Question 1191076: Toss a four-sided die twice. Role-playing games like Dungeons & Dragons use many different types of dice. Suppose that a four-sided die has faces marked 1, 2, 3, and 4. The intelligence of a character is determined by rolling this die twice and adding 1 to the sum of the spots. The faces are equally likely, and the two rolls are independent. What is the average (mean) intelligence for such characters? How spread out are their intelligences, as measured by the standard deviation of the distribution?
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Let's say we had 2 dice. One is red and the other blue.
Each has 4 sides, so it's a tetrahedron.
Here's a table showing all the possible sums of the red and blue dice
| + | 1 | 2 | 3 | 4 | | 1 | 2 | 3 | 4 | 5 | | 2 | 3 | 4 | 5 | 6 | | 3 | 4 | 5 | 6 | 7 | | 4 | 5 | 6 | 7 | 8 |
Eg: Rolling a pair of 4s gets us 4+4 = 8 as shown in the bottom right corner.
The instructions mention that whatever the result of the dice roll, we add on 1. This effectively means we can bump each of the 1,2,3,4 on exactly one die to get 2,3,4,5. I'll do this increment to the red die. The blue die will remain the same.
Here's the updated addition table of all possible sums, aka all possible intelligence values.
| + | 2 | 3 | 4 | 5 | | 1 | 3 | 4 | 5 | 6 | | 2 | 4 | 5 | 6 | 7 | | 3 | 5 | 6 | 7 | 8 | | 4 | 6 | 7 | 8 | 9 |
The 6 unique possible intelligence values for the character are:
3,4,5,6,7,8,9
The result 3 shows up only once. This means it has probability 1/16
The result 4 shows up exactly twice. It has probability 2/16.
And so on.
You should get this probability distribution
| X | P(X) | | 3 | 1/16 | | 4 | 2/16 | | 5 | 3/16 | | 6 | 4/16 | | 7 | 3/16 | | 8 | 2/16 | | 9 | 1/16 |
X = intelligence value for the character.
I recommend not reducing the fractions.
Next, multiply the X and P(X) values together
| X | P(X) | X*P(X) | | 3 | 1/16 | 3/16 | | 4 | 2/16 | 8/16 | | 5 | 3/16 | 15/16 | | 6 | 4/16 | 24/16 | | 7 | 3/16 | 21/16 | | 8 | 2/16 | 16/16 | | 9 | 1/16 | 9/16 |
All of the X*P(X) values have a common denominator of 16.
Adding the numerators will get us: 3+8+15+24+21+16+9 = 96
Then divide that over the common denominator: 96/16 = 6
We've shown that the sum of the X*P(X) values is 6. This is the expected value or the mean.
Long story short: we expect to have an intelligence rating of 6.
This fits with the idea that 6 is right in the middle of the distribution, and it's the most common dice roll (when adding on that extra plus 1)
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To find the standard deviation, we first subtract the mean mu = 6 from each item X, and square the result like so:
| X | P(X) | X*P(X) | (X-mu)^2 | | 3 | 1/16 | 3/16 | 9 | | 4 | 2/16 | 8/16 | 4 | | 5 | 3/16 | 15/16 | 1 | | 6 | 4/16 | 24/16 | 0 | | 7 | 3/16 | 21/16 | 1 | | 8 | 2/16 | 16/16 | 4 | | 9 | 1/16 | 9/16 | 9 |
Eg: X-mu = 3-6 = -3 which squares to 9 in the first row.
Then we multiply that (X-mu)^2 value by its corresponding P(X) value
| X | P(X) | X*P(X) | (X-mu)^2 | (X-mu)^2*P(X) | | 3 | 1/16 | 3/16 | 9 | 9/16 | | 4 | 2/16 | 8/16 | 4 | 8/16 | | 5 | 3/16 | 15/16 | 1 | 3/16 | | 6 | 4/16 | 24/16 | 0 | 0/16 | | 7 | 3/16 | 21/16 | 1 | 3/16 | | 8 | 2/16 | 16/16 | 4 | 8/16 | | 9 | 1/16 | 9/16 | 9 | 9/16 |
Like before, we have a common denominator (16).
Adding the numerators leads to: 9+8+3+0+3+8+9 = 40
Then you divide over the common 16 to get 40/16 = 2.5
The result of adding everything in the (X-mu)^2*P(X) column is exactly 2.5, and that is equivalent to the fraction 5/2.
This represents the variance of the discrete probability distribution.
Apply the square root to get the standard deviation
sqrt(variance) = standard deviation
sqrt(2.5) = 1.5811
which is approximate.
It tells us how spread out the X values are. The larger the standard deviation, the more spread out the X values will be (and vice versa).
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Answers:
Mean = Expected Value = 6 (exact)
Standard Deviation = 1.5811 (approximate)
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