There is 5 players with 5 dices each = 25 throws per game Whats the probability that there is X dices with the value Y(1-6) on the table. for example. probability for 8 three's this is calculated through the binomial theorem right? http://en.wikipedia.org/wiki/Binomial_theorem where n=25, k=8 och p=1/6 calculating the probability of having X dices with the same value could be calculated with the above times six right? so far so good. now it gets a bit more complicated. how to calculate: how many dices of the same value is the most probable result? (assuming all 25 dices has been thrown) Also, conditional probability: i know that i have got Z dices with the same value (1-6 it doesnt matter which value). What is now the most probable result of the number of dices with the same value on the table?(i know my own result, but i dont know the 4 other players result) for instance: i get 4 three's. what is the most likely number of three's on the table totally of the 25dices? please refer to probability theory and distributions if you know them.
First "dices" is the third person singular present tense of a verb, what you do to vegetables; it is not the plural of the noun "dice". "dice" is itself the plural of the noun "die". I say this not as a criticism but because many people here do not have English as a first language- and it is a complicated language! The binomial distribution you describe would be the probability of getting "x" dice to be a specific number. The probability of getting "x" dice the same (but any number between 1 and 6) would be a "multinomial" distribution because you might get, say "x" dice equal to 5, "y" dice equal to 3, etc. The other way, determining the "most likely" number of dice that are the same (and again, there might be "x" dice that are equal to one number, "y" dice that are equal to another, etc.) requires calculating all the probability of each such number and seeing which is the largest.
haha thank you I did not know that about dices. English is not my first language. anyway the multinomial formula being (n!/(x1!...xk!))*p^x1...p^xk in the above case: how many dices of the same value is the most probable result? (assuming all 25 dices has been thrown) n=25 p=1/6 but what exactely is x1 to xk?