You are rolling two dice numbered 1 through 6.
1. The score for a roll is the sum of the two numbers rolled. What is the expected value for the score? What is the value and probability of the most likely score? Show your work.
2. The score for a roll is the product of the two numbers rolled on the dice. What is the expected value for the score?
3. The score for a roll is still the product of the two numbers rolled on the dice, only this time, if you roll doubles, you lose twice the product. Example: If a 2 and 3 is rolled, you gain 6 points. But if you roll two 5's you lose 2*5*5= 50 points. What is the expected value for one turn? Show your work.
+981 Hey Guest! I'm not too familiar with expected value, but I can still give it a go:
NOTE: THIS SOLUTION IS WAY TO DIFICULT, LOOK AT MY SOLUTION TO 2. FOR THE CORRECT ONE.
I'm not sure if this is the best way to do this, but it works.
The possible sums of two dice are \((2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)\)
The probabilty of getting a sum of 2 is: \(\frac{1}{36}\). The numerator is one, because there is one way of getting a sum of 2, 1 + 1.
We do the same for the rest of the sums,
The probabilty of getting a sum of 3 is: \(\frac{2}{36}\), (1,2) ; (2,1)
The probabilty of getting a sum of 4 is: \(\frac{3}{36}\), (1,3) ; (3,1) ; (2,2)
The probabilty of getting a sum of 5 is: \(\frac{4}{36}\), (1,4) ; (4,1) ; (2,3) ; (3,2)
The probabilty of getting a sum of 6 is: \(\frac{5}{36}\), (1,5) ; (5,1) ; (2,4) ; (4,2) ; (3,3)
The probabilty of getting a sum of 7 is: \(\frac{6}{36}\), (1,6) ; (6,1) ; (2,5) ; (5,2) ; (3,4) ; (4,3)
The probabilty of getting a sum of 8 is: \(\frac{5}{36}\), (2,6) ; (6,2) ; (3,5) ; (5,3) ; (4,4)
The probabilty of getting a sum of 9 is: \(\frac{4}{36}\), (3,6) ; (6,3) ; (4,5) ; (5,4)
The probabilty of getting a sum of 10 is: \(\frac{3}{36}\), (4,6) ; (6,4) ; (5,5)
The probabilty of getting a sum of 11 is: \(\frac{2}{36}\), (5,6) ; (6,5)
The probabilty of getting a sum of 12 is: \(\frac{1}{36}\), (6,6)
The expected value is:
\((\frac{1}{36}\cdot2)+(\frac{2}{36}\cdot3)+(\frac{3}{36}\cdot4)+(\frac{4}{36}\cdot5)+(\frac{5}{36}\cdot6)+(\frac{6}{36}\cdot7)+(\frac{5}{36}\cdot8)+(\frac{4}{36}\cdot9)+(\frac{3}{36}\cdot10)+(\frac{2}{36}\cdot11)+(\frac{1}{36}\cdot12)\)
When you simplify:
\(\frac1{18}+\frac16+\frac13+\frac59+\frac56+\frac76+\frac{10}{9}+\frac11+\frac{10}{12}+\frac{11}{18}+\frac13\)
When you add these up, you get 7.
I hope this helped,
Gavin
+981 2.
All we need to do is find the expected value for one die, then do what the problem asks us to do.
The same with the first problem, the expected value for the first die was:
\(\frac{1+2+3+4+5+6}{6}=3.5\),
\(3.5+3.5=\boxed7\), which was what we got after all that work.
This method is way easier and harder to make careless mistakes.
We can do the same with #2. We already got the expected value for one die, the product is just:
\(3.5\cdot3.5=\boxed{12.25}\)
I hope this helped,
Gavin
