If four 4-sided dice are rolled, what is the probability of rolling at least three of the same number? I got 13/64 but Im unsure about it....this is how I did it: The first die does not matter A.Chance of second die matching: 1/4 B.Chance of second die matching and last two dice not matching: 9/64 C.Chance of second die failing and third and fourth dice matching: 3/4 * 2/4 * 1/4 = 6/64 Chance of three of a kind: A-B+C = 13/64

Its correct : 52/256 = 13/64

Could you show me how you did it? Is there a better way to figure it out?

hi

D1 D2 D3 D4 (D stands for dice)

suppose we have 4 place holders, we wanna calc out the number of states which meet our conditions, first we calc the number of states where there are exactly 3 dices with the same number :

4 x 1 x 1 x 3 = 12
first one can be any number, second and third must be just the same as first, the forth must be any number except the first (i.e. 3)
those 3 same numbers with the forth one can make 4 combinations by replacement so we must multiply 12 with 4 = 48

now we calc the states where all dices are the same it is obviuosly 4 states.

adding 48+4=52 this is the number of states we have our conditions met, the propablity it the number of anwers divided by the total number of possible combinations which is 4^4

so p = 52/256 = 13/64

thanks for the help