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Count the possibilities, to get those results, without counting anything more than once. e.g. there are 6 results where all three are the same. Then divide that number by the number of all possible outcomes (which is 6^3)
I do not understand ur reasoning?
It is an Laplace experiment. So in order to get the probability you have to divide the number of positive outcomes by the number of all outcomes.
but when you throw the dice once, you have the probability 1/6 that it is 5 for example but here the dice is thrown 4 times?
do i have to multiply the probabilities?
or is it conditional probability p(A/b)
no, you should take every quadruple-combination as a single outcome.
so in total there are 6^4 results I think?
now how do i find the rest?
There are certainly several ways to count. I would count the disjount situations 1 Pair; 2 Pairs and 1 Tripel seperately. Use binomial to see which of the dices are equal and then multiply with the possible count of numbers.
is it 66/6^4
yes, thats what i did and i think there were 66?
toooooo complicated, nyway thank you
Well I get 6*6*5*4+4*6*5+6*6*5 = 1020 which makes about .787
ah ok, I really made it too complicated you should look at the complementary event: "all 4 numbers are different"
there you can count easily 6*5*4*3 = 360
ah, got it now, thats easier
so the answer must b 6*5*4/6^4
now yes! thanx