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Pascal's Lottery (Posted on 2021-07-01) Difficulty: 3 of 5

Every soul is allotted 6 numbers, chosen at random from the positive integers.

If such numbers are relatively prime, then the soul is admitted to Heaven - otherwise, not.

What is the probability of an eternal reward?

What is the probability of an eternal reward, if such numbers must be pairwise prime?

Note: the integers 30, 42, 70, and 105 are relatively prime, while the numbers 121, 122, and 123 are pairwise prime, see the definitions at https://primes.utm.edu/glossary/xpage/RelativelyPrime.html

No Solution Yet Submitted by broll    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution with research | Comment 1 of 6
I had heard of the simpler problem before:

The probability that two numbers are relatively prime is 1/zeta(2) = 6/pi^2=0.60793

This link shows how to generalize the first question 
https://mathworld.wolfram.com/RelativelyPrime.html
For n numbers the solution is 1/zeta(n)
Here are some exact values of the zeta function
https://mathworld.wolfram.com/RiemannZetaFunction.html

From which the solution to the first question is 1/zeta(6)=945/pi^6=0.98295
(This is pretty high, which is unsurprising, as we'd only be barred from heaven if all 6 numbers are even or all multiples of 3, etc.)

For the second part we have to consider each of the C(6,2)=15 pairs.  (This is not good for out eternal souls.  Now if any two of the six numbers are even etc. we are damned.)
For the probability, assuming the numbers were chosen independently, just do
(6/pi^2)^15=0.00057250


  Posted by Jer on 2021-07-01 10:36:27
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