Two dices are rolled let x be a random variable denoting the absolute value of the different of outcomes in the two dice construct the possible distribution x

For the max to be 4, you could be in the row numbered 4 from column 1 to column 4, or you could be in the column numbered 4 from row 1 to row 4. There are 7 items overall because you need to make sure not to double count 4,4.

Repeating this process, for 2 six sided dice, your pmf would be:

f(x)=2x−162 for x=1,2,3,4,5,6

If you wanted to generalize it for dice having  n  sides, that would become:

f(x)=2x−1n2 for x=1,2,…,n−1,n

I think more people would have answered the question this way if it had been a bit more complex. For example, what if the number of rolls was not necessarily 2, but instead a previously determined constant? For example, what if the question said this: You have  n  fair six sided dice. If you roll all of them, what would be the pmf of the maximum number of pips shown on one of them?

In this more complicated version of the problem, I would define 2 random variables:

X : The result of one dice roll.

fX(x)=16 for x=1,2,3,4,5,6

Y : The maximum of the  n  dice rolls.

Y=max(X1,X2,X3,…,Xn)

This becomes a problem of order statistics, since it is a max function, a good starting point would be the CDF. If it had been a min function, the better starting point would be the inverse of the CDF.

FY(y)=P(Y≤y)=P[max(X1,X2,X3,…,Xn)≤y]

This is the part that might be the hardest if you haven’t done this type of problem before: If one or more rolls are higher than  y , then the max must be higher than  y . Therefore, if the max is less than or equal to  y , ALL of the rolls must be less than or equal to  y .

FY(y)=P(X1≤y,X2≤y,X3≤y,…,Xn≤y)

Since the dice rolls are iid, we can split those probabilities up.

FY(y)=P(X1≤y)⋅P(X2≤y)⋅P(X3≤y)⋯P(Xn≤y)

Again, since those rolls are iid, all of those CDF functions will be identical.

FY(y)=FX(y)⋅FX(y)⋅FX(y)⋯FX(y)

FY(y)=[FX(y)]n

Since  X  is a discrete uniform distribution, the CDF will be:

FX(x)=⎧⎩⎨⎪⎪0⌊x⌋61for x<0for 0≤x<7for x≥7

In case you are not familiar:  ⌊x⌋  is the floor function, it essentially rounds down, it returns the greatest integer less than or equal to  x .

Therefore, the CDF of  Y  will be:

FY(y)=⎧⎩⎨⎪⎪⎪⎪0(⌊y⌋6)n1for y<0for 0≤y<7for y≥7

Since the distribution is discrete, we can find the PMF by taking the CDF value and subtracting the previous CDF value. I will make it into a table.

yfY(y)116n22n−16n33n−2n6n44n−3n6n55n−4n6n66n−5n6n

If the dice are not necessarily 6 sided, but instead  s  sided, that can be written as:

fY(y)=yn−(y−1)nsn for y=1,2,3,…,s

If you put in 2 for  n  and 6 for  s  that becomes:

yfY(y)1136233635364736593661136

This solution does not require you to draw a square grid at all, so some people might like it a bit better.

sorry man i accidentally pressed yes my bad

step-by-step explanation: