ago

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Ah got it. I'ts surprising to see that it simplifies to such a simple expression, when the integral looks so complicated at first glance

Pi Han Goh - 4 years, 1 month ago

I don't understand what you're saying. What does S n = i = 1 n x i S_n = \sum_{i=1}^n x_i represent?

Pi Han Goh - 4 years, 1 month ago

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