# Joint distribution of different sided dice

I've just started to learn probability and statistics and I have difficulties solving the following question;

We have four different dice; one 4-sided, one 6-sided, and two 8-sided. They are all numbered usually starting from 1. One of our friends secretly grabs one of these dice at random and rolls it without showing us.

S = # of sides on the chosen die,

T = result of the roll

The question wants me to determine the joint distribution of S and T by listing all possible values of the pair (S, T) and their probabilities in a 3x8 table.

In order to solve the question, first I wrote the all possible values of S and their probabilities;

$4 -> P(X=4) = 1/4$

$6 -> P(X=6) = 1/4$

$8 -> P(X=8) = 1/2$

and I wrote the values of T = {1,2,3,4,5,6,7,8} and calculated their probabilities (for example, P(T=1) = 1/26, P(T=5) = 1/22, etc.).

Then I took the products of P(S) and P(T) for each element and put everything together in a table; But as you can see it from the table, the marginal total is 537/1144 and I believe it should have been 1 instead of that. What was my mistake calculating the table?

• How the heck did you get those numbers? You want the joint probabilities $\mathsf P(S=4, T=1), \mathsf P(S=4, T=2), \ldots ,\mathsf P(S=8, T=8)$ – Graham Kemp Feb 25 '17 at 13:44

Then I took the products of P(S) and P(T) for each element and put everything together in a table;

No, the random variables are not independent; you cannot obtain the joint probability by multiplying the marginals.   You have to use conditional probability.

For a start, those $1/104$ should be $1/16$. \def\P{\operatorname{\mathsf P}}\begin{align}\P(S=4, T=t) ~&=~ \P(S=4)\P(T=t\mid S=4) \\[1ex] &=~ \tfrac 14\cdot\tfrac 14\cdot\mathbf 1_{t\in\{1,2,3,4\}} \\[1ex] & =~ \tfrac 1{16}\cdot\mathbf 1_{t\in\{1,2,3,4\}}\end{align}

And so forth.   The rest should be recalculated likewise.

• But don't we have 26 numbers in total? 4 from 4-sided, 6 from 6-sided and 16 from two 8-sided, so why don't we consider all of them when choosing a number? – ricster Feb 25 '17 at 13:44
• What do you mean by "consider them all"? You are selecting one die and rolling it to get one result; $S$ the faces on that die, and $T$ the result of the roll. That is all you have to consider. – Graham Kemp Feb 25 '17 at 13:46

The 3X8 table should be filled with each entry equal to the chances of choosing the die for that row times the chances of seeing the result for that column. So, for example, the row for the 4 sided die will have the probabilities 1/16 for columns 1-4, and 0 for columns 5-8. The probabilities will sum to 1 as you thought.

The secrecy in choosing a die and rolling it doesn't make any sense in the context of the 3X8 table asked for. Just saying a die was randomly chosen then rolled would be enough.