Roy Tang

please forward dice game related questions to Zel Teng 's father daw

Di po ba prob ng 4 of a number except 4 = 5/(6^4) Prob ng 3 na 4 = 1/(6^3)?

Okay, I stand corrected. The problem was I was only considering the order mattering for the nonsignificant dice, but I also need to take into account the probability of where the significant rolls would appear. For rolling exactly three fours: Assuming only the first 3 dice are fours, the number of successful outcomes is 1 x 1 x 1 x 5 x 5 x 5 (it's allowed that the last 3 dice are the same). Multiply this by the number of possible placements of the 3 fours: (combination of 6 taken 3 at a time) which is 20, the total number of successful outcomes is 2,500. 2,500 out of 6^6 is roughly 0.05358 (confirmed by wolfram alpha: http://www.wolframalpha.com/input/?i=probability+of+rolling+three+fours+with+six+die ) For rolling exactly four of any nonfour: Assuming only the first 4 dice are the same, the number of successful outcomes are 5 x 1 x 1 x 1 x 5 x 5 (5 choices for die#1, then dies 2-4 are the same value). Multiply this by the number of possible placements of the 4 same values: (combination of 6 taken 2 at a time) which is 15, the total number of successful outcomes is 1,875. 1,875 out of 6^6 is roughly 0.04018 (also confirmed by wolfram alpha: http://www.wolframalpha.com/input/?i=probability+of+rolling+four+ones+with+six+die - this link only gives one fifth of the probability) Above numbers also confirmed by simulation. Matatahimik na ko. I should get some more sleep :p

may combination pa nga po pala. at 5 lang dapat yung choices sa non-4/. waha! :p