For this solution we have to thank Andrei from Tudor Vianu National College, Bucharest, Romania.

(a) Evaluating (x + x2 + x3 + x4)2 I obtained
x2 +2x3+3x4+4x5+3x6+2x7+x8
(b) The sum of two spinners labelled 1, 2, 3 and 4 varies from 2 to 8. There are 16 possibilities in total, giving the following probabilities:
Scores
2
3
4
5
6
7
8
Frequencies
1
2
3
4
3
2
1
Probabilities
0.0625
0.125
0.1875
0.25
0.1875
0.125
0.0625

(c) I observe that the frequency distribution is the same as the coefficients from the expansion of the polynomial from (a).

(d) Using the computer simulation, I took two spinners labelled from 1 to 4, obtaining the following table of frequencies:
Exp. Relative frequency
0 0
1 0
2 0.0631
3 0.1253
4 0.1872
5 0.2493
6 0.188
7 0.1248
8 0.062
The results obtained are very close to the theoretical frequency distribution of the scores.

(e) (x + x2 + x3 + x4)2 could be factorised as follows: x2 (1+x)2 (1+x2)2 and these factors can be re-written as:
1) (1+x)(x+x2) (1+x2)(x+x3)
2) (x+x2)2 (1+x2)2
3) (1+x)2(x+x3)2
The first case corresponds to 4 spinners (0,1), (1,2), (0,2),(1,3), the second to 4 spinners: (1,2), (1,2), (0,2), (0,2) and the third to another 4 spinners (0,1), (0,1), (1,3), (1,3).

(f) Using the simulation, I obtained similar frequency distributions as with two 1,2,3,4 spinners.
Value Frequency distributions
(1) (2) (3)
2 0.0619 0.0617 0.0646
3 0.1264 0.1243 0.1252
4 0.1869 0.187 0.1881
5 0.2495 0.2504 0.2465
6 0.1871 0.1861 0.1875
7 0.1245 0.1269 0.1263
8 0.0634 0.0634
0.0616
(g) Another example of spinners which could be re-labelled in more than one way is: x3 (x+2) (3x+4)2
1) (x4 + 2x3) (3x + 4)2
2) (x2 + 2x) (3x2 + 4x)2
3) (x4 + 2x3) (9x2 + 24x + 16)
4) (x3 + 2x2) (9x3 + 24x2 + 16x)
5) (x2 + 2x) (9x4 + 24x3 + 16x2)
6) (3x4 + 4x3) (3x2 + 10x + 8)
7) (3x3 + 4x2) (3x3 + 10x2 + 8x)
8) (3x2 + 4x) (3x4 + 10x3 + 8x2)
9) (3x+ 4) (3x5 + 10x4 + 8x3)
10) 9x6 + 42x5 + 64x4 +32x3

This example is very interesting, but the numbers are too big to use the simulation! I chose another example, with much smaller numbers:
x2(x+1)(x+2)
1) (x3 + x2)(x+2)
2) (x + 1) (x3 + 2x2)
3) (x2+x) (x2+2x)
Score Frequency distributions
(1) (2) (3)
2 0.3324 0.3312 0.3359
3 0.4996 0.5021 0.4982
4 0.1679 0.1665 0.1658
The theoretical probabilities would be 2/6 = 33.33 %, for 2, 3/6 = 50% for 3, and 1/6 = 16.66%, which are very close to the simulated values.

For a 2-spinner, there are equal probabilities of obtaining the two numbers with which it is labelled. It corresponds to the polynomial xm + xn. If m and n are equal, then the probability ofobtaining that value is 1 (100%).

For a 3-spinner, I have 3 cases:
-All three numbers are different. There is a probability of 1/3 (33.33%) of obtaining any of the three numbers. It corresponds to the polynomial xm + xn + xp, with m ¹ n ¹ p
- Two of the three numbers are equal. There is a probability of 1/3 (33.33%) of obtaining the unique number, and a probability of 66.67% of obtaining the number which appears twice. It corresponds to the polynomial xm + 2xn, m ¹ n.
-All 3 numbers are equal. The probability of obtaining this number is 100%.

Two ordinary dice are equivalent to 2 6-spinners. The theoretical probabilities and the simulated ones are:
Number 2 3 4 5 6 7 8 9 10 11 12
Theoretical 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
Frequency 0.0281 0.0556 0.083 0.1123 0.1397 0.1674 0.1381
0.1093
0.0842
0.0539 0.028
This case corresponds to the polynomial (x + x2 + x3 + x4 + x5 + x6)2. The expansion of this polynomial is:
x2 + 2x3 + 3x4 + 4x5 + 5x6 + 6x7 + 5x8 + 4x9 + 3x10 + 2x11 + x12.
This polynomial could be factorised as follows:
(x + x2 + x3 + x4 + x5 + x6)2 = x2 (1+x)2 (1+x2+x4)2
and from this decomposition different combinations could be obtained, all of them producing the same frequency distribution.