09 Mar 21
@joe-shmo saidTo start with 2 die(as opposed to a million!) the combinations are 6!/2! =15 plus six doubles =21.
Given 10^6 regular ( 6 - sided) dice thrown simultaneously, what is the probability that the product of all the results is even? Express the result as "(a-c)/b" where "(a-c)" and "b" are coprime.
the probability of the product of 21 possibilities being even is 15/21.
I think co prime means consecutive numbers so that takes out another 6 ,so the probability of the product of 21 numbers excluding co primes is 9/21.
I suppose you have to multiply this up by 500k somehow to get the answer.
A bit beyond me I'm afraid my friend but I'll be interested in the answer when some bright spark get's round to it!
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09 Mar 21
@venda saidco prime just means a fraction in lowest terms ( no common factors between numerator and denominator )
To start with 2 die(as opposed to a million!) the combinations are 6!/2! =15 plus six doubles =21.
the probability of the product of 21 possibilities being even is 15/21.
I think co prime means consecutive numbers so that takes out another 6 ,so the probability of the product of 21 numbers excluding co primes is 9/21.
I suppose you have to multiply this up by 500k somehow to ...[text shortened]... I'm afraid my friend but I'll be interested in the answer when some bright spark get's round to it!
2/3 coprime
4/6 not coprime
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10 Mar 21
@venda saidThis problem isn't so bad.
Thanks.I stand corrected(as usual!)
Just to be sure the concept is clear: If we were rolling 2 dice, I'm asking for the probability the product of the resulting rolls is even:
For example: If the rolls were 3,4 we would have 3*4 = 12 = even
So... I'm asking for the ratio of Even Results to ALL results...for 10^6 dice ( not 2 )
10 Mar 21
@joe-shmo saidI realise that Jo.
This problem isn't so bad.
Just to be sure the concept is clear: If we were rolling 2 dice, I'm asking for the probability the product of the resulting rolls is even:
For example: If the rolls were 3,4 we would have 3*4 = 12 = even
So... I'm asking for the ratio of Even Results to ALL results...for 10^6 dice ( not 2 )
Just that a million dice is a bit difficult to imagine.
I might have another think about it.
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10 Mar 21
@venda said"Just that a million dice is a bit difficult to imagine."
I realise that Jo.
Just that a million dice is a bit difficult to imagine.
I might have another think about it.
If I may ( I'm not trying to be an arse hole ) ....start with 1 die, and work your way up. if you do that right for the first few dice, you will have no trouble with the million.
@joe-shmo saidWhatever it is, it will be extremely high, in the high 99%++ probability range because all you need is for one of those dice to roll 2, 4 or 6 to make that product even.
Given 10^6 regular ( 6 - sided) dice thrown simultaneously, what is the probability that the product of all the results is even? Express the result as "(a-c)/b" where "(a-c)" and "b" are coprime.
To get an ODD product, you need all of the rolls to be odd. The chances of that happening are 0.5^(10^6). So the probability of getting an even product will be found by:
1.0 - (0.5)^(10^6)
I'm not sure what to do with this part:
Express the result as "(a-c)/b" where "(a-c)" and "b" are coprime.
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11 Mar 21
4 edits
@athousandyoung saidCorrect ATY!
Whatever it is, it will be extremely high, in the high 99%++ probability range because all you need is for one of those dice to roll 2, 4 or 6 to make that product even.
To get an ODD product, you need all of the rolls to be odd. The chances of that happening are 0.5^(10^6). So the probability of getting an even product will be found by:
1.0 - (0.5)^(10^6)
I ...[text shortened]... ith this part:
Express the result as "(a-c)/b" where "(a-c)" and "b" are coprime.
Don't worry about the last part. I was just looking for you to format the probability as:
P( Even product ) = ( 2^n - 1 ) /2^n
where
n= 10^6
That's not important.