Can you solve this math riddle? • page 6/6 • Off-Topic Discussion • lichess.org

bunyip

#51

I still say the answer is 42...and i have documentary proof of it.

lecctra

#52

@bunyip Go on and post it Douglas. Is your true name ELUBAS by the way?

bunyip

#53

HAHAHA @lecctra

(and no...i don't do bongcloud openings)

HAHAHA @lecctra (and no...i don't do bongcloud openings)

elecctrarevisited edited

#54

@acgusta2
So sad that nobody has answered your question yet!
The answer to your question is that at least an Aleph-1 infinity number of people would suffice for your group to generate 0.3333333... with probability greater than 0.

-PROOF-

One person has probability (1/10)^N of generating 0.3333...3. There are N number of 3's here.
M persons have probability 1-(1-(1/10)^N)^M of generating the same number 0.3333...3 as the late user @lecctra ( RIP ) has shown us.
When M=N, this probability tends to 0 as N goes to infinity, so nothing special there.
But, try to put M=10^N into the formula and you get a limit of 1-1/e as N goes to infinity ( e is the natural logarithm base ). Put M=A*(10^N) for any number A and this probability becomes 1-(1/e)^A as N tends to infinity. When the number A tends to infinity itself the probability is 1!
You can see immediately that for N=Aleph-0 then: M=10^Aleph-0 = 2^Aleph-0 = Aleph-1.

So the answer is: it depends on the kind of infinity you are using for your group of M people to generate an infinite sequence of digits. The probability will be any number ranging from 0 to 1 depending on the kind of infinity you use. To paraphrase @bunyip the probability might as well be 42% !!!!

@acgusta2 So sad that nobody has answered your question yet! The answer to your question is that at least an Aleph-1 infinity number of people would suffice for your group to generate 0.3333333... with probability greater than 0. -PROOF- One person has probability (1/10)^N of generating 0.3333...3. There are N number of 3's here. M persons have probability 1-(1-(1/10)^N)^M of generating the same number 0.3333...3 as the late user @lecctra ( RIP ) has shown us. When M=N, this probability tends to 0 as N goes to infinity, so nothing special there. But, try to put M=10^N into the formula and you get a limit of 1-1/e as N goes to infinity ( e is the natural logarithm base ). Put M=A*(10^N) for any number A and this probability becomes 1-(1/e)^A as N tends to infinity. When the number A tends to infinity itself the probability is 1! You can see immediately that for N=Aleph-0 then: M=10^Aleph-0 = 2^Aleph-0 = Aleph-1. So the answer is: it depends on the kind of infinity you are using for your group of M people to generate an infinite sequence of digits. The probability will be any number ranging from 0 to 1 depending on the kind of infinity you use. To paraphrase @bunyip the probability might as well be 42% !!!!

Savage47

#55

@acgusta2 @NoobBatter

"Dice" is plural. "Die" is singular

Credit for correctly using "math" instead of the incorrect "maths" though.

@acgusta2 @NoobBatter "Dice" is plural. "Die" is singular Credit for correctly using "math" instead of the incorrect "maths" though.

Savage47

#56

The question isn't worded in a way that it can be answered accurately.

"Using this method one individual generates the number 1/3"
Presumably a 10 sided die would have only whole numbers so this statement is nonsensical. I can try and guess what you meant but I shouldn't have to if you're trying to create an interesting question.

The question isn't worded in a way that it can be answered accurately. "Using this method one individual generates the number 1/3" Presumably a 10 sided die would have only whole numbers so this statement is nonsensical. I can try and guess what you meant but I shouldn't have to if you're trying to create an interesting question.

elecctrarevisited

#57

@Savage47 No! The question is worded accurately enough and posts #1 and #8 describe the method used.

Otienimous

#58

@elecctrarevisited I'm sure I have already answered the original question somewhere in this deluge of nonsensical posts, but I'm not surprised you couldn't find it. Moreover, you showed the proof very nicely and concisely.