# Thread: need help to confirm my probability understanding

1. ## need help to confirm my probability understanding

I have 4 unbiased dice with 6 sides.

1) I want dice1 to have value "1" and any values from the other 3 dice.
Ans: 1/6 x 1 x 1 x 1 = 1/6

2) I want at least 3 dice to have values "1"
Ans: 4c3 x 1/6 x 1/6 x 1 x 1 = 1/9

3) I want at least 3 dice with the same values (any value as long as it's the same)
Ans: 4c3 x 1/6 = 2/3

Need some help to check if my understanding of probability is correct. It has been a long time since I worked on this topic. I think I might be wrong with no.3 or if I'm wrong with any, please advise.

Thanks!

2. Originally Posted by zepher
I have 4 unbiased dice with 6 sides.

1) I want dice1 to have value "1" and any values from the other 3 dice.
Ans: 1/6 x 1 x 1 x 1 = 1/6

2) I want at least 3 dice to have values "1"
Ans: 4c3 x 1/6 x 1/6 x 1 x 1 = 1/9
No, for exactly three of them show a "1" so the probability is:

4c3 (1/6)^3 = 1/56

and for four of then to show 1 the probability is:

4c4 (1/6)^4=(1/6)^4

So the probability of three or more 1's is:

4c3 (1/6)^3 + (1/6)^4

3) I want at least 3 dice with the same values (any value as long as it's the same)
Ans: 4c3 x 1/6 = 2/3
No, we gave the probability that three or more show a specified face in 2 above, but
there are 6 possible specifications, so this is 6 times the answer to 2.

RonL

3. Originally Posted by CaptainBlack
No, three of them show a "1" so the probability is:

4c3 (1/6)^3 = 1/56

RonL
Uhhh...

Shouldn't that be 4c3 (1/6)^3 + 4c4 (1/6)^4?

The original post asked for the probability of at least three 1's-- in other words, three 1's or four 1's.

4. Originally Posted by awkward
Uhhh...

Shouldn't that be 4c3 (1/6)^3 + 4c4 (1/6)^4?

The original post asked for the probability of at least three 1's-- in other words, three 1's or four 1's.
Opps, yes. Post corrected.

RonL

5. thanks a bunch!