# Thread: Fair Dice Tossed Independently

1. ## Fair Dice Tossed Independently

Seven fair dice are tossed independently. Find the probability that the number of 1's minus the number of 4's will be three.

Thoughts?

2. ## Re: Fair Dice Tossed Independently

What are all of the possible outcomes that satisfy that condition?

 # of 1's # of 4's # of (not 1's, not 4's) 3 0 4 4 1 2 5 2 0

Those are the only outcomes that you care about. So, what are the probabilities of each?

I will help you out for the first case, see if you can figure out the other two. The probability that three out of the seven dice will roll a 1 and the other four will roll neither a 1 nor a 4: $\displaystyle \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4$

3. ## Re: Fair Dice Tossed Independently

Originally Posted by SlipEternal
I will help you out for the first case, see if you can figure out the other two. The probability that three out of the seven dice will roll a 1 and the other four will roll neither a 1 nor a 4: $\displaystyle \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4$

So is it just: $\displaystyle P = \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4 + \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{7}{1 }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2 + \binom{7}{5}\left(\dfrac{1}{6}\right)^5\binom{7}{2 }\left(\dfrac{1}{6}\right)^2$?

4. ## Re: Fair Dice Tossed Independently

Originally Posted by vidomagru
So is it just: $\displaystyle P = \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4 + \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{7}{1 }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2 + \binom{7}{5}\left(\dfrac{1}{6}\right)^5\binom{7}{2 }\left(\dfrac{1}{6}\right)^2$?
Not quite. You have $\displaystyle \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{7}{1 }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2$
This means out of seven dice, you choose 4 to roll 1's, then you replace those four dice and choose 1 to roll a 4. You don't want to replace the dice. So, it would be:

$\displaystyle P = \binom{7}{3}\left(\dfrac{1}{6}\right)^3\left( \dfrac{4}{6} \right)^4 + \binom{7}{4}\left(\dfrac{1}{6}\right)^4\binom{3}{1 }\left(\dfrac{1}{6}\right)\left( \dfrac{4}{6} \right)^2 + \binom{7}{5}\left(\dfrac{1}{6}\right)^5\binom{2}{2 }\left(\dfrac{1}{6}\right)^2$