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You're not being snarky. Just tell me the probability of at least one six. Don't give me a hard time about the m.o.
Aleksey, imho we're not messing with each other, but we're looking at a specific problem. In which your namesake beat everyone else in a dice game.
I can, of course, bring the precision of the expression (1-p1)(1-p2)... to a value, but I don't understand: what does this have to do with the original DRKNN bike, or its derivatives :)
Well for seven cc's can you get to the probability figure?
And with seven cc's, what would it be? 7/6?
There seems to be a club for adding probabilities - instead of multiplying them :)
There are no interdependencies, of course. But independent events are multiplied by probabilities. Here's how it works: we draw 4 cells. Each of them gets a number when the cube is rolled. The probability that anything other than 6 gets into the first one is 5/6. It is the same for all the others. That's why the 4th degree is 5/6. And then you subtract the result from 1.
Here's a question: the probability of getting the right card from the deck for the first deal is 16%, for the second deal - 32%.
Can we sum the probabilities and say that the probability is 48%?
You're not being clever. Just tell me the probability of at least one 6. Don't give me that m.o. crap.
In short, the probability of the final event in this case is a false notion, like centrifugal force :)
Oh, I see. So you're saying that rolling 4 dice a billion times won't produce some measure of the event "at least one 6 fell" that's close to a theoretical value?
Alexei, what does proximity have to do with anything? I roll four dice a billion times, I get an MO of 666,666,666. Other than that, I wholeheartedly agree with you.
Here's a question: the probability of getting the right card out of the deck for the first deal is 16%, for the second 32%.
Is it possible to sum the probabilities and say that the probability is 48%?
I don't know, the conditions are too strange. You probably can't.
Well, if you really want to, you can.
Incidentally, in our case, a billion times four dice is equivalent to four billion times one dice, or four times a billion dice.
And I get about 518 million. But I won't check it.
By the way, it's easy to check if you simulate in MQL4.
You could also do a field experiment at the weekend :)
Svetlana, excuse me - we've been having a little chat with the namesake. What are you doing this morning?