First, construct the following random variables respectively:
Answer: When the number of dice is A, then E(A)=3.5.
B: Throw a dice, ignore 4, 5 and 6 in the result, and count its points (if it is 4, 5 and 6, it will be discarded and thrown again, and the next one will play.
Same), then E(B)=2.
C: Dice, ignore 1, 2,3 in the result and count its points, then E(C)=5.
D: Throw the dice, ignore 6 in the result and count its points, then E(D)=3.
E: Dice, ignore 1 in the result and count its points, then E(E)=4.
Then, start solving:
I) Let's first discuss the case where n is 1 to 6.
When N=2, 3, 4, 5, just take the random numbers b, d, e and c directly.
When N= 1, we only need to construct the random number e-d.
//Proof: E(E-D)=E(E)-E(D)= 1.
/* Description: You can actually roll dice. If you only take the result of 1, you would expect it to be 1, but the result is
A constant with zero variance is meaningless.
E-D means: roll the dice twice, the first time ignoring 6 in the result, and the second time ignoring 1 in the result.
Subtract the random number obtained by remembering the points twice. */
Similarly, when N=6, a random number 2D is constructed.
Ii) Let's discuss all integer sets N*
For a given integer N=N0 belongs to N*, divide by 7 to get the quotient p and the remainder q, then q is between 1 and 6.
Now construct a random number: 2p*A+T, where t is the random number corresponding to the expected q. Then e (2p * a+t) = 2p * e.
(A)+E(T)=7p+q=N0, which is the expected value.
/* For example: N= 134, it is134 =19 * 7+1.
Then the random number constructed is: 38*A+E-D, that is, roll the dice 38 times first and remember the sum; Then throw it twice, the first time suddenly.
Omit 1 in the result, ignore 6 in the result for the second time, subtract the points remembered twice, record the difference, and sum with the difference phase.
Just add (outline) */
2. About the solution of reference questions (please read the relevant teaching materials first)
1) note that the values of the four dice are w, x, y, z, M=min(W, x, y, z), then w, x, y, z, m.
They are all random numbers.
The result is A=(W+X+Y+Z-M)/3, which is a random number. Now find its expectation.
Yi Zhi E(W)=E(X)=E(Y)=E(Z)=3.5, and the distribution function of w is
FW(w)={
0，w & lt 1
1/6， 1 & lt； = w & lt2
2/6，2 & lt； = w< three
3/6，3 & lt； = w< four
4/6，4 & lt； = w< five
5/6，5 & lt； = w< six
1，w & gt=6
}
According to the related properties, FM (m) =1-[1-fw (m)] 4.
Get FM(m)={
0，m & lt 1
67 1/ 1296， 1 & lt； = m & lt2
65/8 1，2 & lt； = m< three
15/ 16,<= m< four
80/8 1，4 & lt； = m< five
1295/ 1296,<= m< six
1，w & gt=6
}
Then, the distribution law of m is obtained as follows:
1:67 1/ 1296
2:4 1/ 144
3: 175/ 1296
4:65/ 1296
5:5/432
6: 1/ 1296
Then calculate the expected e (m) of m = 1+979/ 1296.
E (a) = [e (w)+e (x)+e (y)+e (z)-e (m)]/3 = 4+317/3888.
2)
//Slight solution
The idea is the same. Remember that the six dice are U, V, W, X, Y, Z and N, which are the smallest three.
Sum of numbers, the result is B=(U+V+W+X+Y+Z-N)/3, which is a random number. Let's find its expectations.
For any given set of values of u, v, w, x, y and z, the following six random numbers are constructed.
M 1: choose four from these six numbers and take the minimum value; N 1: add all the M 1 thus obtained (do not take it repeatedly).
M2: Choose five of these six numbers, and take the minimum value; N2: Add up all M2 thus obtained (do not repeat).
M3: Choose six of these six numbers, and take the minimum value; N3: Add up all the M3s thus obtained (don't repeat them).
There are two points here:
I) m1* * has c (6,4) =15, M2*** has c (6,5) = 6, M3*** has c (6,6) =1.
Although each M 1 is not necessarily independent, the expectation of sum is still equal to the expectation sum.
So e (n1) =15e (m1), e (N2) = 6e (m2), e (n3) = e (m3).
Ii) We assume that the given six random numbers increase from u to z in turn, and now calculate that they are in N 1, N2 and N3.
How many times have each number been added? (Introduction)
xx U V W X Y Z
N 1 10 4 1 0 0 0
N2 5 1 0 0 0
N3 1 0 0 0 0
Then construct a random number: N=N 1-3*N2+6N3 (the coefficient is determined by the undetermined coefficient method).
So in n, these six numbers appear as follows:
N 1 1 1 0 0 0
In other words, n is the sum of the smallest three numbers.
Therefore, e (b) = [e (u)+e (v)+e (w)+e (x)+e (y)+e (z)-e (n)]/3 = [e (u)+e (v)+e (w)+e.
(X)+E(Y)+E(Z)-E(n 1)+3E(N2)-6E(N3)]/3............*
e(n 1)= 15E(m 1)= 15 *( 1+979/ 1296)
In addition, according to the method of 1, the distribution function, distribution law and expectation of M2 and M3 can be obtained respectively.
Now only the distribution law of m and the expectation of n are simply given.
M2:
1:465 1/7776
2:2 10 1/7776
3:78 1/7776
4:2 1 1/7776
5:3 1/7776
6: 1/7776
E(N2)=6E(M2)=32 106/7776
M3:
1:3 103 1/46656
2: 1 1529/46656
3:3367/46656
4:664/46656
5:63/46656
6: 1/46656
N3 = M3 = 67 17 1/46656
Bringing the value into the formula *, the expected value of b is 22 1986/46656.
Complete #