Rolling 6 dice and 4 on the same side

Question

Guys I made myself a question to answer and its this. What is the probability of having 4 dice land on the same side if I toss 6 dice. I got this answer, I don't know if its correct. I dont want to assume they are fair dice, so:

Let the probability of getting a certain number be P.

Probability of getting exactly 4 same sides is:

10C4 x (P^4) x ((1 - P)^2)

Answer 1

Exactly $4$ out of $6$ dice must be picked out as the dice that land on the same side. There are $\left({6\atop 4}\right)=15$ possibilities for that. Then the side on wich these dice land must be picked out: $6$ possibilities. For the remaining $2$ dice there are $5$ possibillities each to land on: $5\times5=25$ possibilities. This leads to $15\times6\times25=2250$ possibilities. Totally there are $6^{6}$ possibilities. So the probability that exactly $4$ land on the same side is $2250\times6^{-6}$.

Answer 2

It generally goes nPr * p^r * q^ n-r So in your case 6 P4 * (4/6)^4 * (2/6)^2

Answer 3

The outcome when you roll six die is an ordered 6 tuplet (a,b,c,d,e) , in which each of a,b,c,d,e can take values from 1 to 6. Hence, there are 6^6 possible outcomes in total.

The number of outcomes in which exactly 4 die land on the same side is C(6,4).6.5.5.

Hence, the probability of you getting exactly 4 die on the same face when you roll 6 die is C(6,4).6.5.5/6^6 = 625/7776 (approximately 0.08).