(a) Evaluating $(x+x^2+x^3+x^4{)}^2$ I obtained

${\displaystyle x^2+2x^3+3x^4+4x^5+3x^6+2x^7+x^8}$

(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:

${\displaystyle \begin{array}{ccccccccc}\multicolumn{1}{c}{}& \mathrm{Scores}& 2& 3& 4& 5& 6& 7& 8\\ \multicolumn{1}{c}{}& \mathrm{Frequencies}& 1& 2& 3& 4& 3& 2& 1\\ \multicolumn{1}{c}{}& \mathrm{Probabilities}& 0.0625& 0.125& 0.1875& 0.25& 0.1875& 0.125& 0.0625\end{array}}$

(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) $(x+x^2+x^3+x^4{)}^2$ could be factorised as follows: $x^2(1+x{)}^2(1+x^2{)}^2$ and these factors can be re-written as:
1) $(1+x)(x+x^2)(1+x^2)(x+x^3)$
2) $(x+x^2{)}^2(1+x^2{)}^2$
3) $(1+x{)}^2(x+x^3{)}^2$
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: $x^3(x+2)(3x+4{)}^2$
1) $(x^4+2x^3)(3x+4{)}^2$
2) $(x^2+2x)(3x^2+4x{)}^2$
3) $(x^4+2x^3)(9x^2+24x+16)$
4) $(x^3+2x^2)(9x^3+24x^2+16x)$
5) $(x^2+2x)(9x^4+24x^3+16x^2)$
6) $(3x^4+4x^3)(3x^2+10x+8)$
7) $(3x^3+4x^2)(3x^3+10x^2+8x)$
8) $(3x^2+4x)(3x^4+10x^3+8x^2)$
9) $(3x+4)(3x^5+10x^4+8x^3)$
10) $9x^6+42x^5+64x^4+32x^3$

This example is very interesting, but the numbers are too big to use the simulation! I chose another example, with much smaller numbers:
$x^2(x+1)(x+2)$
1) $(x^3+x^2)(x+2)$
2) $(x+1)(x^3+2x^2)$
3) $(x^2+x)(x^2+2x)$

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 $x^m+x^n$. 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 $x^m+x^n+x^p$, with $m\ne n\ne p$
- 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 $x^m+2x^n$, $m\ne n$.
-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 $(x+x^2+x^3+x^4+x^5+x^6{)}^2$. The expansion of this polynomial is:

${\displaystyle x^2+2x^3+3x^4+4x^5+5x^6+6x^7+5x^8+4x^9+3x^{10}+2x^{11}+x^{12}.}$

This polynomial could be factorised as follows:

${\displaystyle (x+x^2+x^3+x^4+x^5+x^6{)}^2=x^2(1+x{)}^2(1+x^2+x^4{)}^2}$

and from this decomposition different combinations could be obtained, all of them producing the same frequency distribution.