I'm getting a bit dumb on the probabilities. - page 5

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Go ahead and blurt it out then. Write your whole reasoning for three cubes.

Для трех кубиков вероятность выпадения хотя-бы одной шестерки равна 1-вероятность ее невыпадения

For now.

 
OK, let's keep it simple: there are numbers with 4 digits. The digits are 1...6. How many numbers are there without sixes?
 
Mathemat:

Go ahead and blurt it out then. Write your whole reasoning for three cubes.

For now.


I can do it for 300cc as well :)
 

Alexei, this is no longer a probability, but a mathematical expectation :(

 
You go ahead and write the solution to the three-cube problem. I'll try to criticise.
 

I threw (I mean threw a die) - one outcome. I threw another one, then another... You can roll all at once - no difference...

 

So far, so good. Next...

So you turn a single roll of 4 dice into a series of 4 rolls with p = 1/6? (CHUCKLES) OKAY.

Okay, I get it. Same Bernoulli series. And the answer is the same as mine :)

 

Criticise, don't criticise: the probability of a six is 1/6. For four independent throws the expectation of winning = 4*1/6=2/3. If all four throws occur simultaneously, the situation does not change

 

In short, the probability of the final event in this case is a false concept, like centrifugal force :)

 
You're not being clever. Just tell me the probability of at least one 6. Don't give me that m.o. crap.

Короче, вероятность итогового события в данном случае - ложное понятие, наподобие центробежной силы :)

Ah, I see. So you're saying that by rolling 4 dice a billion times, we don't get some measure of the "at least one 6 fell" event close to a theoretical value?

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