All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
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.)
Some Thoughts partial solution and some thoughts | Comment 2 of 6 |
In the first instance we want the probability that no number divides all 6 of the random numbers. We only need to check the primes as a composite number can't divide all six if none of its prime factors divides all six.

The probability that a given prime divides all six is 1/p^6 as 1/p of all numbers are divisible by p. The probability it does not divide all six is 1-1/p^6. 

The first case converges relatively quickly. Below we show up to primes that are below 300. Larger primes are extremely unlikely to divide all six.

Under the first circumstance you have about a 98.3% probability of getting into heaven.


clearvars, clc

v=1;p=0; 
while p<300
p=nextprime(p+1);
v=v*(1-1/p^6);

disp([v])
end

                  0.984375
         0.983024691358025
         0.982961777777778
         0.982953422740525
         0.982952867888943
         0.982952664244492
         0.982952623521557
         0.982952602628067
         0.982952595988106
         0.982952594335594
         0.982952593228047
         0.982952592844938
         0.982952592638005
         0.982952592482508
         0.982952592391318
          0.98295259234697
         0.982952592323666
         0.982952592304588
         0.982952592293721
         0.982952592286048
         0.982952592279553
         0.982952592275509
         0.982952592272503
         0.982952592270525
         0.982952592269345
         0.982952592268419
         0.982952592267596
         0.982952592266941
         0.982952592266354
         0.982952592265882
         0.982952592265648
         0.982952592265454
         0.982952592265305
         0.982952592265169
         0.982952592265079
         0.982952592264996
          0.98295259226493
         0.982952592264878
         0.982952592264833
         0.982952592264796
         0.982952592264766
         0.982952592264738
         0.982952592264718
         0.982952592264699
         0.982952592264682
         0.982952592264666
         0.982952592264655
         0.982952592264647
          0.98295259226464
         0.982952592264633
         0.982952592264627
         0.982952592264622
         0.982952592264617
         0.982952592264613
          0.98295259226461
         0.982952592264607
         0.982952592264604
         0.982952592264601
         0.982952592264599
         0.982952592264597
         0.982952592264595
         0.982952592264594
         0.982952592264592
         
In the second scenario you are concerned that for every prime either none of the random integers is divisible by that prime or only one is. Their respective probabilities are  (1-1/p)^6 and 6*(1/p)*((p-1)/p)^5. We need to multiply these together over all primes.

clearvars, clc

 p=0;v2=1;
while p<300000
p=nextprime(p+1);

v2=v2*((1-1/p)^6 + 6*(1/p)*((p-1)/p)^5);

disp([v2])
end

At the end of this set of iterations using all the primes under 300,000 the probability still hasn't settled down:

...

        0.0133246974939543
        0.0133246974917327
        0.0133246974895112
        0.01332469748729
        0.0133246974850689
        0.0133246974828478
        0.013324697480627
        0.0133246974784063
        
It seems as if your chances are about 1/75, but it's not outside the realm of possibility the numbers may go down asymptotically to zero when huge numbers with many factors may just inevitably have some that come up twice. It could be that the primes don't get rarer fast enough as the numbers get higher.  At this point you can say to God that the upper limit was not applied and that was not fair. It was an ill-defined problem.        

It's probably significant that the sum of the reciprocals of the primes diverges.

But clearly it's not specified how the random numbers are chosen. The sum of the probabilities of each possible number has to be 1, so it can't be an infinite uniform distribution. Perhaps its a Poisson distribution with some mean--that has infinite possibilities but still a finite mean and presumably the mean of our function would converge and we'd find a probability.

But I see from Jer's post that if God can somehow have a uniform distribution over an infinite range, there is a finite answer to the infinite (i.e. it does not approach zero, but much smaller than the numbers looked in my calculation).

  Posted by Charlie on 2021-07-01 10:45:25
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information