If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?

RektTheNoob
Feb 3, 2018

#1**0 **

If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?

My answer had fatal flaws. Alan fixed it with his answer. :)

Thanks Alan.

Melody
Feb 3, 2018

#3**0 **

**ok so what do you think the correct answer is and what makes you think it is correct?**

**AND if you have the working we would like to see that too. **

**Please do not just say that any answer is wrong without justifying your remark.**

**I, and others, would like to learn too.**

Melody
Feb 4, 2018

#5**+2 **

Solution:

We notice that the probability that he rolls more 1's than 6's must equal the probability that he rolls more 6's than 1's. So, we can find the probability that Greg rolls the same number of 1's and 6's, subtract it from 1, and divide by 2 to find the probability that Greg rolls more 1's than 6's. There are three ways Greg can roll the same number of 1's and 6's: he can roll two of each, one of each, or none of each. If he rolls two of each, there are \(\binom{4}{2}=6 \) ways to choose which two dice roll the 1's. If he rolls one of each, there are \(\binom{4}{1}\binom{3}{1}=12\) ways to choose which dice are the 6 and the 1, and for each of those ways there are \(4\cdot4=16\) ways to choose the values of the other dice. If Greg rolls no 1's or 6's, there are \(4^4=256\) possible values for the dice. In total, there are \(6+12\cdot16+256=454\) ways Greg can roll the same number of 1's and 6's. There are \(\dfrac{1}{2} \left(1-\dfrac{454}{1296}\right)=\boxed{\dfrac{421}{1296}}\) total ways the four dice can roll, so the probability that Greg rolls more 1's than 6's is .

RektTheNoob
Feb 5, 2018

#7**-1 **

RektTheNoob:

If you know the answers to the questions you post on this forum in such detail as the above answer demonstrates, then what is the PURPOSE of posting your questions in the first place?? Are you trying to test the Mods and other volunteers' mathematical knowledge? Or are you trying to show your mathematical superiority??!!

Guest Feb 5, 2018

#10**0 **

**RektTheNoob posted this detailed answer because I asked him too !!**

Presumable he did not know the answer when he asked the question but was given or worked out the answer afterwards.

I am very glad that he did post this answer as I learned from it.

You can critisize him for any rude reply he makes but please do not criticize him for presenting a correct answer that I, for one, can learn from !!

Melody
Feb 5, 2018

#4**+2 **

Modifying Melody's result a little we have:

1 * * * 1*4*4*4 = 64 ways, but 4 possible positions for the 1, so 64*4 = 256 ways

1 1 * * 1*1*4*4 = 16 ways, but 6 possible combinations of the two 1's, so 16*6 = 96 ways

1 1 6 * 1*1*1*4 = 4 ways, but 6 possible positions for the 1's

and two possible remaining positions for the 6, so 4*6*2 = 48 ways

1, 1, 1, any 1*1*1*5 = 5 ways, but 4 possible positions for the three 1's, so 5*4 = 20 ways

total = 256 + 96 + 48 + 20 = 420 ways

Hence probability = 420/1296 → 35/108 ≈ 0.324

.

Alan
Feb 4, 2018

#9**+2 **

Thanks you Alan and Rekt for your detailed answers.

Firstly I am embarrased by the fundamental error I made.

Thank you Alan for correcting it :)

However, Alan you made one small omission.

1, 1, 1, any 1*1*1*5 = 5 ways, but 4 possible positions for the three 1's, so 5*4 = 20 ways

This should have been 1,1,1, not 1 and there are indeed 20 ways to get this result

But the last one is 1,1,1,1, there is 1 way to get this result so the number of ways is

total = 256 + 96 + 48 + 20 + **1** = 42**1** ways

Hence probability = 421/1296 ≈ 0.325

**Now Alan's answer is the same as Rekt's answer. **

I really like the way you did this question Rekt, it makes total sense and I am very glad you have shown me.

Melody
Feb 5, 2018