Can you solve this math riddle?

Pssst.... geniuses, put 1,000,000 people in a row with a 10 random number generator each to get the first 1,000,000 digits of pi at once. The probability is (1/10)^1,000,000.
Do the same with 1 person in a sequential draw and the probability is still the same: (1/10)^1,000,000. The sample space of the experiment is the same either with 1 or N persons. The only thing that changes is the time to completion. Multi-tasking can reduce the time to almost 0.
Now let N be as big as you like. What is the size of the sample space? ANSWER 10^N. The question a maths student would ask is: if I repeat this experiment with N people N times over what the probability of getting the first N digits of pi would be? But I guess none of you knows any maths to ask such questions here exists, hence the only conclusion is: ELUBAS of chess.com fame, HAHAHA

3 is rolled every time. The die is weighted.

At a more serious note, let us try and answer the question posed in post #41. The answer would be: 1-(1-(1/10)^N)^N . That is if N people were to draw each to find the first N digits of Pi and if this experiment were to be repeated N times then the probability of hitting the first N digits of Pi randomly would be exactly: 1-(1-(1/10)^N)^N. Same with 1 person. Same with any real or rational number! But it would require an enormous effort and amount of time.
Now you can let N tend to infinity and evaluate the chances... Is it 1 or 0 this time? Answer with limits specifically designed for ELUBAS.

Oh, I see that you already edited the post to cover your stupid mistake in writing down the formula. Impressive, very impressive! As for the limes, for N going to infinity the value of this formula goes to 0.

Yet, it has no meaning, because you don't seem to understand the problem. The people in our group don't "roll once and pass the dice". Everybody has his or her dice and rolls it an infinite number of times, so that finally every member of our group has one, personal, random real number from <0, 1>.

Moreover, our group is also infinite (e. g. one person for every natural number). What is the probability that at least one of the members will have a particular number, for example 1/3? Is this probability 0? 1? Something in between? Are you able to cope with this problem?

PS. I don't know who is Elubas, but it is meaningless in a serious mathematical discussion, isn't it?

@Otienimous 0 man, you just said it, and i proved it in post #43. So what's your story and what's the big deal about it?

Yes, it is 0, I mentioned it already indeed. As for the proof, #43 doesn't seem very precise, but look at https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument - I think that you will like the method.

@Otienimous Oh, but it's very precise. Very precise indeed! And don't flash Cantor on me... ok?

I had to go through 4 pages to see what u mean. And there is not a regular decahedron with equal sides in nature to be a fair dice.

There is a problem with your proof: you assume that number of people and decimal length of approximation go to infinity hand-in-hand, whereas it could also be that one of them tends there much quicker - the "proof" doesn't take into consideration this possibility.

What's more, I don't understand your "don't flash Cantor on me". His diagonal argument is really brilliant and I remember I was deeply moved by it when I saw it for the first time. Try yourself!

And, of course, nothing like a fair dice exists in nature. Maybe, just maybe, quantum physics will give us some way of having true randomness? I don't know, but it isn't that meaningful. Let's play with mathematics, who cares about nature?

@Otienimous You're right. The dice procedure samples from the reals, as the probability of rolling an infinitely repeating sequence of numbers (to get a rational) is zero.

@Otienimous Such stupidities that my proof doesn't take all possibilities into consideration, you go discuss with ELUBAS, BOTS and all the stupidos in this thread. This is cruel math, I draw them out and hit you on the head, all of you! You just gave a result, you read somewhere but never proved anything!
Well, read on...
A probability is defined as the ratio of the number of times something occured over the size of the sample space (all possible outcomes of the experiment.)
So take N=2 and you are looking for 00 say. How many times did 2 people draw a 0 each ( 0 and 1 are the only allowable digits for simplicity), if the experiment is to be repeated 2 times. So, the sample space is:
{ 00 , 00 }, { 00 , 01 }, { 00 , 10 }, { 00 , 11 }, { 01 , 00 }, { 01 , 01 }, { 01 , 10 }, { 01 , 11 }, { 10 , 00 }, { 10 , 01 }, { 10 , 10 }, { 10 , 11 }, { 11 , 00 }, { 11 , 01 }, { 11 , 10 }, { 11 , 11 }. As you can see for N=2 are 16 distinct and different outcomes. How many times did you find a 00 in one of them? Answer 7. Please don't count 00 twice in { 00 , 00 }, it has no meaning but if you are stupid enough to insist +1 won't make much difference for large N.
Is 7/16 the answer I gave you in my formula in post #43? YES. Do the same for all N to convince yourself until you reach infinity. What is a more quick way to reach infinity? N people and 2N repititions? Adjust the formula accordingly: 1-(1-(1/p)^N)^2N. How stupid can one be as not to get this at once? But for your information, there is no way to reach infinity quicker, because infinity is never reached! You just work with bigger N until you satisfy yourself.
Ok, son? Are you happy now? Kiss ELUBAS for me and tell him to stick his philosophies in his ass.