# Rolling Dice - Discrete Statistics

Let $$X$$ be the random variable that is defined as the smaller of the two numbers that appear when a pair of dice is rolled. For example, if you roll 2 and 5, then $$X = 2$$.

a) Determine the expected value and the standard deviation of $$X$$ for two fair dice whose results are independent of each other.

What I've got so far: $$s={1, 2, 3, 4, 5, 6}$$ $$P=1/6$$ for all outcomes

36 possibilities for all outcome considering a pair of two dice $$(a,b)$$. I need to find Expected value $$E(X)$$, and the Standard deviation $$= \sqrt{\operatorname{Var}(X)}$$ I've got $$E(X) = 1(1) + 3(2) + 5(3) + 7(4) + 9(5) + 11(6)=\frac{161}{36}$$

b) Someone inadvertently steps on one of the two dice, thereby flattening it so that from now on it can only show the numbers one and six (with equal probability). Determine the expected value (no standard deviation) of $$X$$ after this accident.

c) To make things worse, suddenly a mysterious mechanism comes into effect: The sum of the numbers of the two dice (the flat and the normal one) is never seven anymore, while all other possible outcomes, i.e., pairs of numbers, occur with equal probability. What is the new expected value of $$X$$?

$$E(X) = \frac{5(6) + 3(5)}{10}$$

d) In the final, mysterious state of the dice, are the numbers that appear on the flat die ($$X_1$$) and on the normal die ($$X_2$$) independent? Prove your answer.

• There are only $36$ equiprobable combinations for the dice (fewer in your later problems). Just list them and do the calculations by hand if nothing else comes to mind.
– lulu
Commented Apr 12, 2018 at 11:10
• @lulu equiprobable... If only I knew of that word, that would have made my verbal questions easier to understand, hahah :) Commented Apr 12, 2018 at 16:37

For (a),

$$X=\min(X_1,X_2)$$ where $X_i$ is the outcome from each die.

So the possible outcomes are

\begin{array}{c|cccccc} &1&2&3&4&5&6\\ \hline 1&1&1&1&1&1&1\\ 2&1&2&2&2&2&2\\ 3&1&2&3&3&3&3\\ 4&1&2&3&4&4&4\\ 5&1&2&3&4&5&5\\ 6&1&2&3&4&5&6\\ \end{array}

So as you have done (but somehow with max instead), we have $$E(X)=\frac{1(11)+2(9)+3(7)+4(5)+5(3)+6(1)}{36}=\frac{91}{36}$$

and

$$E(X^2)=\frac{1^2(11)+2^2(9)+3^2(7)+4^2(5)+5^2(3)+6^2(1)}{36}=\frac{301}{36}$$

So $$\sigma=\sqrt{E(X^2)-E(X)^2}=\frac{\sqrt{2555}}{36}$$

For (b),

Now the possible outcomes become

\begin{array}{c|cccccc} &1&2&3&4&5&6\\ \hline 1&1&1&1&1&1&1\\ 6&1&2&3&4&5&6\\ \end{array}

So we have $$E(X)=\frac{1(7)+2(1)+3(1)+4(1)+5(1)+6(1)}{12}=\frac{9}{4}$$

For (c),

We simply cross out the last column

\begin{array}{c|cccccc} &1&2&3&4&5&6\\ \hline 1&1&1&1&1&1&\times\\ 1&\times&2&3&4&5&6\\ \end{array}

So we have $$E(X)=\frac{1(5)+2(1)+3(1)+4(1)+5(1)+6(1)}{10}=2.5$$

For (d),

We simply check if $$P(X_1=x_1)P(X_2=x_2)=P(X_1=x_1,X_2=x_2)$$

holds.

We know that the probability distribution is simply

\begin{array}{c|ccccc} &1&2&3&4&5\\ \hline 1&\frac1{10}&\frac1{10}&\frac1{10}&\frac1{10}&\frac1{10}\\ 6&\frac1{10}&\frac1{10}&\frac1{10}&\frac1{10}&\frac1{10}\\ \end{array}

while $$P(X_1=1)=P(X_1=6)=\frac12$$ and $$P(X_2=1)=P(X_2=2)=\ldots=P(X_2=6)=\frac16$$ are still in effect.

Very clearly,

$$P(X_1=1,X_2=6)=0\ne\frac1{12}=P(X_1=1)P(X_2=6)$$

Hence, $X_1,X_2$ are no longer independent.

• Beautifully explained! $(+1)$ if I didn't reach my daily voting limit :) Commented Apr 12, 2018 at 16:41
• Hi, I have a question about 1c. The initial question says: Let X be the random variable that is defined as the smaller of the two numbers that appear when a pair of dice is rolled. For example, if you roll 2 and 5, then X=2. Wouldn't we eliminate 1 from first row last column, and 1 from second row first column. E(x) = 25/10=2.5 I am confused.
– Fara
Commented Apr 17, 2018 at 17:30
• @Fara (2,5) isn't a possible combination in (c). Commented Apr 17, 2018 at 17:59
• For 1c 1 1 1 1 1 x x 2 3 4 5 6 E(X) = 5/10⋅1 + 1/10⋅2 + 1/10⋅3 + 1/10⋅4 + 1/10⋅5 + 1/10⋅6 = 25/10 = 2.5
– Fara
Commented Apr 17, 2018 at 22:26
• @Fara Oh i see now my bad! Commented Apr 18, 2018 at 3:09