Question

1) A traditional die is a cube with each of its six sides representing the numbers from 1 to 6. Mr. Dicey customized a die by replacing each number with its reciprocal (reciprocal of x is 1/x ). What is the expected value if you roll the customized die for 6 times?
2) A game is played by throwing 3 dice. You will win in this game if the summation of the scores of these 3 dice is 3, 4, 17, 18.

What is the variance of your expected values?

3) Two dice are rolled. [A die has six sides, each representing the score from 1 to 6]

What is the expected value if you roll these two dice for 6 times?

Answer 1

Answer:

1) $\frac{2}{5}$

2) 49.225

3) $\frac{7}{2}$

Step-by-step explanation:

1) To find the expected value of the dice we can use the following equation:

$E(x)=x_{1}*P(x_{1})+x_{2}*P(x_{2})+...+x_{n}*P(x_{n})$

So in our problem the values x will be: 1/1, 1/2, 1/3, 1/4, 1/5 and 1/6 and the probavility for all values is 1/6 so the expected values will be:$E(x)=(\frac{1}{1} *\frac{1}{6}) +(\frac{1}{2} *\frac{1}{6}) +(\frac{1}{3} *\frac{1}{6})+(\frac{1}{4} *\frac{1}{6})+(\frac{1}{5} *\frac{1}{6})+(\frac{1}{6} *\frac{1}{6})$

$E(x)=0.167+0.083+0.056+0.042+0.033+0.028=0.409\approx \frac{2}{5}$

2) To find the variance of the expected values we can use the equation:

$Var(x)=\frac{\sum_{i=1}^{n}(x_{i}-\overline{x})^{2} }{n}$

So for our problem will be:

$Var(x)=\frac{(3-10.5)^2+(4-10.5)^2+(17-10.5)^2+(18-10.5)^2}{4}$

$Var(x)=\frac{56.25+42.25+42.25+56.25}{4}$$Var(x)=\frac{196.9}{4}=49.225$

3) To find the expected value of the dice we can use the following equation:

$E(x)=x_{1}*P(x_{1})+x_{2}*P(x_{2})+...+x_{n}*P(x_{n})$

So in our problem the values x will be: 1, 2, 3, 4, 5 and 6 and the probavility for all values is 1/6 so the expected values will be:$E(x)=(1*\frac{1}{6}) +(2 *\frac{1}{6}) +(3 *\frac{1}{6})+(4 *\frac{1}{6})+(5 *\frac{1}{6})+(6 *\frac{1}{6})$

$E(x)=0.17+0.33+0.5+0.67+0.83+1=3.5\approx \frac{7}{2}$